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AP Physics C Mechanics

Comprehensive AP Physics C Mechanics curriculum covering circular dynamics, work, energy, conservation laws, and systems of particles.

CMChuguna Murugesan

Circular Dynamics

A comprehensive AP Physics C mechanics lesson on circular dynamics, covering horizontal curves, banked tracks, vertical circles, conical pendulums, and rotating drum rides.

Work and Energy

An in-depth AP Physics C Mechanics unit covering the mathematical definitions of work, the work-energy theorem, conservative forces, potential energy functions, and power.

AP Physics C Mechanics

Comprehensive AP Physics C Mechanics curriculum covering circular dynamics, work, energy, conservation laws, and systems of particles.

CMChuguna Murugesan

Circular Dynamics Reading Guide

Cornell Notes Comprehensive Text • AP Physics C: Mechanics

Circular Motion: Kinematics & Net Radial Dynamics

Linear Formulation Page 1 of 7

Cornell Notes Essential Question:

How do physical interactions constrain an object to a curved trajectory, and how does Newton's second law apply along the radial axis without treating centripetal force as an independent physical force?

1. In-Depth Explanation: The Physics of Circular Motion

Under Newton's first law, an isolated object naturally continues along a straight line at constant speed. To deviate from this rectilinear path and trace a circle of radius \(r\), an external force must continuously deflect the body toward the center of curvature. In Uniform Circular Motion (UCM), the scalar tangential speed \(v = |\vec{v}|\) remains constant, but the velocity vector \(\vec{v}\) changes direction at every infinitesimal instant. Because acceleration is \(\vec{a} = \frac{d\vec{v}}{dt}\), this continuous redirection requires an acceleration. Since speed does not change, tangential acceleration is zero (\(a_t = \frac{dv}{dt} = 0\)). The entire acceleration vector points strictly perpendicular to instantaneous velocity, directed inward toward the center: \(\vec{a}_c \perp \vec{v}\).

Radius (\(r\))

\(r\) [m]

Center-to-orbit distance

Period (\(T\))

\(T = \frac{2\pi r}{v}\)

Duration of 1 revolution (s)

Frequency (\(f\))

\(f = \frac{1}{T} = \frac{v}{2\pi r}\)

Revolutions per sec (Hz)

Centripetal Accel (\(a_c\))

\(a_c = \frac{v^2}{r}\)

Inward rate of velocity change

Center O r v a_c Velocity Triangle v₁ v₂ Δv

Geometric Derivation of \(a_c = \frac{v^2}{r}\)

As a particle traverses arc \(\Delta s = v\,\Delta t\), the velocity vector changes by \(\Delta\vec{v}\). The position triangle (sides \(r\), base \(\Delta s\)) is similar to the velocity triangle (sides \(v\), base \(|\Delta\vec{v}|\)):

\(\frac{|\Delta\vec{v}|}{v} \approx \frac{\Delta s}{r} \implies |\Delta\vec{v}| = \frac{v}{r}\,\Delta s\)

Dividing by \(\Delta t\) as \(\Delta t \to 0\): \(a_c = \frac{v}{r}\left(\lim_{\Delta t \to 0}\frac{\Delta s}{\Delta t}\right) = \frac{v^2}{r}\).

The "Centripetal Role" Principle: In \(\Sigma\vec{F} = m\vec{a}\), the term \(m\frac{v^2}{r}\) is the mass times acceleration side. Never sketch "\(F_c\)" on an FBD. Instead, determine which physical force (gravity, normal force, tension, friction) supplies this inward acceleration.

Checkpoint 1.1: Work in UCM Using \(W = \int \vec{F}_{net}\cdot d\vec{r}\), explain why the net centripetal force does zero work on an object in uniform circular motion.

Checkpoint 1.2: Jet Acceleration A fighter jet flies at \(v = 240\text{ m/s}\) in a turn of radius \(r = 1200\text{ m}\). Calculate its centripetal acceleration in \(\text{m/s}^2\) and in \(g\)'s.

Checkpoint 1.3: Scaling Proportions If a vehicle triples its speed (\(v \to 3v\)) while halving turn radius (\(r \to r/2\)), by what factor does centripetal acceleration change?

Checkpoint 1.4: Velocity Rate of Change In UCM, \(\frac{d|\vec{v}|}{dt} = 0\) while \(|\frac{d\vec{v}}{dt}| \neq 0\). Explain the physical difference between these two quantities.

Cornell Notes Study Text • AP Physics C: Mechanics Page 1 of 7

Cornell Notes Page 2: Vertical Circular Motion Page 2 of 7

Motion in Vertical Circle: Strings & Roller Coaster Loops

In a vertical plane, gravity acts parallel to the plane of motion. Because gravity always points downward toward the Earth, its component along the radial axis varies continuously with angular position. As a result, the tension or normal force required to sustain the circular path changes at every point.

Example 1: Objects Spinning on the End of a String

Critical Speed: \(v_{top} \ge \sqrt{gR}\)

T_top ↓ mg ↓ T_bot ↑ mg ↓

A. Force Types: Cord Tension (\(T\)) along the string, and Downward Gravity (\(mg\)).

B. At highest point centripetal force is: Both tension and gravity pull directly toward the center: \(\Sigma F_r = T_{top} + mg = m\,\frac{v_{top}^2}{R} \implies T_{top} = m\left(\frac{v_{top}^2}{R} - g\right)\) Because a flexible cord cannot push (\(T \ge 0\)), tension drops to zero at the critical threshold: \(v_{crit} = \sqrt{gR}\). If speed is lower, the cord slacks and the mass enters free fall.

C. At lowest point centripetal force is: Tension pulls toward center while gravity opposes it: \(\Sigma F_r = T_{bot} - mg = m\,\frac{v_{bot}^2}{R} \implies T_{bot} = m\left(\frac{v_{bot}^2}{R} + g\right)\)

Checkpoint 2.1: Slack Threshold What physical behavior occurs if a rock on a string has speed \(v_{top} < \sqrt{gR}\) at the apex?

Checkpoint 2.2: Apex Calculation A \(0.40\text{ kg}\) ball on a \(0.80\text{ m}\) string moves at \(v = 4.0\text{ m/s}\) at apex. Calculate cord tension \(T_{top}\).

Example 2: Roller Coaster Loops

Release: \(h_{min} = 2.5 R\)

F_N ↓ mg ↓ F_N ↑ mg ↓

A. Force Types: Downward Track Normal Force (\(F_N\)) and Downward Gravity (\(mg\)).

B. At highest point centripetal force is: Track push and gravity act downward: \(\Sigma F_r = F_{N,top} + mg = m\,\frac{v_{top}^2}{R} \implies F_{N,top} = m\left(\frac{v_{top}^2}{R} - g\right)\) Contact requires \(F_N \ge 0 \implies v_{top} \ge \sqrt{gR}\). Energy conservation (\(mgh = mg(2R) + \frac{1}{2}mv_{top}^2\)) requires a minimum release height of \(h_{min} = 2.5R\).

C. At lowest point centripetal force is: Upward track push supports weight and accelerates cart: \(\Sigma F_r = F_{N,bot} - mg = m\,\frac{v_{bot}^2}{R} \implies F_{N,bot} = m\left(\frac{v_{bot}^2}{R} + g\right)\) For release from \(h = 2.5R\), \(v_{bot}^2 = 5gR\), producing a heavy load of \(F_{N,bot} = 6\,mg\).

Checkpoint 2.3: Elevated Release A cart of mass \(m\) is released from height \(h = 3.5R\) on a circular loop. Find \(F_{N,top}\) in terms of \(mg\).

Checkpoint 2.4: Clothoid Loops Why do modern roller coaster designers use teardrop clothoid loops rather than pure circular loops?

Cornell Notes Study Text • AP Physics C: Mechanics Page 2 of 7

Cornell Notes Pages 2 & 3: Ferris Wheel & Flat Turns Page 3 of 7

The Ferris Wheel & Horizontal Flat Turns

A comparative analysis of uniform vertical circular motion (where a motor holds linear speed \(v\) constant) and planar horizontal curves (where static friction provides the sole centripetal acceleration).

Example 3: The Ferris Wheel

Apparent Weight: \(F_N\)

F_N,top ↑ mg ↓ F_N,bot ↑ mg ↓

A. Force Types: Upward Chair Normal Force (\(F_N\)) and Downward Gravity (\(mg\)).

B. At highest point centripetal force is: Inward acceleration points downward toward center: \(\Sigma F_r = mg - F_{N,top} = m\,\frac{v^2}{R} \implies F_{N,top} = m\left(g - \frac{v^2}{R}\right)\) Because \(F_{N,top} < mg\), rider feels lighter. Apparent weightlessness occurs at \(v = \sqrt{gR}\).

C. At lowest point centripetal force is: Inward acceleration points upward toward center: \(\Sigma F_r = F_{N,bot} - mg = m\,\frac{v^2}{R} \implies F_{N,bot} = m\left(g + \frac{v^2}{R}\right)\) Because \(F_{N,bot} > mg\), rider feels heavier, compressed into the seat.

Checkpoint 3.1: Scale Reading A \(55\text{ kg}\) rider moves at \(v = 6.0\text{ m/s}\) on a wheel of \(R = 12\text{ m}\). Find the scale reading at apex and nadir.

Checkpoint 3.2: True Zero Normal What wheel rotation period \(T\) produces true zero normal force at the top for radius \(R = 16.0\text{ m}\)?

Horizontal Circular Motion: Flat Curve Without Slipping

\(v_{max} = \sqrt{\mu_s g r}\)

Center O F_N mg f_s (inward)

Force Types: Ground Normal Force (\(F_N\)), Downward Gravity (\(mg\)), Static Friction (\(f_s\)).

Force equation: Y direction: \(\Sigma F_y = F_N - mg = 0 \implies F_N = mg\)

Force equation: X direction: \(\Sigma F_x = f_s = m\,\frac{v^2}{r}\)

Static friction condition: \(f_s \le \mu_s F_N = \mu_s mg\). Setting \(f_s = \mu_s mg\) yields: \(m\,\frac{v_{max}^2}{r} = \mu_s mg \implies v_{max} = \sqrt{\mu_s g r}\) Mass cancels completely: safe turning speed is independent of vehicle weight.

Checkpoint 3.3: Rain vs Dry Asphalt A flat curve has radius \(r = 55.0\text{ m}\). Find \(v_{max}\) on dry road (\(\mu_s = 0.75\)) and icy road (\(\mu_s = 0.15\)).

Checkpoint 3.4: Mass Invariance Explain why a loaded semi-truck and a lightweight sports car have the identical speed limit on a flat curve.

Cornell Notes Study Text • AP Physics C: Mechanics Page 3 of 7

Cornell Notes Pages 3 & 4: Banked Curves & Turntables Page 4 of 7

Banked Curves & Rotating Turntables

Banking tilts the roadway surface so that the horizontal projection of normal force supplies centripetal acceleration, whereas horizontal platforms rely entirely on static friction to hold rotating objects in place.

Banking Curve Dynamics (Frictionless & With Friction)

Ideal: \(\tan\theta = \frac{v^2}{gr}\)

θ F_N F_N cos θ F_N sin θ mg

Force Types: Normal Force perpendicular to slope (\(F_N\)) and Gravity (\(mg\)).

Force equation: Y direction: \(\Sigma F_y = F_N\cos\theta - mg = 0 \implies F_N = \frac{mg}{\cos\theta}\)

Force equation: X direction: \(\Sigma F_x = F_N\sin\theta = m\,\frac{v^2}{r}\)

Dividing radial by vertical equation: \(\tan\theta = \frac{v^2}{gr} \implies v_{ideal} = \sqrt{gr\tan\theta}\) With friction: \(v_{max} = \sqrt{gr\frac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}}\); \(v_{min} = \sqrt{gr\frac{\sin\theta-\mu_s\cos\theta}{\cos\theta+\mu_s\sin\theta}}\).

Checkpoint 4.1: Banking Angle Calculate ideal angle \(\theta\) for radius \(r = 150\text{ m}\) and rated speed \(v = 25.0\text{ m/s}\).

Checkpoint 4.2: Ice Incline If a vehicle drives at \(v < v_{ideal}\) on a frictionless banked curve, what is the net force direction?

Object Sitting on a Rotating Turntable

\(f_s \le \mu_s mg\)

r f_s v

Force Types: Normal force (\(F_N = mg\)), Gravity (\(mg\)), Inward Static Friction (\(f_s\)).

Dynamics: \(\Sigma F_r = f_s = m\,\frac{v^2}{r} = \frac{4\pi^2 m r}{T^2}\).

Because required friction increases linearly with placement radius \(r\), the maximum radius before slipping occurs is: \(\frac{4\pi^2 m r_{max}}{T^2} \le \mu_s mg \implies r_{max} = \frac{\mu_s g T^2}{4\pi^2}\)

Checkpoint 4.3: Radial Limit A turntable has period \(T = 1.60\text{ s}\) and \(\mu_s = 0.32\). Calculate the maximum distance \(r_{max}\) before slipping.

Checkpoint 4.4: Friction Vector If a coin is placed at \(r < r_{max}\), does static friction equal \(\mu_s mg\)? Explain the distinction.

Cornell Notes Study Text • AP Physics C: Mechanics Page 4 of 7

Cornell Notes Page 4: Conical Pendulum & Rotor Page 5 of 7

The Conical Pendulum & The Rotating Drum Ride

Two classic AP Physics C configurations where the body's motion is strictly confined to a horizontal circle, but geometry dictates unconventional roles for the contact forces.

Conical Pendulum Dynamics

\(T_p = 2\pi\sqrt{\frac{L\cos\theta}{g}}\)

L θ m r T mg

Force Types: Cord Tension (\(T\)) along string, Downward Gravity (\(mg\)).

Force equation: Y direction: \(\Sigma F_y = T\cos\theta - mg = 0 \implies T = \frac{mg}{\cos\theta}\) (\(T > mg\) always!)

Force equation: X direction: \(\Sigma F_x = T\sin\theta = m\,\frac{v^2}{r}\)

Geometry: \(r = L\sin\theta\). Dividing equations: \(\tan\theta = \frac{v^2}{gr} \implies v = \sqrt{gr\tan\theta}\).
Revolution Period: Substituting \(v = \frac{2\pi r}{T_p}\): \(T_p = \frac{2\pi (L\sin\theta)}{\sqrt{g(L\sin\theta)\tan\theta}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}\)

Checkpoint 5.1: String Limit A string snaps if \(T > 2.5mg\). Find the maximum cone angle \(\theta_{max}\) and corresponding speed \(v\).

Checkpoint 5.2: Angle Limit Why can a conical pendulum never spin fast enough to become horizontal (\(\theta = 90^\circ\))?

Rotating Drum Ride ("The Rotor / Gravitron")

\(v_{min} = \sqrt{\frac{g R}{\mu_s}}\)

Floor drops down F_N (wall) f_s (up) mg

Force Types: Inward Normal Force from wall (\(F_N\)), Upward Static Friction (\(f_s\)), Gravity (\(mg\)).

Force equation: X direction: \(\Sigma F_x = F_N = m\,\frac{v^2}{R}\) (wall push supplies centripetal acceleration)

Force equation: Y direction: \(\Sigma F_y = f_s - mg = 0 \implies f_s = mg\)

Slipping prevented if: \(f_s \le \mu_s F_N \implies mg \le \mu_s\left(m\frac{v^2}{R}\right)\). \(v_{min} = \sqrt{\frac{g R}{\mu_s}} \iff T_{max} = 2\pi\sqrt{\frac{\mu_s R}{g}}\) Mass cancels completely: a child and adult stick with identical safety margins.

Checkpoint 5.3: Rotor Calculations A drum of radius \(R = 4.0\text{ m}\) has \(\mu_s = 0.40\). Calculate minimum speed \(v_{min}\) and period \(T_{max}\).

Checkpoint 5.4: Friction Halved If a rider wears slick clothing with half the friction coefficient (\(\mu_s' = \mu_s / 2\)), by what factor must \(v_{min}\) change?

Cornell Notes Study Text • AP Physics C: Mechanics Page 5 of 7

Cornell Notes Page 4: Centripetal Force as a New Role Page 6 of 7

Centripetal Force – New Role – Complete Examples

The central theme of circular dynamics: Centripetal force is not an independent physical force, but a mathematical role filled by tangible physical forces. Here are all seven situations from your Cornell Notes:

F_g v

Earth-Sun Motion: Centripetal Force \(\equiv\) Universal Gravitation (\(F_g\)) \(\Sigma F_r = \frac{G M_S M_E}{r^2} = \frac{M_E v^2}{r} \implies v = \sqrt{\frac{G M_S}{r}}\) Gravitational attraction between solar mass and planet is the sole radial agent. Kepler's third law follows directly: \(T^2 = \frac{4\pi^2}{G M_S} r^3\).

Cornell Notes System	Centripetal Force Identity (\(\Sigma F_r\))	Governing Equation
Earth-Sun Motion	Gravitational Force (\(F_g\))	\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Object on Turntable	Static Friction (\(f_s\))	\(f_s = \frac{m v^2}{r} \le \mu_s mg\)
Rock-String (Horizontal)	Horizontal Component of Tension (\(T_x\))	\(T\sin\theta = \frac{m v^2}{r}\)
Wall-Person (Rotating Room)	Wall Normal Force (\(F_N\))	\(F_N = \frac{m v^2}{R};\; mg \le \mu_s F_N\)
Ferris Wheel (Lowest Point)	Normal Force minus Gravity (\(F_N - mg\))	\(F_{N,bot} = m\left(g + \frac{v^2}{R}\right)\)
Ferris Wheel (Highest Point)	Gravity minus Normal Force (\(mg - F_N\))	\(F_{N,top} = m\left(g - \frac{v^2}{R}\right)\)
Rock-String (Vertical: Top)	Tension plus Gravity (\(T + mg\))	\(T_{top} = m\left(\frac{v^2}{R} - g\right)\)
Rock-String (Vertical: Bottom)	Tension minus Gravity (\(T - mg\))	\(T_{bot} = m\left(\frac{v^2}{R} + g\right)\)

AP Free-Body Diagram Mastery Protocol

Identify Center of Curvature: Mentally designate the center point of the circular path.
Sign Convention: Forces pointing directly toward the center are positive (\(+\)); forces pointing away are negative (\(-\)).
Never Sketch Inertial Forces: Centrifugal force does not exist in an inertial laboratory reference frame.

Checkpoint 6.1: Higher Orbit Scaling If a satellite's orbital radius is quadrupled (\(r \to 4r\)), by what factor does its linear speed \(v\) and revolution period \(T\) change?

Checkpoint 6.2: Spherical Bowl A marble rolls in a horizontal circle at depth \(d\) inside a bowl of radius \(R\). Identify the force component acting as centripetal force.

Checkpoint 6.3: FBD Verification List every true interaction force acting on a passenger in a Gravitron with the floor dropped. State the agent of each force.

Checkpoint 6.4: Net Force Sign At the top of a vertical circle, why are both tension \(T\) and weight \(mg\) given positive signs in \(\Sigma F_r = m a_c\)?

Cornell Notes Study Text • AP Physics C: Mechanics Page 6 of 7

Cornell Notes: Unit Summary & AP Practice Page 7 of 7

Summary & AP Physics C Practice Set

Consolidating all circular dynamics principles into three governing theorems, followed by four multi-part AP Physics C practice problems.

Cornell Notes Core Summary

1. Kinematic Reality: Any body traversing a circle of radius \(r\) with linear speed \(v\) experiences centripetal acceleration \(a_c = \frac{v^2}{r}\) directed strictly toward the center of curvature.

2. Dynamic Reality: Centripetal acceleration requires a real net inward radial force: \(\Sigma F_r = m a_c = m \frac{v^2}{r}\). This role is filled by genuine physical interactions: gravity, friction, tension, or normal force.

3. Vertical vs Horizontal Duality: Horizontal systems have zero vertical acceleration (\(\Sigma F_y = 0\)). Vertical systems have gravitational contributions directly along the radial axis, causing apparent weight variations: \(F_{N,bot} > mg > F_{N,top}\).

Problem 1: Banked Ramp Analysis

A frictionless curve of radius \(r = 120\text{ m}\) is banked at \(\theta = 20.0^\circ\). (a) Derive and compute design speed \(v_{ideal}\). (b) Calculate normal force on a \(1400\text{ kg}\) car at this speed.

Problem 2: Gravitron Dropped Floor

A carnival rotor of radius \(R = 5.0\text{ m}\) operates with \(\mu_s = 0.35\). Derive the symbolic formula for \(v_{min}\) and determine the maximum period \(T_{max}\) per revolution to safely drop the floor.

Problem 3: Vertical Loop Tension

A \(0.40\text{ kg}\) tethered ball is whirled in a vertical circle of radius \(R = 0.75\text{ m}\). If \(v_{top} = 4.0\text{ m/s}\), calculate: (a) cord tension \(T_{top}\), (b) critical speed \(v_{crit}\), and (c) nadir tension \(T_{bot}\).

Problem 4: Breaking Cord Limit

A conical pendulum uses a string that breaks if tension exceeds \(T_{max} = 3.0\,mg\). Derive the maximum allowable cone angle \(\theta_{max}\) and corresponding speed \(v\) in terms of cord length \(L\) and \(g\).

Student Self-Check: Have you drawn a complete FBD for each problem with no "\(F_c\)" vectors? Cornell Notes Complete • 7/7 Topics

Cornell Notes Study Text • AP Physics C: Mechanics Page 7 of 7

Circular Dynamics Practice Assignment

AP Physics C: Mechanics • Practice Assignment

Circular Dynamics Problem Set

Page 1 of 5

Student Name:

Date:

Class Period:

Score / Grade:

Part 1: Horizontal Planar Dynamics Flat Roadways & Spinning Platforms

Problem 1: Sports Car on an Unbanked Turn

[10 Points Total]

A \(1500\text{ kg}\) car travels on a level, unbanked circular track of radius \(r = 75.0\text{ m}\). The static coefficient of friction between the tires and dry pavement is \(\mu_s = 0.80\). The car moves at constant linear speed \(v\).

Free-Body Diagram (Rear View):

Draw all forces with labels

(a) Write Newton's 2nd Law equations for both horizontal (\(x\)) and vertical (\(y\)) directions:

(b) Derive the algebraic expression for maximum safe speed \(v_{max}\) and compute its numerical value:

(c) Conceptual Evaluation: If the mass of the car is doubled to \(3000\text{ kg}\), how does \(v_{max}\) change? Justify using your algebraic result from part (b).

Problem 2: Laboratory Turntable Placement Limit

[10 Points Total]

A small steel coin of mass \(m = 0.050\text{ kg}\) rests on a horizontal rotating turntable. The turntable rotates with a constant revolution period of \(T = 1.40\text{ s}\). The coefficient of static friction is \(\mu_s = 0.45\).

(a) Derive the maximum radial distance \(r_{max}\) from the center spindle at which the coin can rest without slipping:

(b) If the coin is placed at \(r = 10.0\text{ cm}\), calculate the magnitude of the actual static friction force \(f_s\) acting on it:

AP Physics C: Mechanics • Unit 3: Circular Motion Page 1 of 5

AP Physics C: Mechanics • Practice Assignment Page 2 of 5

Part 2: Banked Turn Engineering Normal Force Tilting & Friction Envelopes

Problem 3: Superhighway Exit Ramp Design

[12 Points Total]

A civil engineer designs an exit ramp with radius of curvature \(r = 140\text{ m}\). The ramp is banked at angle \(\theta\) so that a car traveling at design speed \(v_{design} = 22.0\text{ m/s}\) can negotiate the curve safely even if ice covers the pavement (\(\mu = 0\)).

Free-Body Diagram (Car on Incline):

Draw F_N, mg, and angle θ

(a) Resolve Newton's 2nd Law along horizontal and vertical axes to derive \(\tan\theta = \frac{v^2}{gr}\):

(b) Calculate the required banking angle \(\theta\) in degrees:

(c) What is the normal force \(F_N\) on a \(1200\text{ kg}\) car driving at this speed?

(d) If the car moves at \(v = 12.0\text{ m/s}\) on ice, describe what happens and explain using force components:

Circular Dynamics Teacher Answer Key

AP Physics C: Mechanics • Teacher Solution Guide

Circular Dynamics Answer Key & Rubric

Key • Page 1 of 5

Total Assignment Points: 70 Points Strict AP Exam Grading Criteria & Full Vector Solutions

Problem 1 Solution: Sports Car on Unbanked Turn [10 Points]

10 Pts

Scored FBD [2 Pts]: F_N mg f_s 1 pt: 3 correct forces; 1 pt: no "\(F_c\)" drawn

(a) Equations of Motion [3 Pts]:

\(\Sigma F_y = F_N - mg = 0 \implies F_N = mg\) [1 pt]
\(\Sigma F_x = f_s = m\,\frac{v^2}{r}\) [1 pt]; \(f_s \le \mu_s F_N\) [1 pt]

(b) Maximum Speed Derivation [3 Pts]:

\(m\,\frac{v_{max}^2}{r} = \mu_s mg \implies v_{max} = \sqrt{\mu_s g r}\) [1.5 pts]
\(v_{max} = \sqrt{(0.80)(9.8\text{ m/s}^2)(75.0\text{ m})} = \sqrt{588} \approx \mathbf{24.2\text{ m/s}}\) (\(87.3\text{ km/h}\)) [1.5 pts]

(c) Conceptual Mass Invariance [2 Pts]: Doubling the mass to \(3000\text{ kg}\) produces no change in \(v_{max}\) (\(v_{max} = 24.2\text{ m/s}\)). Mass \(m\) appears linearly on both the inertial side (\(m v^2/r\)) and the gravitational normal side (\(\mu_s mg\)) and cancels out completely: \(v_{max} = \sqrt{\mu_s gr}\). [2 pts]

Problem 2 Solution: Laboratory Turntable [10 Points]

10 Pts

(a) Maximum Radial Distance \(r_{max}\) [5 Pts]:

In terms of linear speed \(v = 2\pi r / T\):

\(f_s = m\,\frac{v^2}{r} = m\,\frac{(2\pi r / T)^2}{r} = \frac{4\pi^2 m r}{T^2}\) [2 pts]
Setting \(f_{s,max} = \mu_s mg\):
\(\frac{4\pi^2 m r_{max}}{T^2} = \mu_s mg \implies r_{max} = \frac{\mu_s g T^2}{4\pi^2}\) [1.5 pts]
\(r_{max} = \frac{(0.45)(9.8)(1.40)^2}{4\pi^2} = \frac{8.6436}{39.478} \approx \mathbf{0.219\text{ m}} = \mathbf{21.9\text{ cm}}\) [1.5 pts]

(b) Actual Friction at \(r = 10.0\text{ cm}\) [5 Pts]:

At \(r = 0.100\text{ m} < r_{max}\), the coin does not slip:

\(f_s = \frac{4\pi^2 m r}{T^2}\) [2 pts]
\(f_s = \frac{4\pi^2 (0.050\text{ kg})(0.100\text{ m})}{(1.40\text{ s})^2}\) [1.5 pts]
\(f_s = \frac{0.1974}{1.96} \approx \mathbf{0.101\text{ N}}\) [1.5 pts]
Note: Full available friction is \(\mu_s mg = (0.45)(0.05)(9.8) = 0.221\text{ N}\). Since \(0.101\text{ N} < 0.221\text{ N}\), static friction easily holds.

AP Physics C: Mechanics • Teacher Solution Key Page 1 of 5

AP Physics C: Mechanics • Teacher Solution Guide Key • Page 2 of 5

Part 2 Solutions: Banked Turn Engineering Problems 3 & 4 Rubric [22 Points Total]

Problem 3 Solution: Superhighway Exit Ramp [12 Points]

Circular Dynamics MCQ Review

AP Physics C: Mechanics • Unit 3 Assessment

Circular Dynamics MCQ Review

Page 1 of 5

Name:

Date:

Period:

Score: ____ / 20

1. Kinematic Velocity & Acceleration Vectors [1 Pt]

An object moves along a horizontal circular path of radius \(r\) at constant linear speed \(v\). Which statement regarding velocity \(\vec{v}\) and acceleration \(\vec{a}\) must be true at every instant?

O v a_c

(A) \(\vec{a}\) is tangent and parallel to \(\vec{v}\).

(B) \(\vec{a}\) is radially outward, opposite to \(\vec{v}\).

(C) \(\vec{a}\) is directed radially inward toward the center, such that \(\vec{a}\cdot\vec{v} = 0\).

(D) \(\vec{a} = 0\) because scalar speed is constant.

(E) \(\vec{a}\) is at \(45^\circ\) between radial & tangent.

2. Work-Energy Theorem in Uniform Circular Motion [1 Pt]

A satellite of mass \(m\) executes a circular orbit around a planet of mass \(M\) at constant speed \(v\) and radius \(r\). How much net work is done by gravity on the satellite over a semicircle (\(\Delta s = \pi r\))?

M F_g v

(A) \(W = \pi m v^2\)

(B) \(W = 2 m v^2\)

(C) \(W = \frac{G M m}{r}\)

(D) \(W = 0\), since \(\vec{F}_g \perp d\vec{r}\)

(E) \(W = -\frac{2 G M m}{\pi r}\)

3. Proportional Scaling of Centripetal Acceleration [1 Pt]

Particle 1 moves on a circle of radius \(r\) with speed \(v\). Particle 2 moves on radius \(r/3\) with speed \(2v\). What is the ratio \(a_{c2} / a_{c1}\)?

Radius r Speed v r/3 2v

(A) \(2/3\)

(B) \(4/3\)

(C) \(6\)

(D) \(12\)

(E) \(18\)

4. Non-Uniform Circular Motion Acceleration Magnitude [1 Pt]

A particle starts from rest on radius \(R\) with constant tangential acceleration \(a_t\). What is total acceleration magnitude \(|\vec{a}_{net}|\) at time \(t\)?

a_t a_c a_net

(A) \(|\vec{a}_{net}| = a_t\)

(B) \(|\vec{a}_{net}| = a_t^2 t^2 / R\)

(C) \(|\vec{a}_{net}| = \sqrt{a_t^2 + (a_t^2 t^2 / R)^2} = a_t\sqrt{1 + a_t^2 t^4 / R^2}\)

Circular Dynamics MCQ Answer Key

AP Physics C: Mechanics • Unit 3 Solution Guide

Circular Dynamics MCQ Answer Key

Key • Page 1 of 5

Page 1 Answers: 1. (C) • 2. (D) • 3. (D) • 4. (C) Detailed Rationales & AP Exam Pitfalls

1. Kinematic Velocity & Acceleration Vectors Correct: (C)

Physical Rationale: In Uniform Circular Motion, scalar speed is constant (\(v = \text{const}\)), meaning tangential acceleration is zero (\(a_t = \frac{dv}{dt} = 0\)). The acceleration is purely centripetal (\(a_c = v^2/r\)), directed radially inward toward the center of curvature. Since velocity \(\vec{v}\) is tangent to the circle and acceleration \(\vec{a}\) is radial, the vectors are perpendicular: \(\vec{a} \perp \vec{v} \implies \vec{a}\cdot\vec{v} = |\vec{a}||\vec{v}|\cos(90^\circ) = 0\).

• Teacher Note / Common Trap: Students often mistakenly pick (D) thinking constant speed means zero acceleration, forgetting velocity is a vector whose direction changes.

2. Work-Energy Theorem in Uniform Circular Motion Correct: (D)

Physical Rationale: The gravitational force \(\vec{F}_g\) points radially inward toward the planet's center, while instantaneous displacement \(d\vec{r}\) is tangent to the circular orbit. At every infinitesimal step, \(\vec{F}_g \perp d\vec{r}\), giving \(dW = \vec{F}_g \cdot d\vec{r} = F_g dr\cos(90^\circ) = 0\). Integrating around any arc or full revolution yields \(W = 0\text{ J}\).

• Teacher Note: Centripetal forces change trajectory direction but perform strictly zero work.

3. Proportional Scaling of Centripetal Acceleration Correct: (D)

Mathematical Derivation: \(a_{c1} = \frac{v^2}{r}\).
For Particle 2: \(a_{c2} = \frac{(2v)^2}{(r/3)} = \frac{4v^2}{r/3} = 12\left(\frac{v^2}{r}\right) = 12\,a_{c1} \implies \frac{a_{c2}}{a_{c1}} = \mathbf{12}\).

• Teacher Note: Make sure students square the factor of 2 in numerator (\(2^2 = 4\)) and divide by \(1/3\) to get \(4 \times 3 = 12\).

4. Non-Uniform Circular Motion Acceleration Magnitude Correct: (C)

Mathematical Derivation: Since \(a_t = \text{const}\), tangential velocity is \(v(t) = a_t t\).
Radial acceleration is \(a_c(t) = \frac{v(t)^2}{R} = \frac{(a_t t)^2}{R} = \frac{a_t^2 t^2}{R}\).
Since \(\vec{a}_t \perp \vec{a}_c\): \(|\vec{a}_{net}| = \sqrt{a_t^2 + a_c^2} = \sqrt{a_t^2 + \left(\frac{a_t^2 t^2}{R}\right)^2} = a_t\sqrt{1 + \frac{a_t^2 t^4}{R^2}}\).

• Teacher Note: This tests orthogonal vector addition in planar kinematics.

AP Physics C: Mechanics • Circular Dynamics MCQ Answer Key Page 1 of 5

AP Physics C: Mechanics • Unit 3 Solution Guide Key • Page 2 of 5

Page 2 Answers: 5. (C) • 6. (A) • 7. (B) • 8. (B) Detailed Rationales & AP Exam Pitfalls

Work and Energy Reading Guide

AP Physics C: Mechanics • Chapter 6

Work, Energy & Force Resolution

Theoretical Text Page 1 of 11

Chapter 6 Core Concept:

In classical mechanics, work is the fundamental operational mechanism by which an applied force transfers energy into or out of a physical system. Evaluated through the vector dot product, work is a scalar quantity possessing magnitude and sign, but zero spatial direction.

1. The Physics of Work: Definition & Force Components

In everyday language, "work" implies physical or mental exertion. However, in scientific physics, work carries an exact, quantitative meaning: work is performed on an object only when a force exerts a push or pull that produces a displacement along the direction of that force. If an object does not move, or if the force is directed perpendicular to motion, zero work is performed regardless of how exhausting the effort feels.

Mathematically, when a constant force \(\vec{F}\) acts upon an object as it undergoes a straight-line displacement of magnitude \(s\), the work \(W\) is defined by the geometric dot product:

\(W = \vec{F} \cdot \vec{s} = |\vec{F}|\,|\vec{s}|\cos\theta = F\,s\cos\theta\)

Here, \(\theta\) represents the angle between the applied force vector \(\vec{F}\) and the displacement vector \(\vec{s}\). The SI unit of work is the Joule (J), defined as one Newton-meter (\(1\text{ J} = 1\text{ N}\cdot\text{m} = 1\text{ kg}\cdot\text{m}^2/\text{s}^2\)).

F F cos θ F sin θ θ Distance Moved (s)

Case 1: Force Parallel to Surface (\(\theta = 0^\circ\)) When force acts in the exact direction of motion, \(\cos(0^\circ) = 1\), giving maximum work: \(W = F\,s\)

Case 2: Force Applied at an Angle \(\theta\) We resolve \(\vec{F}\) into perpendicular axes. The vertical component \(F\sin\theta\) is perpendicular to displacement and does zero work. The horizontal component \(F\cos\theta\) acts along displacement: \(W = (F\cos\theta)\,s\)

Checkpoint 1.1: Physical Work Condition A weightlifter holds a \(150\text{ kg}\) barbell motionless overhead for 5 seconds. Does the lifter do mechanical work on the barbell? Explain.

Checkpoint 1.2: Angled Pull Work A wagon is pulled \(18.0\text{ m}\) across a floor by a tension of \(35.0\text{ N}\) oriented at \(40.0^\circ\) above horizontal. Compute the work done.

Checkpoint 1.3: Vertical Component Role In Case 2, does the vertical component \(F\sin\theta\) do any work on the cart? Explain why or why not using the dot product angle.

Checkpoint 1.4: Unit Dimensional Proof Express the Joule (\(\text{J}\)) entirely in terms of fundamental SI base units (kilogram, meter, second). Show step-by-step algebra.

Work and Energy Practice Assignment

AP Physics C: Mechanics • Chapter 6 Assignment

Work, Energy & Power Problem Set

Page 1 of 5

Student Name:

Date:

Class Period:

Score / Grade:

Part 1: Constant Forces & Friction Work Dot Products & Normal Force Modulation

Problem 1: Angled Pull & Friction Work (Slide 8 & 9)

[12 Points Total]

A crate of mass \(m = 5.0\text{ kg}\) is pulled along a horizontal floor by an applied tension force \(F = 40.0\text{ N}\) oriented at an angle \(\theta = 37.0^\circ\) above the horizontal. The crate moves a displacement \(s = 12.0\text{ m}\). The coefficient of kinetic friction is \(\mu_k = 0.25\).

Free-Body Diagram (Show all 4 forces):

Draw F_N, mg, F, and f_k

(a) Calculate the work done by the applied tension force \(W_F\):

(b) Write the vertical force balance equation and compute the normal force \(F_N\):

(c) Compute the work done by kinetic friction \(W_f\):

(d) Calculate the net mechanical work \(W_{net}\) done on the crate:

Problem 2: Zero Work Justifications (Slide 3)

[8 Points Total]

(a) Explain using \(\vec{F}\cdot\vec{s}\) why a person carrying a \(20\text{ kg}\) box horizontally at constant speed does zero work:

(b) Explain why a satellite in a circular orbit experiences zero net work from gravity over an entire full revolution:

AP Physics C: Mechanics • Chapter 6: Work & Energy Page 1 of 5

AP Physics C: Mechanics • Chapter 6 Assignment Page 2 of 5

Part 2: Variable Forces & Work-Energy Theorem Calculus Line Integrals & \(\Delta K\)

Problem 3: Variable Force 1D Integration (Slide 7 & 10)

[10 Points Total]

A particle of mass \(m = 2.0\text{ kg}\) moves along the \(x\)-axis subject to a position-dependent net force given by \(F(x) = (6.0 x^2 - 4.0 x + 2.0)\text{ N}\), where \(x\) is in meters. The particle starts from rest at \(x = 0\).

(a) Set up and evaluate the definite integral to find total work done from \(x = 0\) to \(x = 3.0\text{ m}\):

(b) Using the Work-Energy Theorem (\(W_{net} = \Delta K\)), determine the particle's final linear speed at \(x = 3.0\text{ m}\):

(c) What is the particle's instantaneous kinetic energy at \(x = 2.0\text{ m}\)?

Problem 4: Elastic Spring Launch & Friction Runout (Slide 11 & 22)

[10 Points Total]

A horizontal spring of stiffness \(k = 600\text{ N/m}\) is compressed by a distance \(x = 0.20\text{ m}\). A block of mass \(m = 0.50\text{ kg}\) placed against the spring is released from rest. Upon leaving the spring at \(x = 0\), the block slides across a rough horizontal floor with \(\mu_k = 0.30\).

Work and Energy Teacher Answer Key

AP Physics C: Mechanics • Chapter 6 Solution Key

Work, Energy & Power Answer Key

Key • Page 1 of 5

Total Assignment Points: 70 Points Strict AP Grading Criteria & Step-by-Step Scoring

Problem 1 Solution: Angled Pull & Friction Work [12 Points]

12 Pts

Scored FBD [2 Pts]: F_N mg F f_k 1 pt: 4 correct forces; 1 pt: correct angles

(a) Work of Applied Force \(W_F\) [2 Pts]: \(W_F = F\,s\cos\theta = (40.0\text{ N})(12.0\text{ m})\cos(37.0^\circ)\) [1 pt]
\(W_F = (480.0)(0.7986) \approx \mathbf{383.3\text{ J}}\) [1 pt]

(b) Normal Force \(F_N\) [3 Pts]: \(\Sigma F_y = F_N + F\sin\theta - mg = 0 \implies F_N = mg - F\sin\theta\) [1.5 pts]
\(F_N = (5.0\text{ kg})(9.8\text{ m/s}^2) - (40.0)\sin(37.0^\circ) = 49.0 - 24.07 \approx \mathbf{24.9\text{ N}}\) [1.5 pts]

(c) Work of Friction \(W_f\) [3 Pts]: \(W_f = -f_k\,s = -\mu_k F_N\,s\) [1.5 pts]
\(W_f = -(0.25)(24.93\text{ N})(12.0\text{ m}) \approx \mathbf{-74.8\text{ J}}\) [1.5 pts]

(d) Net Mechanical Work \(W_{net}\) [2 Pts]: \(W_{net} = W_F + W_f = 383.33\text{ J} - 74.79\text{ J}\)
\(W_{net} \approx \mathbf{+308.5\text{ J}}\) [2 pts]

Problem 2 Solution: Zero Work Justifications [8 Points]

8 Pts

(a) Carrying Load Horizontally [4 Pts]: The upward lifting force \(\vec{F}_{lift}\) acts at \(\theta = 90.0^\circ\) relative to horizontal displacement \(\vec{s}\). By the dot product definition:

\(W = F\,s\cos(90^\circ) = F\,s(0) = \mathbf{0\text{ J}}\)

The force has zero component along the displacement axis. [4 pts]

(b) Satellite in Circular Orbit [4 Pts]: The gravitational force \(\vec{F}_g\) points radially inward toward Earth's center, while the instantaneous displacement \(d\vec{r}\) is tangent to the circle. Thus \(\vec{F}_g \perp d\vec{r}\) at every instant, giving:

\(dW = \vec{F}_g \cdot d\vec{r} = 0 \implies W_{orbit} = \mathbf{0\text{ J}}\)

Centripetal forces do strictly zero work. [4 pts]

AP Physics C: Mechanics • Teacher Solution Key Page 1 of 5

AP Physics C: Mechanics • Chapter 6 Solution Key Key • Page 2 of 5

Part 2 Solutions: Variable Forces & Work-Energy Problems 3 & 4 Rubric [20 Points Total]

Problem 3 Solution: Variable Force 1D Integration [10 Points]

10 Pts

(a) Definite Integral for Work [4 Pts]: \(W = \int_0^3 (6.0 x^2 - 4.0 x + 2.0)\,dx\) [1.5 pts]
\(W = \left[ 2.0 x^3 - 2.0 x^2 + 2.0 x \right]_0^3\) [1.5 pts] \(W = [2(27) - 2(9) + 2(3)] - 0\) \(W = 54.0 - 18.0 + 6.0 = \mathbf{42.0\text{ J}}\) [1 pt]

Work and Energy AP Review

AP Physics C: Mechanics • Chapter 6 Review

Work & Energy AP Multiple-Choice Review

Page 1 of 5

Name:

Date:

Period:

Score: ____ / 20

1. Geometric Dot Product Definition [1 Pt]

A constant force \(\vec{F}\) acts on a block as it undergoes a straight-line displacement \(\vec{s}\). Under which condition is the work done by \(\vec{F}\) strictly zero?

(A) Force acts parallel to displacement (\(\theta = 0^\circ\)).

(B) Force acts antiparallel (\(\theta = 180^\circ\)).

(C) Force acts perpendicular to displacement vector (\(\vec{F} \perp \vec{s}\)).

(D) Non-zero force acts at \(\theta = 45^\circ\).

(E) Block accelerates at non-zero rate.

2. Angled Pull Work Expression [1 Pt]

A rope pulls a sled across a frictionless surface with tension \(T\) at angle \(\theta\) above horizontal. Over horizontal displacement \(d\), the work done by tension is:

T θ

(A) \(W = T\,d\)

(B) \(W = T\,d\sin\theta\)

(C) \(W = T\,d\cos\theta\)

(D) \(W = T\,d\tan\theta\)

(E) \(W = \frac{T\,d}{\cos\theta}\)

3. Centripetal Work in Planetary Motion [1 Pt]

A satellite orbits Earth in a circular path of radius \(r\) at constant speed \(v\). The net work done by gravity on the satellite over one complete orbit is:

M F_g v

(A) \(W = 2\pi r (G M m / r^2)\)

(B) \(W = 0\text{ J}\)

(C) \(W = \pi m v^2\)

(D) \(W = \frac{1}{2}m v^2\)

(E) \(W = -G M m / r\)

4. Gravitational Work: Ascent vs Descent [1 Pt]

An apple of mass \(m\) is tossed vertically upward to peak height \(h\) and caught upon returning. What are the signs of work done by gravity during ascent and descent?

d ↑ mg ↓ d, mg ↓

(A) Positive on ascent; positive on descent.

(B) Negative on ascent; negative on descent.

(C) Positive on ascent; negative on descent.

(D) Negative on ascent; positive on descent; net round-trip work is zero.

Zero on ascent; negative on descent.

Work and Energy Review Answer Key

AP Physics C: Mechanics • Chapter 6 Review Guide

Work & Energy Review Answer Key

Key • Page 1 of 5

Page 1 Answers: 1. (C) • 2. (C) • 3. (B) • 4. (D) Detailed Rationales & AP Exam Pitfalls

1. Kinematic Velocity & Acceleration Vectors Correct: (C)

Physical Rationale: In Uniform Circular Motion (UCM), scalar speed is constant (\(v = \text{const}\)), meaning tangential acceleration is zero (\(a_t = \frac{dv}{dt} = 0\)). The acceleration is purely centripetal (\(a_c = v^2/r\)), directed radially inward toward the center of curvature. Since velocity \(\vec{v}\) is tangent to the circle and acceleration \(\vec{a}\) is radial, the vectors are perpendicular: \(\vec{a} \perp \vec{v} \implies \vec{a}\cdot\vec{v} = |\vec{a}||\vec{v}|\cos(90^\circ) = 0\).

• Teacher Note / Common Trap: Students often mistakenly pick (D) thinking constant speed means zero acceleration, forgetting velocity is a vector whose direction changes.

2. Angled Pull Work Expression Correct: (C)

Physical Rationale: By the definition of the dot product, \(W = \vec{T}\cdot\vec{d} = T d\cos\theta\). Resolving \(\vec{T}\) into components, the horizontal component \(T\cos\theta\) acts parallel to displacement, while the vertical component \(T\sin\theta\) acts perpendicular to displacement and does zero work. Thus, \(W = (T\cos\theta)d\).

• Teacher Note: Emphasize that only the parallel component of force performs mechanical work.

3. Centripetal Work in Planetary Motion Correct: (B)

Physical Rationale: In a circular orbit, the gravitational force \(\vec{F}_g\) points radially inward toward the planet's center, while the instantaneous displacement \(d\vec{r}\) is tangent to the orbit. At every infinitesimal step, \(\vec{F}_g \perp d\vec{r}\), which gives \(dW = \vec{F}_g \cdot d\vec{r} = 0\). Integrating around any arc or full revolution yields \(W = 0\text{ J}\).

• Teacher Note: This illustrates why satellites in circular orbits maintain constant kinetic energy without engine fuel.

4. Gravitational Work: Ascent vs Descent Correct: (D)

Physical Rationale: On the way up, displacement is upward while gravity is downward (\(\theta = 180^\circ \implies W_g = -mgh < 0\)), extracting kinetic energy. On the way down, displacement and gravity are both downward (\(\theta = 0^\circ \implies W_g = +mgh > 0\)), restoring kinetic energy. The net round-trip work is \(W_{net} = -mgh + mgh = 0\text{ J}\), confirming gravity is a conservative force.

• Teacher Note: Connecting this to the closed-loop integral \(\oint \vec{F}\cdot d\vec{r} = 0\) reinforces conservative force theory.

AP Physics C: Mechanics • Chapter 6 Solution Key Page 1 of 5

Problem 4: Safe Speed Envelope with Static Friction

[10 Points Total]

A track of radius \(r = 90.0\text{ m}\) is banked at \(\theta = 15.0^\circ\). Tires on the roadway provide a static friction coefficient of \(\mu_s = 0.30\).

(a) Sketch the FBD when the car travels at maximum speed \(v_{max}\) (impending slip up the incline):

Show direction of f_s

(b) Write the coupled force balance equations along \(x\) and \(y\) for the \(v_{max}\) condition:

(c) Compute the maximum safe speed \(v_{max}\) without skidding up the incline:

AP Physics C: Mechanics • Unit 3: Circular Motion Page 2 of 5

AP Physics C: Mechanics • Practice Assignment Page 3 of 5

Part 3: Tethered & Cylindrical Systems Conical Pendulums & The Gravitron Rotor

Problem 5: Conical Pendulum Dynamics

[12 Points Total]

A pendulum bob of mass \(m = 0.80\text{ kg}\) is suspended by a light string of length \(L = 1.60\text{ m}\). The bob moves at constant linear speed \(v\) in a horizontal circle, with the string maintaining a steady angle of \(\theta = 35.0^\circ\) with the vertical.

Free-Body Diagram of Bob:

Draw T, mg, and angle θ

(a) Calculate the tension \(T\) in the suspension string:

(b) Determine the radius of the circular path \(r\) and the linear speed \(v\) of the bob:

(c) Derive and compute the revolution period \(T_p\):

(d) If the string snaps when \(T = 3.0\,mg\), find the maximum cone angle \(\theta_{max}\):

Problem 6: Carnival Gravitron Ride

[10 Points Total]

In a Gravitron ride, riders stand against the inner vertical wall of a hollow cylinder of radius \(R = 4.80\text{ m}\). The static friction coefficient between clothing and the padded wall is \(\mu_s = 0.38\). The cylinder spins until its linear speed is sufficient to allow the floor to drop away without passengers slipping down.

FBD of Rider against Wall:

Show F_N, f_s, and mg

(a) State the force equations along the horizontal (\(x\)) and vertical (\(y\)) directions:

(b) Derive the symbolic formula for minimum tangential speed \(v_{min}\) and compute its value:

(c) Calculate the maximum allowable time period \(T_{max}\) for one complete revolution:

AP Physics C: Mechanics • Unit 3: Circular Motion Page 3 of 5

AP Physics C: Mechanics • Practice Assignment Page 4 of 5

Part 4: Vertical Circular Dynamics & Energy Ferris Wheels & Inverted Roller Coaster Loops

Problem 7: Observation Ferris Wheel Apparent Weight

[10 Points Total]

A \(60.0\text{ kg}\) passenger rides a large Ferris wheel of radius \(R = 15.0\text{ m}\). An electric motor maintains a constant linear speed of \(v = 6.50\text{ m/s}\).

(a) Draw FBD and calculate normal force \(F_{N,top}\) on the passenger at the highest point (apex):

(b) Draw FBD and calculate normal force \(F_{N,bot}\) on the passenger at the lowest point (nadir):

(c) What tangential speed \(v\) would cause the passenger to experience momentary weightlessness (\(F_N = 0\)) at the apex?

Problem 8: Roller Coaster Loop-the-Loop & Hill Crest

[12 Points Total]

A coaster cart of mass \(m = 400\text{ kg}\) is released from rest at height \(h\) above the base of a frictionless circular loop of radius \(R = 12.0\text{ m}\).

FBD at Apex of Inside Loop:

Both forces point down

(a) Derive the minimum release height \(h_{min}\) (in terms of \(R\)) necessary to maintain track contact at apex:

(b) If released from \(h = 2.5R\), prove that the normal force at the bottom of the loop is \(F_{N,bot} = 6\,mg\):

(c) After exiting the loop, the cart goes over a convex circular hill of radius \(r_{hill} = 20.0\text{ m}\). What maximum speed can it have at the hill crest without becoming airborne?

AP Physics C: Mechanics • Unit 3: Circular Motion Page 4 of 5

AP Physics C: Mechanics • Practice Assignment Page 5 of 5

Part 5: AP Conceptual Analysis & Justifications Five Rigorous Theoretical Explanations

Question 1: The Outward Fling Sensation in a Turning Automobile

[5 Points]

A passenger in a car making a sharp left turn claims: "A strong centrifugal force threw me outward to the right against the car door." Explain from the perspective of an inertial ground observer why the passenger presses against the door, and clearly identify the actual physical force acting on the passenger and its direction.

Question 2: Total Acceleration Vector in a Vertical Circle

[5 Points]

A mass attached to a string is swung in a vertical circle. At the instant the string is exactly horizontal (the 3 o'clock position while ascending): (a) Identify which force supplies the centripetal acceleration, and (b) identify which force supplies the tangential acceleration, and sketch the direction of the total acceleration vector \(\vec{a}_{net}\).

Question 3: Mechanical Work Done by Normal Force on a Banked Turn

[5 Points]

A car traverses a frictionless banked curve of radius \(r\) at the exact ideal speed \(v_{ideal}\). Does the normal force \(F_N\) exerted by the tilted road do positive, negative, or zero mechanical work on the vehicle over a quarter-turn? Justify using the vector definition \(dW = \vec{F}_N \cdot d\vec{r}\).

Question 4: Impossibility of a Horizontal Conical Pendulum

[5 Points]

A student claims that with a sufficiently high motor speed, a conical pendulum bob can be made to spin so that the suspension string becomes completely horizontal (\(\theta = 90.0^\circ\)). Using Newton's second law along the vertical axis, prove mathematically why this condition is physically impossible.

Question 5: Dynamic Scaling in the Carnival Gravitron

[5 Points]

A person is pinned safely against the wall of a rotating Gravitron cylinder with the floor dropped. If the cylinder's linear speed \(v\) is doubled (\(v \to 2v\)): (a) by what factor does the wall normal force \(F_N\) change, and (b) by what factor does the vertical static friction force \(f_s\) change? Explain carefully.

AP Physics C: Mechanics • Unit 3: Circular Motion Page 5 of 5

12 Pts

Scored FBD [2 Pts]: F_N mg 1 pt: \(F_N \perp\) slope; 1 pt: \(mg \downarrow\)

(a) Derivation of \(\tan\theta = \frac{v^2}{gr}\) [4 Pts]: \(\Sigma F_y = F_N\cos\theta - mg = 0 \implies F_N = \frac{mg}{\cos\theta}\) [1.5 pts]
\(\Sigma F_x = F_N\sin\theta = m\,\frac{v^2}{r}\) [1.5 pts]
\(\frac{F_N\sin\theta}{F_N\cos\theta} = \frac{m v^2 / r}{mg} \implies \tan\theta = \frac{v^2}{gr}\) [1 pt]

(b) Angle Calculation [2 Pts]: \(\tan\theta = \frac{22.0^2}{(9.8)(140)} = \frac{484}{1372} \approx 0.3528 \implies \theta = \arctan(0.3528) \approx \mathbf{19.4^\circ}\) [2 pts]

(c) Normal Force \(F_N\) [2 Pts]: \(F_N = \frac{mg}{\cos\theta} = \frac{(1200\text{ kg})(9.8\text{ m/s}^2)}{\cos(19.4^\circ)} = \frac{11760}{0.9432} \approx \mathbf{12,470\text{ N}}\) [2 pts]

(d) Slow Speed on Ice (\(v = 12\text{ m/s}\)) [2 Pts]: At \(12\text{ m/s}\), the required centripetal force \(m v^2 / r\) is less than the inward normal force component \(F_N\sin\theta\). The excess inward force accelerates the car down the icy incline into the median. [2 pts]

Problem 4 Solution: Banked Turn with Static Friction [10 Points]

10 Pts

(a) FBD at \(v_{max}\) [3 Pts]:

Impending slip is up the slope, so static friction \(f_s\) points down the incline:

1 pt: \(F_N \perp\) slope
1 pt: \(mg \downarrow\)
1 pt: \(f_s\) parallel to slope pointing downward

(b) Component Equations [3 Pts]: \(\Sigma F_x = F_N\sin\theta + f_s\cos\theta = m\,\frac{v_{max}^2}{r}\) [1.5 pts]
\(\Sigma F_y = F_N\cos\theta - f_s\sin\theta - mg = 0\) [1.5 pts]

(c) Computation of \(v_{max}\) [4 Pts]: With \(f_s = \mu_s F_N\): \(v_{max} = \sqrt{gr\,\left(\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}\right)}\) [2 pts]
\(\sin(15^\circ) = 0.2588\), \(\cos(15^\circ) = 0.9659\). Numerator: \(0.2588 + (0.30)(0.9659) = 0.5486\).
Denominator: \(0.9659 - (0.30)(0.2588) = 0.8883\). Ratio: \(0.5486 / 0.8883 \approx 0.6176\).
\(v_{max} = \sqrt{(9.8)(90.0)(0.6176)} = \sqrt{544.7} \approx \mathbf{23.3\text{ m/s}}\) (\(84.0\text{ km/h}\)) [2 pts]

AP Physics C: Mechanics • Teacher Solution Key Page 2 of 5

AP Physics C: Mechanics • Teacher Solution Guide Key • Page 3 of 5

Part 3 Solutions: Conical Pendulum & The Rotor Problems 5 & 6 Rubric [22 Points Total]

Problem 5 Solution: Conical Pendulum Dynamics [12 Points]

12 Pts

Scored FBD [2 Pts]: T mg 1 pt: \(T\) along string; 1 pt: \(mg \downarrow\)

(a) Cord Tension \(T\) [3 Pts]: \(\Sigma F_y = T\cos\theta - mg = 0 \implies T = \frac{mg}{\cos\theta}\) [1.5 pts]
\(T = \frac{(0.80\text{ kg})(9.8\text{ m/s}^2)}{\cos(35.0^\circ)} = \frac{7.84}{0.8192} \approx \mathbf{9.57\text{ N}}\) [1.5 pts]

(b) Radius \(r\) and Linear Speed \(v\) [3 Pts]: \(r = L\sin\theta = (1.60\text{ m})\sin(35.0^\circ) = \mathbf{0.918\text{ m}}\) [1 pt]
\(\Sigma F_x = T\sin\theta = m\,\frac{v^2}{r} \implies v = \sqrt{gr\tan\theta}\) [1 pt]
\(v = \sqrt{(9.8)(0.918)\tan(35.0^\circ)} = \sqrt{8.996 \times 0.7002} \approx \mathbf{2.51\text{ m/s}}\) [1 pt]

(c) Period \(T_p\) [2 Pts]: \(T_p = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{(1.60)(0.8192)}{9.8}} = 2\pi\sqrt{0.1337} \approx \mathbf{2.30\text{ s}}\) [2 pts]

(d) Breaking Angle Limit [2 Pts]: \(T_{max} = \frac{mg}{\cos\theta_{max}} = 3.0\,mg \implies \cos\theta_{max} = \frac{1}{3} \approx 0.3333\)
\(\theta_{max} = \arccos(0.3333) \approx \mathbf{70.5^\circ}\) [2 pts]

Problem 6 Solution: Carnival Gravitron Ride [10 Points]

10 Pts

Scored FBD [2 Pts]: F_N f_s mg 1 pt: \(F_N\) inward; 1 pt: \(f_s \uparrow\) and \(mg \downarrow\)

(a) Equations of Motion [3 Pts]: \(\Sigma F_x = F_N = m\,\frac{v^2}{R}\) [1.5 pts]; \(\Sigma F_y = f_s - mg = 0 \implies f_s = mg\) [1.5 pts]

(b) Minimum Linear Speed \(v_{min}\) [3 Pts]: \(f_s \le \mu_s F_N \implies mg \le \mu_s\left(m\,\frac{v^2}{R}\right) \implies v_{min} = \sqrt{\frac{gR}{\mu_s}}\) [1.5 pts]
\(v_{min} = \sqrt{\frac{(9.8)(4.80)}{0.38}} = \sqrt{123.79} \approx \mathbf{11.1\text{ m/s}}\) [1.5 pts]

(c) Maximum Revolution Period \(T_{max}\) [2 Pts]: \(T_{max} = \frac{2\pi R}{v_{min}} = \frac{2\pi(4.80\text{ m})}{11.13\text{ m/s}} \approx \mathbf{2.71\text{ s}}\) per revolution. [2 pts]

AP Physics C: Mechanics • Teacher Solution Key Page 3 of 5

AP Physics C: Mechanics • Teacher Solution Guide Key • Page 4 of 5

Part 4 Solutions: Vertical Circular Dynamics Problems 7 & 8 Rubric [22 Points Total]

Problem 7 Solution: Ferris Wheel Apparent Weight [10 Points]

10 Pts

(a) Normal Force at Apex \(F_{N,top}\) [4 Pts]: FBD: \(F_N \uparrow\), \(mg \downarrow\), \(\vec{a}_c \downarrow\) toward center. [1 pt]
\(\Sigma F_r = mg - F_{N,top} = m\,\frac{v^2}{R}\) [1.5 pts]
\(F_{N,top} = m\left(g - \frac{v^2}{R}\right) = 60.0\left(9.8 - \frac{6.50^2}{15.0}\right)\)
\(F_{N,top} = 60.0(9.8 - 2.817) = 60.0(6.983) \approx \mathbf{419\text{ N}}\) [1.5 pts]

(b) Normal Force at Nadir \(F_{N,bot}\) [4 Pts]: FBD: \(F_N \uparrow\), \(mg \downarrow\), \(\vec{a}_c \uparrow\) toward center. [1 pt]
\(\Sigma F_r = F_{N,bot} - mg = m\,\frac{v^2}{R}\) [1.5 pts]
\(F_{N,bot} = m\left(g + \frac{v^2}{R}\right) = 60.0(9.8 + 2.817)\)
\(F_{N,bot} = 60.0(12.617) \approx \mathbf{757\text{ N}}\) [1.5 pts]

(c) Apparent Weightlessness Condition [2 Pts]: \(F_{N,top} = 0 \implies m\left(g - \frac{v^2}{R}\right) = 0 \implies v = \sqrt{gR} = \sqrt{(9.8)(15.0)} = \sqrt{147} \approx \mathbf{12.1\text{ m/s}}\) [2 pts]

Problem 8 Solution: Roller Coaster Loop & Hill [12 Points]

12 Pts

Apex Loop FBD [2 Pts]: F_N mg 1 pt: \(F_N \downarrow\); 1 pt: \(mg \downarrow\) (both down)

(a) Minimum Release Height \(h_{min}\) [4 Pts]: At loop apex: \(\Sigma F_r = F_{N,top} + mg = m\,\frac{v_{top}^2}{R} \implies F_{N,top} \ge 0 \iff v_{top}^2 \ge gR\) [2 pts]
Conservation of Energy: \(mgh = mg(2R) + \frac{1}{2}m v_{top}^2 = 2mgR + \frac{1}{2}mgR \implies \mathbf{h_{min} = 2.5\,R}\) [2 pts]

(b) The "6g" Bottom Normal Force [3 Pts]: From \(h = 2.5R\): \(v_{bot}^2 = 2gh = 2g(2.5R) = 5gR\) [1 pt]
\(F_{N,bot} - mg = m\,\frac{v_{bot}^2}{R} = \frac{m(5gR)}{R} = 5mg\) [1 pt]
\(F_{N,bot} = 5mg + mg = \mathbf{6\,mg}\) [1 pt] (Q.E.D.)

(c) Hill Crest Airborne Threshold [3 Pts]: Over convex hill: \(mg - F_N = m\,\frac{v^2}{r_{hill}}\) [1.5 pts]
Stay on track: \(F_N \ge 0 \implies v_{max} = \sqrt{g r_{hill}}\)
\(v_{max} = \sqrt{(9.8)(20.0)} = \sqrt{196} = \mathbf{14.0\text{ m/s}}\) [1.5 pts]

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AP Physics C: Mechanics • Teacher Solution Guide Key • Page 5 of 5

Part 5 Solutions: Conceptual Analysis & Rubrics Questions 1 – 5 Rubric [25 Points Total • 5 Pts Each]

Question 1: Inertial Reference Frame & Door Contact [5 Pts] 5 Pts

Ideal AP Response: According to Newton's first law, the passenger's body possesses inertia and tends to maintain its straight-line velocity tangent to the road. As the automobile initiates a left turn, the car body rotates underneath the passenger. The passenger continues forward along a straight line until the right door intercepts them. The door exerts an inward normal force directed to the left toward the turn center. This real contact force provides the required centripetal acceleration. There is no real outward force; the outward sensation is merely the subjective feeling of one's own inertia resisting curved motion in an accelerating reference frame.

• Rubric: 2 pts for Newton's 1st law / inertia explanation; 2 pts for identifying normal force of door to the left; 1 pt for denying centrifugal force as an agent.

Question 2: Total Acceleration Vector at 3 o'clock [5 Pts] 5 Pts

Ideal AP Response: At the 3 o'clock position (string horizontal, moving upward):
(a) The string tension \(\vec{T}\) points horizontally to the left toward the center, supplying the centripetal acceleration \(a_r = T/m = v^2/R\).
(b) The gravitational force \(m\vec{g}\) acts straight down, perpendicular to the string, supplying the tangential acceleration \(a_t = g\).
The total acceleration vector is the vector sum: \(\vec{a}_{net} = \vec{a}_r + \vec{a}_t = -\frac{v^2}{R}\,\hat{i} - g\,\hat{j}\), which points diagonally inward and downward.

• Rubric: 2 pts for identifying string tension as radial; 2 pts for gravity as tangential; 1 pt for vector sum pointing inward/downward.

Question 3: Mechanical Work Done by Normal Force [5 Pts] 5 Pts

Ideal AP Response: The normal force \(F_N\) does zero mechanical work (\(W = 0\)). By definition, the normal force is perpendicular to the surface of the banked track. The instantaneous velocity vector \(\vec{v}\) and displacement element \(d\vec{r}\) are tangent to the horizontal circular curve, lying entirely within the local plane of the track surface. Consequently, \(\vec{F}_N \perp d\vec{r}\) at every instant, giving an angle \(\phi = 90^\circ\) between force and displacement. Thus, \(dW = \vec{F}_N \cdot d\vec{r} = |\vec{F}_N|\,|d\vec{r}|\cos(90^\circ) = 0\).

• Rubric: 2 pts for stating work is zero; 3 pts for demonstrating \(\vec{F}_N \perp d\vec{r}\) via dot product.

Question 4: Horizontal Conical Pendulum Impossibility [5 Pts] 5 Pts

Ideal AP Response: For a conical pendulum, the bob undergoes zero vertical acceleration (\(\Sigma F_y = 0\)). The only vertical forces are the upward component of string tension \(T\cos\theta\) and downward gravity \(mg\), requiring \(T\cos\theta = mg \implies T = \frac{mg}{\cos\theta}\). If the string were completely horizontal (\(\theta = 90.0^\circ\)), \(\cos(90^\circ) = 0\), which yields \(T = \frac{mg}{0} \to \infty\). Because no physical string can sustain infinite tension, a purely horizontal string has zero vertical force component to counteract gravity. The string must always hang at an angle \(\theta < 90^\circ\).

• Rubric: 2 pts for vertical equilibrium equation; 2 pts for showing \(\cos(90^\circ) = 0 \implies T \to \infty\); 1 pt for physical interpretation.

Question 5: Dynamic Scaling in the Gravitron [5 Pts] 5 Pts

Ideal AP Response:
(a) The wall normal force provides centripetal acceleration: \(F_N = m\,\frac{v^2}{R}\). Because \(F_N \propto v^2\), doubling speed (\(v \to 2v\)) causes the normal force to quadruple (increase by a factor of 4): \(F_N' = 4F_N\).
(b) The vertical static friction force satisfies vertical equilibrium: \(\Sigma F_y = f_s - mg = 0 \implies f_s = mg\). Because the passenger's mass and gravitational acceleration remain unchanged, the actual static friction force remains unchanged (factor of 1). The maximum allowable friction capacity \(\mu_s F_N\) quadruples, providing a fourfold greater safety margin against sliding, but the actual force exerted by the wall remains exactly \(mg\).

• Rubric: 2.5 pts for proving \(F_N \propto v^2 \implies 4\times\); 2.5 pts for explaining that actual friction \(f_s = mg\) is unchanged (\(1\times\)).

AP Physics C: Mechanics • Teacher Solution Key Page 5 of 5

(D) \(|\vec{a}_{net}| = \sqrt{a_t^2 + a_t t^2 / R}\)

(E) \(|\vec{a}_{net}| = a_t + a_t^2 t^2 / R\)

AP Physics C: Mechanics • Circular Dynamics MCQ Review Page 1 of 5

AP Physics C: Mechanics • Unit 3 Assessment Page 2 of 5

5. Automobile Turning on a Flat Unbanked Curve [1 Pt]

A car of mass \(m\) negotiates an unbanked curve of radius \(r\) at speed \(v\). The static friction coefficient is \(\mu_s\). Which interaction directly supplies the inward centripetal acceleration?

F_N mg f_s (inward)

(A) Downward gravity of Earth.

(B) Upward road normal force.

(C) Static friction directed radially inward toward center of curvature.

(D) Backward kinetic friction.

(E) Outward centrifugal force.

6. Mass Comparison on Flat Circular Track [1 Pt]

Car A has mass \(m\) and Car B has mass \(2m\). Both vehicles enter the exact same unbanked turn of radius \(r\) at identical speed \(v\). Which statement correctly compares their motion?

m 2m

(A) Car B needs twice the friction force of Car A, but both share identical \(v_{max}\).

(B) Car B has higher maximum speed due to greater normal force.

(C) Car A requires twice the centripetal acceleration of Car B.

(D) Car A has twice the maximum safe speed because it is lighter.

(E) Both vehicles experience identical friction forces regardless of mass.

7. Critical Slipping Radius on a Rotating Turntable [1 Pt]

A turntable completes one revolution in period \(T\). A small coin rests on the turntable with static friction coefficient \(\mu_s\). What is the maximum radial distance \(r_{max}\) before slipping?

r_max

(A) \(r_{max} = \frac{\mu_s g T}{2\pi}\)

(B) \(r_{max} = \frac{\mu_s g T^2}{4\pi^2}\)

(C) \(r_{max} = \frac{4\pi^2 \mu_s g}{T^2}\)

(D) \(r_{max} = 2\pi\sqrt{\frac{\mu_s g}{T}}\)

(E) \(r_{max} = \frac{\mu_s T^2}{g}\)

8. Comparative Slipping of Two Coins on a Turntable [1 Pt]

Identical coins \(C_1\) and \(C_2\) are on a turntable at \(r_1 = d\) and \(r_2 = 2d\). As rotation speed gradually increases (period \(T\) decreases), which coin slips first and why?

C₁ (d) C₂ (2d)

(A) \(C_1\) slips first (tighter turn).

(B) \(C_2\) slips first because required \(a_c = 4\pi^2 r / T^2\) is twice as large.

(C) Both coins slip simultaneously.

(D) \(C_2\) slips first because \(\mu_s\) drops.

(E) \(C_1\) slips first due to speed.

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9. Frictionless Banked Turn Design Speed [1 Pt]

A roadway curve of radius \(r\) is banked at angle \(\theta\). At what linear speed \(v_{ideal}\) can a vehicle negotiate the turn without relying on any friction?

θ F_N

(A) \(v_{ideal} = \sqrt{g r \sin\theta}\)

(B) \(v_{ideal} = \sqrt{g r \cos\theta}\)

(C) \(v_{ideal} = \sqrt{g r \tan\theta}\)

(D) \(v_{ideal} = \sqrt{\frac{g r}{\tan\theta}}\)

(E) \(v_{ideal} = g r \tan\theta\)

10. Direction of Static Friction on a Banked Curve [1 Pt]

A car traverses a banked curve of angle \(\theta\) at a speed \(v\) greater than ideal design speed (\(v > v_{ideal}\)). In which direction does static friction \(\vec{f}_s\) act?

f_s (down)

(A) Parallel to the road surface, pointing down the incline toward the inside of the curve.

(B) Parallel to the road surface, pointing up the incline toward the outside of the curve.

(C) Strictly vertically upward, directly opposing gravity.

(D) Perpendicular to the road surface, parallel to the normal force.

(E) Zero, because normal force alone supplies the centripetal acceleration.

11. Normal Force Magnitude on a Banked Road [1 Pt]

A vehicle of mass \(m\) travels at ideal speed \(v_{ideal}\) around a curve banked at angle \(\theta\). How does normal force \(F_N\) compare to rest weight \(mg\)?

F_N mg m v²/r

(A) \(F_N = mg\cos\theta < mg\)

(B) \(F_N = mg\)

(C) \(F_N = \frac{mg}{\cos\theta} > mg\)

(D) \(F_N = mg\tan\theta\)

(E) \(F_N = \frac{mg}{\sin\theta}\)

12. Slow Vehicle on a Frictionless Banked Track [1 Pt]

A car drives at speed \(v < v_{ideal}\) on a frictionless banked curve. Which describes the resulting motion and net force?

Slide ↓

(A) The car continues smoothly because inertia automatically increases to balance the turn.

(B) Inward normal force \(F_N\sin\theta\) exceeds required \(m v^2/r\), accelerating the car down the slope.

(C) Outward centrifugal force exceeds the inward force, pushing the car up the incline.

(D) The normal force drops to zero and the car flips over backwards.

(E) The car lifts vertically upward off the roadway surface.

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13. Cord Tension in a Conical Pendulum [1 Pt]

A bob of mass \(m\) on a string of length \(L\) executes horizontal circular motion at cone angle \(\theta\) from vertical. What is the tension \(T\) in the string?

T mg

(A) \(T = mg\cos\theta\)

(B) \(T = mg\sin\theta\)

(C) \(T = \frac{mg}{\cos\theta}\)

(D) \(T = mg\tan\theta\)

(E) \(T = mg\)

14. Period Scaling of a Conical Pendulum [1 Pt]

The period is \(T_p = 2\pi\sqrt{\frac{L\cos\theta}{g}}\). If string length is quadrupled (\(L \to 4L\)) at constant angle \(\theta\), the period:

L 4L

(A) Decreases by factor 4.

(B) Decreases by factor 2.

(C) Remains completely unchanged.

(D) Doubles (increases by 2).

(E) Quadruples (increases by 4).

15. Minimum Safe Speed in the Carnival Rotor [1 Pt]

In a Gravitron cylinder of radius \(R\), riders are held by wall friction (\(\mu_s\)). What is minimum safe speed \(v_{min}\) when floor drops?

F_N f_s mg

(A) \(v_{min} = \sqrt{\mu_s g R}\)

(B) \(v_{min} = \sqrt{\frac{g R}{\mu_s}}\)

(C) \(v_{min} = \sqrt{\frac{\mu_s g}{R}}\)

(D) \(v_{min} = \frac{g R}{\mu_s}\)

(E) \(v_{min} = \mu_s \sqrt{g R}\)

16. Dynamic Scaling in the Rotor Ride [1 Pt]

A rider is pinned against the wall of a Gravitron. If cylinder speed \(v\) is doubled (\(v \to 2v\)), how do \(F_N\) and \(f_s\) change?

v F_N ∝ v² f_s = mg (const)

(A) Both \(F_N\) & \(f_s\) quadruple.

(B) Both \(F_N\) & \(f_s\) double.

(C) \(F_N\) quadruples, while \(f_s\) remains unchanged at \(mg\).

(D) \(F_N\) doubles; \(f_s\) is halved.

(E) Neither force changes.

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17. Vertical String Tension Difference [1 Pt]

A mass \(m\) is whirled in a vertical circle of radius \(R\) on an unpowered string. Assuming energy conservation, the difference \(T_{bot} - T_{top}\) is:

T_top ↓ T_bot ↑

(A) \(2\,mg\)

(B) \(4\,mg\)

(C) \(6\,mg\)

(D) \(8\,mg\)

(E) Dependent on the specific apex speed \(v_{top}\).

18. Critical Speed to Prevent String Slacking [1 Pt]

What minimum speed \(v_{crit}\) must an object have at the apex of a vertical circle of radius \(R\) to complete the loop without string slacking?

Center O v mg

(A) \(v_{crit} = 0\)

(B) \(v_{crit} = \frac{1}{2}\sqrt{g R}\)

(C) \(v_{crit} = \sqrt{g R}\)

(D) \(v_{crit} = \sqrt{2 g R}\)

(E) \(v_{crit} = \sqrt{5 g R}\)

19. Ferris Wheel Apparent Weight Ratio [1 Pt]

A motorized Ferris wheel of radius \(R\) turns at \(v = \sqrt{gR / 2}\). What is the ratio of apparent weight at bottom to top (\(F_{N,bot} / F_{N,top}\))?

(A) \(1.5\)

(B) \(2.0\)

(C) \(3.0\)

(D) \(4.0\)

(E) \(1.0\) (apparent weight is identical since speed is constant)

20. Inverted Roller Coaster Normal Force [1 Pt]

A cart is released from rest at \(h = 2.5R\) on a frictionless circular loop of radius \(R\). What normal force acts on the cart at the lowest point?

2.5R F_N

(A) \(mg\)

(B) \(3\,mg\)

(C) \(5\,mg\)

(D) \(6\,mg\)

(E) \(7\,mg\)

AP Physics C: Mechanics • Circular Dynamics MCQ Review Page 5 of 5

5. Automobile Turning on a Flat Unbanked Curve Correct: (C)

Physical Rationale: On a level surface, gravity and normal force are vertical and balance (\(F_N = mg\)). To accelerate horizontally toward the center of curvature, an unbalanced horizontal force must exist. Static friction between tire contact patches and pavement points radially inward, supplying \(\Sigma F_r = f_s = m v^2/r\).

• Teacher Note: Reiterate that tires roll without slipping laterally, so static (not kinetic) friction governs safe turning.

6. Mass Comparison on Flat Circular Track Correct: (A)

Mathematical Derivation: Required force is \(f_s = m v^2/r\). For Car B with mass \(2m\), required friction is \(2m v^2/r = 2 f_{s,A}\) (twice as large).
However, maximum static friction is \(f_{s,max} = \mu_s F_N = \mu_s mg\). Equating: \(m v^2/r = \mu_s mg \implies v_{max} = \sqrt{\mu_s gr}\). Mass cancels completely; both cars share the exact same speed limit!

• Teacher Note: This is an essential AP concept: vehicle mass does not change the maximum safe turning speed.

7. Critical Slipping Radius on a Rotating Turntable Correct: (B)

Mathematical Derivation: In terms of period \(T\) (\(v = 2\pi r / T\)), the required centripetal force is \(f_s = m \frac{v^2}{r} = \frac{4\pi^2 m r}{T^2}\).
Setting \(f_s \le \mu_s mg\): \(\frac{4\pi^2 m r_{max}}{T^2} = \mu_s mg \implies r_{max} = \frac{\mu_s g T^2}{4\pi^2}\).

• Teacher Note: If placed further out (\(r > r_{max}\)), static friction cannot meet demand and the coin slips outward.

8. Comparative Slipping of Two Coins on a Turntable Correct: (B)

Physical Rationale: Both coins have the same period \(T\). The required centripetal acceleration is \(a_c = 4\pi^2 r / T^2 \propto r\). Coin \(C_2\) at \(r = 2d\) requires twice as much centripetal acceleration (and twice as much static friction) as Coin \(C_1\) at \(r = d\). As \(T\) decreases, Coin \(C_2\) reaches its static friction limit (\(\mu_s g\)) first and slips.

• Teacher Note: This reinforces that at fixed rotation period, larger radius demands higher centripetal force.

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AP Physics C: Mechanics • Unit 3 Solution Guide Key • Page 3 of 5

Page 3 Answers: 9. (C) • 10. (A) • 11. (C) • 12. (B) Detailed Rationales & AP Exam Pitfalls

9. Frictionless Banked Turn Design Speed Correct: (C)

Mathematical Derivation: Vertical equilibrium: \(\Sigma F_y = F_N\cos\theta - mg = 0 \implies F_N = \frac{mg}{\cos\theta}\).
Horizontal centripetal balance: \(\Sigma F_x = F_N\sin\theta = m\,\frac{v^2}{r}\).
Dividing: \(\frac{F_N\sin\theta}{F_N\cos\theta} = \frac{m v^2 / r}{mg} \implies \tan\theta = \frac{v^2}{gr} \implies v_{ideal} = \sqrt{gr\tan\theta}\).

• Teacher Note: This is the rated design speed: the normal force alone provides the required centripetal acceleration.

10. Direction of Static Friction on a Banked Curve Correct: (A)

Physical Rationale: When traveling faster than ideal design speed (\(v > v_{ideal}\)), the horizontal normal force component \(F_N\sin\theta\) is insufficient to supply \(m v^2/r\). The vehicle tends to slide upward along the incline toward the outside of the turn. Static friction opposes this impending slip by pointing down the incline toward the inside of the curve.

• Teacher Note: Conversely, if \(v < v_{ideal}\), the car tends to slide down the slope, so friction points up the incline.

11. Normal Force Magnitude on a Banked Road Correct: (C)

Mathematical Derivation: Since there is zero vertical acceleration (\(a_y = 0\)):
\(\Sigma F_y = F_N\cos\theta - mg = 0 \implies F_N = \frac{mg}{\cos\theta}\).
Because \(\cos\theta < 1\) for any angle \(\theta > 0\), \(F_N > mg\). The road supports more than the car's rest weight!

• Teacher Note: Common trap: students confuse this with an object on a stationary inclined plane where \(F_N = mg\cos\theta\). On a flat horizontal turn, \(a\) is horizontal, requiring \(F_N = mg/\cos\theta\).

12. Slow Vehicle on a Frictionless Banked Track Correct: (B)

Physical Rationale: On a frictionless track with \(v < v_{ideal}\), the centripetal demand \(m v^2/r\) is less than the inward horizontal component \(F_N\sin\theta\). The net horizontal force exceeds what is needed to maintain the circular radius, accelerating the vehicle down the incline toward the inside of the curve.

• Teacher Note: Real-world engineering context: this explains why icy banked turns can cause slow cars to slide down into the median.

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AP Physics C: Mechanics • Unit 3 Solution Guide Key • Page 4 of 5

Page 4 Answers: 13. (C) • 14. (D) • 15. (B) • 16. (C) Detailed Rationales & AP Exam Pitfalls

13. Cord Tension in a Conical Pendulum Correct: (C)

Mathematical Derivation: The bob moves in a horizontal plane, so vertical acceleration is zero (\(a_y = 0\)).
\(\Sigma F_y = T\cos\theta - mg = 0 \implies T = \frac{mg}{\cos\theta}\).
Because \(\cos\theta < 1\), the cord tension always strictly exceeds the bob's rest weight: \(T > mg\).

• Teacher Note: As \(\theta \to 90^\circ\), \(\cos\theta \to 0\), meaning \(T \to \infty\). A conical pendulum string can never become perfectly horizontal.

14. Period Scaling of a Conical Pendulum Correct: (D)

Mathematical Derivation: The period is \(T_p = 2\pi\sqrt{\frac{L\cos\theta}{g}}\).
Since period scales with the square root of length (\(T_p \propto \sqrt{L}\)), quadrupling length (\(4L\)) gives:
\(T_p' = 2\pi\sqrt{\frac{(4L)\cos\theta}{g}} = \sqrt{4}\left(2\pi\sqrt{\frac{L\cos\theta}{g}}\right) = 2\,T_p\). The period doubles.

• Teacher Note: Notice the mathematical identity with a simple planar pendulum of length \(h = L\cos\theta\).

15. Minimum Safe Speed in the Carnival Rotor Correct: (B)

Mathematical Derivation: The wall normal force provides centripetal acceleration: \(F_N = m v^2/R\).
Vertical equilibrium prevents slipping: \(f_s = mg\).
Static friction limit: \(f_s \le \mu_s F_N \implies mg \le \mu_s\left(m\,\frac{v^2}{R}\right) \implies v_{min} = \sqrt{\frac{gR}{\mu_s}}\).

• Teacher Note: Mass \(m\) cancels completely: riders of any weight stick with identical safety margins.

16. Dynamic Scaling in the Rotor Ride Correct: (C)

Physical Rationale: The wall normal force scales quadratically with speed (\(F_N = m v^2/R \propto v^2\)), so doubling \(v\) quadruples \(F_N\) (\(4\times\)). However, vertical equilibrium strictly requires \(f_s - mg = 0 \implies f_s = mg\). Because the rider's weight is constant, the actual static friction force remains unchanged (\(1\times\)). The maximum possible friction capacity \(\mu_s F_N\) quadruples, but the actual force needed to hold the rider up remains \(mg\).

• Teacher Note: Classic AP trap: students often assume \(f_s = \mu_s F_N\) at all times, rather than \(f_s \le \mu_s F_N\).

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AP Physics C: Mechanics • Unit 3 Solution Guide Key • Page 5 of 5

Page 5 Answers: 17. (C) • 18. (C) • 19. (C) • 20. (D) Detailed Rationales & AP Exam Pitfalls

17. Vertical String Tension Difference Correct: (C)

Mathematical Derivation:
At apex: \(T_{top} + mg = m\,\frac{v_{top}^2}{R} \implies T_{top} = \frac{m v_{top}^2}{R} - mg\).
At nadir: \(T_{bot} - mg = m\,\frac{v_{bot}^2}{R} \implies T_{bot} = \frac{m v_{bot}^2}{R} + mg\).
Energy conservation: \(\frac{1}{2}m v_{bot}^2 = \frac{1}{2}m v_{top}^2 + mg(2R) \implies v_{bot}^2 - v_{top}^2 = 4gR\).
\(T_{bot} - T_{top} = \frac{m(v_{bot}^2 - v_{top}^2)}{R} + 2mg = \frac{m(4gR)}{R} + 2mg = 4mg + 2mg = \mathbf{6\,mg}\).

• Teacher Note: This \(6mg\) tension difference is constant for any free vertical circular motion, independent of radius or speed.

18. Critical Speed to Prevent String Slacking Correct: (C)

Mathematical Derivation: At the apex, \(\Sigma F_r = T_{top} + mg = m v_{top}^2 / R\).
The string remains taut as long as \(T_{top} \ge 0\). The minimum critical speed occurs when tension drops to zero (\(T_{top} = 0\)):
\(0 + mg = m\,\frac{v_{crit}^2}{R} \implies v_{crit} = \sqrt{gR}\).

• Teacher Note: At \(v_{crit}\), gravity alone provides the exact inward acceleration needed to maintain the circular arc.

19. Ferris Wheel Apparent Weight Ratio Correct: (C)

Mathematical Derivation: Here \(v^2 / R = (gR/2)/R = g/2\).
At top: \(mg - F_{N,top} = m(v^2/R) = mg/2 \implies F_{N,top} = mg - mg/2 = 0.5\,mg\).
At bottom: \(F_{N,bot} - mg = m(v^2/R) = mg/2 \implies F_{N,bot} = mg + mg/2 = 1.5\,mg\).
Ratio: \(\frac{F_{N,bot}}{F_{N,top}} = \frac{1.5\,mg}{0.5\,mg} = \mathbf{3.0}\).

• Teacher Note: The passenger feels three times heavier at the bottom than at the top!

20. Inverted Roller Coaster Normal Force Correct: (D)

Mathematical Derivation: Released from rest at \(h = 2.5R\):
Speed at bottom: \(\frac{1}{2}m v_{bot}^2 = mgh = mg(2.5R) \implies v_{bot}^2 = 5gR\).
Newton's 2nd law at lowest point: \(F_{N,bot} - mg = m\,\frac{v_{bot}^2}{R} = \frac{m(5gR)}{R} = 5mg\).
\(F_{N,bot} = 5mg + mg = \mathbf{6\,mg}\).

• Teacher Note: This severe \(6g\) load is why roller coaster designers use clothoid loops rather than pure circular loops.

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2. Types of Work & Zero Work Conditions

Because the dot product contains \(\cos\theta\), the algebraic sign of mechanical work is governed entirely by the relative angle \(\theta\) between the applied force vector \(\vec{F}\) and the displacement vector \(\vec{d}\).

Geometric Regimes of the Dot Product (Slide 4)

(a) Positive Work (\(W > 0\)) d F θ \(-90^\circ < \theta < 90^\circ\) \(\cos\theta > 0\); energy added; speed increases.

(b) Zero Work (\(W = 0\)) d F \(\theta = \pm 90^\circ\) (\(\vec{F} \perp \vec{d}\)) \(\cos(90^\circ) = 0\); zero energy transferred.

(c) Negative Work (\(W < 0\)) d F θ \(90^\circ < \theta \le 180^\circ\) \(\cos\theta < 0\); energy removed; speed decreases.

Case 3: When Force and Displacement Exist, Yet No Work Is Done

Example 1: Carrying an Object Horizontally

When a person walks across a room carrying an object, their hands exert an upward supporting force (\(\vec{F}_{up}\)) to balance gravity. However, displacement \(\vec{d}\) is horizontal (\(\theta = 90^\circ\)): \(W = F\,d\cos(90^\circ) = 0\text{ J}\) The upward lifting force does no work on the object because it has zero component along the displacement.

Example 2: Centripetal Forces in Orbit

For a satellite or celestial body in a circular orbit, gravity acts radially inward toward the central mass, while instantaneous velocity is tangent to the circle (\(\vec{F}_g \perp \vec{v}\)): \(dW = \vec{F}_c \cdot d\vec{r} = F_c\,dr\cos(90^\circ) = 0\text{ J}\) Centripetal forces continually alter the trajectory direction, but do strictly zero work.

Apple Falling Downward (\(W_g > 0\)): Force \(m\vec{g}\) and displacement \(\vec{d}\) both point downward (\(\theta = 0^\circ\)). \(W_g = (mg)d\cos(0^\circ) = +mgh > 0\) Positive work adds kinetic energy; the falling apple speeds up.

Apple Tossed Upward (\(W_g < 0\)): Force \(m\vec{g}\) points downward while displacement \(\vec{d}\) is upward (\(\theta = 180^\circ\)). \(W_g = (mg)d\cos(180^\circ) = -mgh < 0\) Negative work extracts kinetic energy; the rising apple slows down.

Checkpoint 2.1: Horizontal Book Walk A student walks \(15.0\text{ m}\) at constant speed carrying a \(4.0\text{ kg}\) backpack horizontally. How much work do their arms do?

Checkpoint 2.2: Magnetic Field Work A proton enters a uniform magnetic field where the magnetic force is \(\vec{F}_B = q(\vec{v}\times\vec{B})\). Does the magnetic field do work?

Checkpoint 2.3: Upward Toss Work A \(0.15\text{ kg}\) baseball is tossed straight up to a peak height of \(12.0\text{ m}\). Calculate the work done by gravity during the ascent.

Checkpoint 2.4: Lowered Elevator Cable An elevator descends at constant speed. What is the algebraic sign (\(+\), \(-\), or \(0\)) of the work done by the upward cable tension?

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3. Work Done by Friction & Normal Force Modulation

Kinetic friction \(\vec{f}_k\) invariably opposes the instantaneous relative sliding motion between contacting surfaces. Because the angle between kinetic friction and displacement is always \(\theta = 180^\circ\), the work done by kinetic friction is inherently negative: \(W_f = f_k s\cos(180^\circ) = -f_k s\).

Normal Force Modulation & Net Work Formulations (Slide 5)

Recall that the magnitude of friction depends on the perpendicular contact normal force: \(f_k = \mu_k F_N\). When applied external forces possess vertical components, the normal force is altered, directly modifying the friction force and the mechanical energy dissipated:

Force Applied Parallel to Surface: Vertical balance: \(F_N = mg\). Friction: \(f_k = \mu_k mg\).

\(W_T = W_{app} + W_f = F\,s - \mu_k mg\,s\)

Friction work scales strictly with object rest weight \(mg\).

Force Applied at Angle \(\theta\) Above Horizontal: Vertical balance: \(F_N + F\sin\theta = mg \implies F_N = mg - F\sin\theta\).

\(W_T = (F\cos\theta)s - \mu_k(mg - F\sin\theta)s\)

The upward component lifts the box, reducing normal force and friction!

Motion (s) Applied F Friction f_k F_N mg

Dissipation of Mechanical Energy

When friction does negative work on an object, mechanical energy is not conserved; it is dissipated into random thermal vibrations of the contacting atoms (internal thermal energy, or heat \(Q\)):

\(\Delta E_{thermal} = -W_f = +f_k\,s = +\mu_k F_N\,s\)

This thermal energy cannot be spontaneously recovered to do macroscopic mechanical work, illustrating the Second Law of Thermodynamics.

Checkpoint 3.1: Friction Angle Why is the angle \(\theta\) between kinetic friction and displacement always exactly \(180^\circ\) for any sliding motion?

Checkpoint 3.2: Downward Push Effect A box is pushed with force \(F\) angled at \(\theta\) below the horizontal. Express the normal force \(F_N\) and friction work \(W_f\).

Checkpoint 3.3: Incline Friction A crate of mass \(m\) slides distance \(d\) down a ramp of angle \(\phi\) with friction coefficient \(\mu_k\). Express \(W_f\) symbolically.

Checkpoint 3.4: Static Friction Work Can static friction ever perform positive mechanical work on an object? Give a clear real-world physical example.

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4. Work Done by a Variable Force: Calculus Integrals

In real physical systems, forces rarely remain constant over distance. When a force varies with spatial position \(F = F(x)\), simple multiplication fails. Work must be calculated as the definite integral—the geometric area under the curve of a Force versus Displacement graph.

Constant Force: Rectangle W = F · d x_m \(W = F \Delta x\)

Linear Spring: Triangle W = ½ k x² x_m \(W = \frac{1}{2}(F_0 + F)d\)

Arbitrary Force: Integral W = ∫ F(x)dx x_m \(W = \int_{x_0}^x F(x')dx'\)

Definite Integration Formulation (Slide 7)

Dividing the total path into infinitesimal displacements \(dx\), during which the force \(F(x)\) is approximately constant, the incremental work is \(dW = F(x)\,dx\). Integrating over the full path from initial position \(x_0\) to final position \(x\):

\(W = \int_{x_0}^x F(x')\,dx'\)

If the force changes at a constant linear rate from \(F_0\) to \(F\), the area is a trapezoid: \(W = \bar{F}\,d = \frac{1}{2}(F_0 + F)d\).

Work Done to Stretch an Ideal Spring (Hooke's Law)

According to Hooke's Law, an ideal spring exerts a restoring force \(F_s = -kx\). An external agent stretching the spring must apply an opposing force \(F_{ext} = +kx\). The work done by this external agent from rest position (\(x = 0\)) to extension \(x\) is:

\(W_{ext} = \int_0^x (kx')\,dx' = \left[\frac{1}{2}k(x')^2\right]_0^x = \frac{1}{2}k x^2\)

This work is stored completely within the spring as elastic potential energy (\(U_s = \frac{1}{2}kx^2\)).

Checkpoint 4.1: Polynomial Force A 1D force \(F(x) = (6.0 x^2 - 4.0 x)\text{ N}\) acts from \(x = 0\) to \(x = 3.0\text{ m}\). Calculate the total work done.

Checkpoint 4.2: Spring Work Ratio How much more work is required to stretch a spring from \(x\) to \(3x\) compared to stretching it from \(0\) to \(x\)?

Checkpoint 4.3: Negative Area If an \(F\) vs \(x\) graph dips below the horizontal axis (\(F < 0\)), what does the enclosed area represent physically?

Checkpoint 4.4: Variable Spring A non-linear spring exerts \(F(x) = k_1 x + k_2 x^3\). Derive the work done by an external agent to stretch it to \(x = D\).

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5. The Work-Energy Theorem & Kinetic Energy

The Work-Energy Theorem forms the central bridge between Newtonian forces and energy conservation: the net mechanical work performed by all combined forces acting upon a body equals the net change in its kinetic energy: \(W_{net} = \Delta K\).

AP Calculus Derivation of the Work-Energy Theorem (Slide 10)

Using Newton's second law \(F_{net} = m a\) and expressing acceleration via the chain rule as \(a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}\):

\(W_{net} = \int_{x_0}^x F_{net}\,dx' = \int_{x_0}^x (m a)\,dx'\)

\(W_{net} = \int m\left(\frac{dv}{dt}\right)dx' = \int m\left(\frac{dx'}{dt}\right)dv = \int_{v_0}^v m v'\,dv'\)

\(W_{net} = \left[\frac{1}{2}m(v')^2\right]_{v_0}^v = \frac{1}{2}m v^2 - \frac{1}{2}m v_0^2 = K_f - K_i = \Delta K\)

This proves that Kinetic Energy is defined as \(K = \frac{1}{2}m v^2\). Kinetic energy is always non-negative (\(K \ge 0\)) and is a scalar.

v Kinetic Energy (K) v → K 2v → 4K

The Quadratic Velocity Relationship (Slide 18)

Because kinetic energy scales with the square of speed (\(K \propto v^2\)), doubling an automobile's speed requires four times as much work to accelerate and four times as much distance to stop under constant braking friction.

\(W_{brake} = -f_k\,d = 0 - \frac{1}{2}mv^2 \implies d = \frac{v^2}{2\mu_k g}\)

Checkpoint 5.1: Braking Distance A car traveling at speed \(v\) stops in distance \(D\). If the driver increases speed to \(3v\), what is the new stopping distance?

Checkpoint 5.2: Speed from Work A \(1200\text{ kg}\) car at rest has \(+150\text{ kJ}\) of net work done on it. Compute its final speed \(v\).

Checkpoint 5.3: Friction Applicability Does the Work-Energy Theorem hold true when non-conservative forces like friction or drag are present? Explain why.

Checkpoint 5.4: Base Runner Stopping A \(70\text{ kg}\) player slides at \(v_0 = 4.0\text{ m/s}\) to rest with \(\mu_k = 0.70\). Calculate the stopping distance \(d\).

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6. Mechanical Energy & Energy Lost Due to Friction

Energy is defined as the capability of a physical system to perform mechanical work (Slide 14). In mechanics, we distinguish between Mechanical Energy—which can be directly recovered—and non-mechanical energy dissipated by friction into microscopic heat.

9. Mechanical Energy Definition & Composition (Slide 15)

Mechanical Energy (\(E_{mech}\)) is the total macroscopic energy possessed by an object due to its motion or its spatial position relative to an interacting field. It consists of two constituent components:

\(E_{mech} = \text{Kinetic Energy} + \text{Potential Energy} = K + U\)

Kinetic Energy (\(K = \frac{1}{2}mv^2\)): Energy of motion; dependent purely on object mass and scalar speed.

Potential Energy (\(U\)): Stored energy of position; exists only when conservative interaction forces act.

Mechanical Energy Lost Due to Friction (Slide 26)

When an object slides against a rough surface, kinetic friction performs negative work (\(W_f < 0\)). This negative work extracts macroscopic mechanical energy and converts it into non-mechanical thermal energy:

\(\Delta E_{mech} = E_{final} - E_{initial} = W_f = -f_k\,s = -\mu_k F_N\,s\)

The mechanical energy lost is quantitatively equal to the thermal energy generated: \(\Delta E_{thermal} = -W_f = +f_k s\). Consequently, the final total mechanical energy is always strictly less than initial: \((TME)_{final} = (TME)_{initial} - \mu_k F_N\,s\)

Checkpoint 6.1: Mechanical Energy Definition What two distinct physical forms comprise total mechanical energy? Express the governing formula.

Checkpoint 6.2: Friction Dissipation A \(4.0\text{ kg}\) block with \(50\text{ J}\) of mechanical energy slides \(3.0\text{ m}\) against \(f_k = 8.0\text{ N}\). Find final \(E_{mech}\).

Checkpoint 6.3: Energy Irreversibility Why cannot the thermal energy produced by friction spontaneously turn back into macroscopic kinetic energy?

Checkpoint 6.4: Sled Stopping Work A sled loses \(600\text{ J}\) of kinetic energy while sliding to rest on snow. How much thermal energy was generated?

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7. Gravitational & Elastic Potential Energy

Potential energy (\(U\)) is the stored energy of configuration possessed by an object due to its position within a conservative force field. Work done against a conservative force increases the system's potential energy: \(\Delta U = -W_{cons}\).

Gravitational Potential Energy (Slide 16)

For an object of mass \(m\) near Earth's surface lifted vertically through height \(h\):

\(U_g(h) = m\,g\,h\)

Height (h) U_g

Linear relation: \(U_g \propto h\). Doubling height strictly doubles stored energy.

Elastic Spring Potential Energy (Slide 17, 19)

For an ideal spring of stiffness constant \(k\) stretched or compressed by distance \(x\):

\(U_s(x) = \frac{1}{2}\,k\,x^2\)

x U_s

Parabolic relation: \(U_s \propto x^2\). Doubling stretch quadruples stored energy.

Energy Exchange in Simple Harmonic Motion (Slide 20)

In an ideal mass-spring system oscillating with amplitude \(A\), mechanical energy continuously trades between potential and kinetic forms:

At \(x = \pm A\) \(E = U_{max} = \frac{1}{2}kA^2\)
\(K = 0\) (Turning points)

At \(x = 0\) \(E = K_{max} = \frac{1}{2}mv_{max}^2\)
\(U = 0\) (Equilibrium)

Between \(0 < |x| < A\) \(E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2\)
\(E = \text{Constant}\)

Checkpoint 7.1: Datum Arbitrariness Does changing the reference height where \(U_g = 0\) alter the calculated physical acceleration or final speed of a falling object?

Checkpoint 7.2: Half-Amplitude Energy In SHM with amplitude \(A\), what percentage of total mechanical energy is kinetic when the position is \(x = A/2\)?

Checkpoint 7.3: Spring Constant Calculation A \(0.50\text{ kg}\) mass hanging vertically stretches a spring by \(0.10\text{ m}\). Calculate the spring constant \(k\) and stored energy.

Checkpoint 7.4: Compression vs Extension Does compressing a spring by distance \(x\) store more, less, or equal potential energy compared to stretching it by \(x\)?

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8. Conservation of Mechanical Energy: Segmented Energy Bar Charts

In conservative systems, total mechanical energy \(E = K + U\) remains constant. Bar charts provide a powerful visual representation of how energy transfers between potential, kinetic, and dissipative friction channels.

Multi-State Energy Bar Analysis on a Curved Track (Slide 24 & 26)

State 1 (Top Left) PE KE = TME State 2 (Bottom) PE KE = TME State 3 (Conservative) PE KE = TME

Key Visual Takeaways: At State 1 (top), energy is \(100\%\) potential. As the cart descends into State 2 (bottom), all potential energy transforms into kinetic energy (\(PE \to 0\), \(KE \to \text{Max}\)). In a conservative system, the cart reaches the exact same height at State 3. The total bar (yellow) remains completely constant.

Accounting for Non-Conservative Friction Work (Slide 26)

When track friction acts along distance \(s\), the total mechanical energy is reduced: \((PE + KE)_{initial} = (PE + KE)_{final} + |W_f| \iff E_{i} = E_{f} + \mu_k F_N\,s\) On the bar chart, the final total mechanical energy bar is shorter than the initial bar by an amount exactly equal to the thermal dissipation bar \(W_f\).

Checkpoint 8.1: Midpoint Bar Height At what height relative to the bottom is the kinetic energy bar exactly equal to the potential energy bar?

Checkpoint 8.2: Friction Runout Height If friction removes \(20\%\) of initial mechanical energy, what fraction of the release height \(h\) does the cart reach on the right?

Checkpoint 8.3: Bar Conservation Does the sum of \(PE + KE\) change when an object accelerates under gravity alone? Explain using the bar diagram.

Checkpoint 8.4: Bottom Speed A \(3.0\text{ kg}\) cart drops from height \(h = 4.9\text{ m}\) without friction. Calculate its speed at the bottom.

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9. Potential Energy Curves: Calculus, Turning Points & Stability

Potential energy curves \(U(x)\) provide a complete, powerful graphical landscape of 1D particle dynamics. By analyzing slopes, local extrema, and curvature, one can determine instantaneous forces, velocities, and stability.

Differential Relation \(F(x) = -\frac{dU}{dx}\)

Integral Relation \(U(x) = -\int_{x_0}^x F(x')\,dx'\)

The Negative Slope Principle: The physical force acting on a particle is the negative slope of the potential energy curve. If \(U(x)\) slopes downward to the right (\(\frac{dU}{dx} < 0\)), the force pushes to the right (\(F > 0\)). If \(U(x)\) slopes upward (\(\frac{dU}{dx} > 0\)), the force pushes to the left (\(F < 0\)).

Total Energy Line \(E\), Kinetic Energy & Turning Points (Slide 29)

When a particle with constant total mechanical energy \(E\) moves in a potential well:

Kinetic Energy \(K(x) = E - U(x)\)
Vertical gap between \(E\) and \(U\)

Turning Points \(E = U(x) \implies K = 0\)
Particle stops and reverses

Forbidden Zone \(E < U(x) \implies K < 0\)
Physically impossible classically

The particle's linear speed at any coordinate \(x\) is given directly by: \(v(x) = \sqrt{\frac{2(E - U(x))}{m}}\). Speed is maximal where the potential energy curve dips to its lowest point (deepest well).

Stable Equilibrium \(\frac{dU}{dx} = 0,\; \frac{d^2U}{dx^2} > 0\) Local Minimum. Small displacement creates restoring force.

Unstable Equilibrium \(\frac{dU}{dx} = 0,\; \frac{d^2U}{dx^2} < 0\) Local Maximum. Small displacement creates repelling force.

Neutral Equilibrium \(\frac{dU}{dx} = 0,\; \frac{d^2U}{dx^2} = 0\) Flat Plateau. Small displacement produces zero net force.

Checkpoint 9.1: Force from Potential Given \(U(x) = (4.0 x^3 - 12.0 x)\text{ J}\), derive the force function \(F(x)\) and find the equilibrium positions.

Checkpoint 9.2: Curvature Classification For the equilibrium coordinates in Checkpoint 9.1, evaluate \(\frac{d^2U}{dx^2}\) to classify each as stable or unstable.

Checkpoint 9.3: Kinetic Energy from Curve A \(0.50\text{ kg}\) particle has \(E = 25\text{ J}\). At coordinate \(x_1\), \(U = 9\text{ J}\). Find its linear speed at \(x_1\).

Checkpoint 9.4: Classical Turning Point Why cannot a classical particle enter a spatial region where the potential energy curve exceeds total energy (\(U(x) > E\))?

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10. Deep-Dive Case Study: Detailed Analysis of Potential Curve 5.8c

Slide 30 (Example 5.8c) presents a multi-well potential energy function \(U(x)\). Here is the full-page, comprehensive analysis evaluating force signs, zero-force points, and stability at every labeled location.

Potential Energy Graph (Slide 30: Example 5.8c) Grid units: PE in Joules, x in meters

0 10 20 30 40 50 60 PE (J) 0 1.0 2.0 3.0 4.0 x (m) Release point (x = 0.25 m) A B C D E F G H I

a) Force is Negative (\(F < 0\)): Occurs where slope is positive (\(dU/dx > 0\)):
• Points D, H, I. The force pushes to the left (\(-\hat{i}\)).

b) Force is Positive (\(F > 0\)): Occurs where slope is negative (\(dU/dx < 0\)):
• Points A, B, F. The force pushes to the right (\(+\hat{i}\)).

c) Force is Zero (\(F = 0\)): Occurs at horizontal tangents (\(dU/dx = 0\)):
• Points C, E, G. These are equilibrium points.

d & e) Stability Classifications: • Stable Points: C and G (Local minima, concave up).
• Unstable Point: E (Local maximum, concave down).

Checkpoint 10.1: Release Motion A particle is released from rest at Point A (\(U = 45\text{ J}\)). In which direction does it accelerate? Explain using slope.

Checkpoint 10.2: Barrier Crossing Can the particle released at Point A (\(E = 45\text{ J}\)) cross the hill at Point E (\(U = 20\text{ J}\))? Explain.

Checkpoint 10.3: Maximum Speed Coordinate At which labeled point on the entire curve will the particle achieve its absolute maximum linear speed?

Checkpoint 10.4: Right Turning Point At what approximate coordinate \(x\) on the right slope will the particle come to a momentary stop?

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11. Power Dynamics & Chapter Synthesis

Power measures the instantaneous rate at which work is performed or energy is transformed across time (Slide 31). Running up stairs requires the exact same change in gravitational potential energy (\(\Delta U_g = mgh\)) as walking, but requires far greater power because the time interval \(\Delta t\) is smaller.

Average Power (Slide 31): \(\bar{P} = \frac{\Delta W}{\Delta t} = \frac{\Delta E}{\Delta t}\)
SI Unit: Watt (\(1\text{ W} = 1\text{ J/s} = 1\text{ kg}\cdot\text{m}^2/\text{s}^3\)); \(1\text{ hp} = 746\text{ W}\).

Instantaneous Power (Slide 32): \(P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v} = F v\cos\theta\)
Dot product of applied force vector and instantaneous velocity vector.

Vehicle Power on an Incline (Slide 32)

F_drive F_R mg

When a car climbs an incline of angle \(\alpha\) at constant velocity \(v\), the engine must overcome both the gravitational component \(mg\sin\alpha\) and rolling resistance \(F_R\):

\(P = (mg\sin\alpha + F_R)\,v\)

Physical Concept	Governing AP Calculus Formula	Key Constraint / Condition
Work (1D Variable)	\(W = \int F(x)\,dx\)	Area under \(F\) vs \(x\) curve
Work-Energy Theorem	\(W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\)	Applies to all forces (conservative & non-cons)
Conservative Force	\(F(x) = -\frac{dU}{dx} \iff \Delta U = -W_{cons}\)	\(\oint \vec{F}\cdot d\vec{r} = 0\); path-independent
Non-Conservative Energy	\(W_{nc} = \Delta E_{mech} = \Delta K + \Delta U\)	Friction, drag; dissipates into thermal heat
Instantaneous Power	\(P = \vec{F} \cdot \vec{v} = F v\cos\theta\)	Scalar rate of mechanical work transfer

Checkpoint 11.1: Elevator Power An elevator of mass \(1200\text{ kg}\) rises vertically at \(2.5\text{ m/s}\) constant speed. Calculate the motor power in kW.

Checkpoint 11.2: Car Hill Climb Power A \(1500\text{ kg}\) car climbs a \(10.0^\circ\) incline at \(22.0\text{ m/s}\) against \(F_R = 350\text{ N}\). Compute required power in hp.

Checkpoint 11.3: Air Drag Scaling If aerodynamic drag force scales as \(F_{drag} \propto v^2\), prove that the power required to overcome drag scales as \(P \propto v^3\).

Checkpoint 11.4: Accelerating Engine A \(1000\text{ kg}\) car accelerates from rest at constant acceleration \(a = 2.5\text{ m/s}^2\). Express output power as a function of time \(P(t)\).

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(a) Calculate the elastic potential energy stored in the compressed spring:

(b) Find the speed \(v\) of the block at the exact moment it leaves the spring:

(c) Using the Work-Energy Theorem, calculate how far \(d\) the block slides along the rough floor before coming to rest:

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AP Physics C: Mechanics • Chapter 6 Assignment Page 3 of 5

Part 3: Energy Conservation & Simple Harmonic Motion Curved Tracks & Oscillator Energy Partitioning

Problem 5: Halfpipe Track & Dissipative Runout (Slide 24 & 26)

[12 Points Total]

A sled and rider of combined mass \(m = 80\text{ kg}\) start from rest at height \(h = 45.0\text{ m}\) on a frictionless curved snow slope. At the bottom, the slope transitions smoothly into a level runout zone covered in unpacked snow with \(\mu_k = 0.25\).

(a) Using conservation of mechanical energy, calculate the sled's speed at the bottom of the slope:

(b) Sketch the segmented energy bars (PE, KE, Total) at the release point and at the bottom:

Draw bar charts for State 1 and State 2

(c) Using \(W_{nc} = \Delta E_{mech}\), determine the stopping distance \(s\) in the unpacked snow zone:

Problem 6: Simple Harmonic Motion Energy Partitioning (Slide 20)

[10 Points Total]

A block of mass \(m = 0.40\text{ kg}\) attached to a horizontal spring oscillates frictionlessly with amplitude \(A = 0.25\text{ m}\). The spring constant is \(k = 160\text{ N/m}\).

(a) Compute the total mechanical energy \(E\) of the oscillator:

(b) Find the maximum linear speed \(v_{max}\) at the equilibrium position (\(x = 0\)):

(c) At position \(x = A / 2\), calculate: (1) the potential energy \(U\), (2) the kinetic energy \(K\), and (3) the instantaneous speed \(v\):

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AP Physics C: Mechanics • Chapter 6 Assignment Page 4 of 5

Part 4: Potential Energy Landscapes & Stability Calculus Derivations & Graph Interpretation

Problem 7: Analytical Force & Stability Derivation [10 Points]

[10 Pts]

A particle moves in a conservative field with potential energy \(U(x) = (x^4 - 8.0 x^2 + 16.0)\text{ J}\), where \(x\) is in meters.

(a) Using \(F(x) = -\frac{dU}{dx}\), find all equilibrium positions:

(b) Evaluate \(\frac{d^2U}{dx^2}\) at each point to classify stability:

Problem 8: Potential Energy Graph Analysis (Slide 30) [12 Points]

[12 Pts]

0 10 20 30 40 50 60 PE (J) 0 1.0 2.0 3.0 4.0 x (m) A B C D E F G H I

(a) Identify points where force is negative (\(F < 0\)) and positive (\(F > 0\)):

(b) Identify stable and unstable equilibrium points:

(c) A \(2.0\text{ kg}\) particle is released from rest at Point A (\(U = 45\text{ J}\)). Calculate its speed as it passes Point G (\(U = 5\text{ J}\)):

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Part 5: Power Dynamics & Conceptual Justifications Instantaneous Rates & AP Deep-Dive

Problem 9: Vehicle Power on an Incline (Slide 32) [8 Points]

[8 Points]

A \(1400\text{ kg}\) automobile climbs a \(10.0^\circ\) incline at a constant speed of \(v = 25.0\text{ m/s}\). A constant rolling friction and air resistance force \(F_R = 450\text{ N}\) opposes motion.

(a) Draw an FBD of the vehicle and calculate the total engine driving force \(F_{drive}\):

(b) Using \(P = \vec{F}\cdot\vec{v}\), calculate the engine's power output in kilowatts (\(\text{kW}\)) and horsepower (\(\text{hp}\)):

Conceptual 1: Work by Internal Forces When a person jumps straight up from the floor, does the normal force from the floor do positive work on the jumper? Explain carefully.

Conceptual 2: Frame Dependence of Kinetic Energy A passenger on a train moving at \(v\) drops a ball. Is the work done on the ball identical in the train frame and ground frame? Justify.

Conceptual 3: Negative Potential Energy Can the potential energy \(U\) of a system be negative? What is the physical significance of a negative total mechanical energy (\(E < 0\))?

Conceptual 4: Aerodynamic Drag Power Scaling High-speed air drag satisfies \(F_{drag} = c\,v^2\). Prove that the engine power required to overcome drag scales as \(P \propto v^3\).

Conceptual 5: Spring Compression Asymmetry Why does compressing a spring by \(2x\) require four times the work of compressing it by \(x\), rather than just double the work?

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(b) Final Speed via Work-Energy [3 Pts]: \(W_{net} = \Delta K = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2\) [1 pt]
\(42.0\text{ J} = \frac{1}{2}(2.0\text{ kg}) v_f^2 - 0\) [1 pt]
\(v_f^2 = 42.0 \implies v_f = \sqrt{42.0} \approx \mathbf{6.48\text{ m/s}}\) [1 pt]

(c) Kinetic Energy at \(x = 2.0\text{ m}\) [3 Pts]: \(K(2) = W(0 \to 2) = \left[ 2.0 x^3 - 2.0 x^2 + 2.0 x \right]_0^2 = 2.0(8) - 2.0(4) + 2.0(2) = 16.0 - 8.0 + 4.0 = \mathbf{12.0\text{ J}}\) [3 pts]

Problem 4 Solution: Spring Launch & Friction Runout [10 Points]

10 Pts

(a) Spring Potential Energy [3 Pts]: \(U_s = \frac{1}{2}k x^2 = \frac{1}{2}(600\text{ N/m})(0.20\text{ m})^2\) [1.5 pts]
\(U_s = 300(0.04) = \mathbf{12.0\text{ J}}\) [1.5 pts]

(b) Release Speed \(v\) [3 Pts]: \(\frac{1}{2}m v^2 = U_s \implies \frac{1}{2}(0.50\text{ kg}) v^2 = 12.0\text{ J}\) [1.5 pts]
\(0.25 v^2 = 12.0 \implies v = \sqrt{48.0} \approx \mathbf{6.93\text{ m/s}}\) [1.5 pts]

(c) Sliding Distance to Rest [4 Pts]: Work-Energy: \(W_{nc} = \Delta E_{mech} \implies -f_k\,d = 0 - U_s = -12.0\text{ J}\) [1.5 pts]
\(f_k = \mu_k mg = (0.30)(0.50\text{ kg})(9.8\text{ m/s}^2) = 1.47\text{ N}\) [1 pt]
\(d = \frac{12.0\text{ J}}{1.47\text{ N}} \approx \mathbf{8.16\text{ m}}\) [1.5 pts]

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Part 3 Solutions: Energy Conservation & SHM Problems 5 & 6 Rubric [22 Points Total]

Problem 5 Solution: Curved Track & Runout [12 Points]

12 Pts

(a) Bottom Speed [4 Pts]: \(mgh = \frac{1}{2}m v^2 \implies v = \sqrt{2gh}\) [2 pts]
\(v = \sqrt{2(9.8\text{ m/s}^2)(45.0\text{ m})} = \sqrt{882}\)
\(v \approx \mathbf{29.7\text{ m/s}}\) [2 pts]

(b) Scored Bar Charts [4 Pts]: • Top: PE = 100%, KE = 0, TME = 100% [2 pts]
• Bottom: PE = 0, KE = 100%, TME = 100% [2 pts]

(c) Snow Runout Stopping Distance [4 Pts]: \(W_{nc} = \Delta E_{mech} \implies -f_k\,s = 0 - mgh \implies \mu_k mg\,s = mgh\) [2 pts]
\(s = \frac{h}{\mu_k} = \frac{45.0\text{ m}}{0.25} = \mathbf{180\text{ m}}\) [2 pts]

Problem 6 Solution: Simple Harmonic Motion Partitioning [10 Points]

10 Pts

(a) Total Mechanical Energy \(E\) [3 Pts]: \(E = \frac{1}{2}k A^2 = \frac{1}{2}(160\text{ N/m})(0.25\text{ m})^2\) [1.5 pts]
\(E = 80(0.0625) = \mathbf{5.00\text{ J}}\) [1.5 pts]

(b) Maximum Speed \(v_{max}\) at \(x=0\) [3 Pts]: \(\frac{1}{2}m v_{max}^2 = E \implies \frac{1}{2}(0.40) v_{max}^2 = 5.0\) [1.5 pts]
\(v_{max} = \sqrt{\frac{5.0}{0.20}} = \sqrt{25.0} = \mathbf{5.00\text{ m/s}}\) [1.5 pts]

(c) Energy & Speed at \(x = A / 2\) [4 Pts]: (1) \(U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}\left(\frac{1}{2}kA^2\right) = \frac{1}{4}(5.0\text{ J}) = \mathbf{1.25\text{ J}}\) [1 pt]
(2) \(K = E - U = 5.00 - 1.25 = \mathbf{3.75\text{ J}}\) (\(75\%\) of total energy!) [1.5 pts]
(3) \(\frac{1}{2}(0.40) v^2 = 3.75 \implies v = \sqrt{18.75} \approx \mathbf{4.33\text{ m/s}}\) [1.5 pts]

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Part 4 Solutions: Potential Energy Curves & Stability Problems 7 & 8 Rubric [22 Points Total]

Problem 7 Solution: Analytical Potential Function [12 Points]

12 Pts

(a) Force & Equilibria [4 Pts]: \(F(x) = -\frac{dU}{dx} = -(4x^3 - 16x) = -4x(x^2 - 4)\)
Equilibria: \(\mathbf{x = 0},\; \mathbf{x = +2.0\text{ m}},\; \mathbf{x = -2.0\text{ m}}\) [4 pts]

(b) Second Derivative Test [4 Pts]: \(\frac{d^2U}{dx^2} = 12x^2 - 16\)
• At \(x = 0\): \(-16 < 0 \implies\) Unstable [2 pts]
• At \(x = \pm 2\): \(12(4) - 16 = 32 > 0 \implies\) Stable [2 pts]

(c) Turning Points for \(E = 25.0\text{ J}\) [4 Pts]: \(U(x) = 25.0 \implies (x^2 - 4)^2 = 25 \implies x^2 - 4 = +5 \implies x^2 = 9 \implies \mathbf{x = \pm 3.0\text{ m}}\) [4 pts]

Problem 8 Solution: Slide 30 Graph Interpretation [10 Points]

10 Pts

A B C D E F G H I

(a) Force Directions [3 Pts]: • \(F > 0\) (pushing right): Points A, B, F (negative slope).
• \(F < 0\) (pushing left): Points D, H, I (positive slope).

(b) Stability Identifications [3 Pts]: • Stable Equilibrium: Minima C and G.
• Unstable Equilibrium: Maximum E.

(c) Speed at Point G [4 Pts]: \(E = U_A = 45\text{ J} \implies K_G = E - U_G = 45 - 5 = 40.0\text{ J} \implies v_G = \sqrt{\frac{2(40.0)}{2.0}} = \sqrt{40.0} \approx \mathbf{6.32\text{ m/s}}\) [4 pts]

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Part 5 Solutions: Power & Conceptual Justifications Problem 9 (8 Pts) & Conceptual Questions (10 Pts) [18 Pts Total]

Problem 9 Solution: Vehicle Power on an Incline [8 Points]

8 Pts

(a) Engine Driving Force \(F_{drive}\) [4 Pts]: \(F_{drive} = mg\sin\alpha + F_R\) [2 pts]
\(F_{drive} = (1400\text{ kg})(9.8)\sin(10.0^\circ) + 450\text{ N}\)
\(F_{drive} = 13720(0.1736) + 450 \approx \mathbf{2832\text{ N}}\) [2 pts]

(b) Engine Power Output [4 Pts]: \(P = F_{drive}\,v = (2832.4\text{ N})(25.0\text{ m/s})\)
\(P = 70,810\text{ W} \approx \mathbf{70.8\text{ kW}}\) [2 pts]
\(P = \frac{70810}{746} \approx \mathbf{94.9\text{ hp}}\) [2 pts]

Conceptual 1: Normal Force on a Jumper [2 Pts]: The floor normal force does zero work (\(W_N = 0\)) because the contact point between feet and ground does not displace while the force is applied (\(d\vec{r}_{contact} = 0\)). The jumper's center of mass accelerates upward due to work done by internal leg muscles converting chemical energy into kinetic energy.

Conceptual 2: Frame Dependence of Work & Kinetic Energy [2 Pts]: Work is frame-dependent. In the train frame, the ball moves purely vertically (\(W = mgh\)). In the ground frame, the ball has a horizontal displacement component \(v t\), so observers record different displacements and different calculated work values; however, \(\Delta K = W_{net}\) holds true within each respective frame.

Conceptual 3: Physical Meaning of Negative Potential Energy [2 Pts]: Negative potential energy simply reflects the choice of datum (\(U = 0\) at infinity for gravitation). A negative total mechanical energy (\(E < 0\)) signifies a bound state: the particle cannot escape the potential well to infinity without receiving external energy input.

Conceptual 4: Aerodynamic Drag Power Scaling [2 Pts]: Instantaneous power is \(P = \vec{F}_{drag}\cdot\vec{v} = (c\,v^2)v = c\,v^3\). Therefore, engine power required to overcome drag scales as the cube of velocity (\(P \propto v^3\)). Doubling speed demands an eightfold increase in engine power!

Conceptual 5: Spring Compression Work Asymmetry [2 Pts]: Because the spring force increases linearly with distance (\(F = kx\)), compressing from \(x\) to \(2x\) requires pushing against a much larger average force (\(\frac{3}{2}kx\)) than compressing from \(0\) to \(x\) (\(\frac{1}{2}kx\)). Integrating yields \(W \propto x^2\), producing a fourfold increase in work.

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(E)

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5. Sign of Kinetic Friction Work [1 Pt]

When an object slides across a stationary rough floor, the work done by kinetic friction on the sliding object is always:

v → ← f_k

(A) Positive (along plane).

(B) Zero (interlocking cancel).

(C) Negative, because friction is antiparallel (\(\theta = 180^\circ\)) to displacement.

(D) Depends on applied force.

(E) Stored as potential energy.

6. Normal Force Modulation Under Upward Pull [1 Pt]

A box of mass \(m\) is pulled along a rough floor (\(\mu_k\)) by force \(F\) angled at \(\theta\) above horizontal. The work done by friction over displacement \(s\) is:

F_N mg F

(A) \(W_f = -\mu_k m g\,s\)

(B) \(W_f = -\mu_k(mg - F\sin\theta)\,s\)

(C) \(W_f = -\mu_k(mg + F\sin\theta)\,s\)

(D) \(W_f = -\mu_k(mg - F\cos\theta)\,s\)

(E) \(W_f = -(F\cos\theta)\,s\)

7. Downward Pushing Force vs Upward Pull [1 Pt]

A worker pushes a crate with force \(F\) angled at \(\theta\) below horizontal. How does friction work compare to pulling at \(\theta\) above horizontal?

F (push) f_k

(A) Pushing produces less friction work.

(B) Both produce identical friction work.

(C) Pushing produces greater frictional dissipation because downward force increases \(F_N\).

(D) Pushing produces zero friction work.

(E) Pushing reverses the sign of friction work.

8. Thermal Dissipation Accounting [1 Pt]

A block of mass \(m\) slides across a rough surface through distance \(s\) and stops. The mechanical energy lost by the block is equal to:

Heat Q

(A) \(\mu_k m g s\) (thermal energy generated)

(B) \(m g s\)

(C) \(\frac{1}{2}\mu_k m g s\)

(D) Zero, mechanical energy is conserved

(E) \(\mu_k^2 m g s\)

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9. Definite Integral for 1D Variable Force [1 Pt]

A net force \(F(x) = c\,x^3\) acts on a particle as it moves from \(x = 0\) to \(x = d\). The work done by this force (shaded area) is:

d F

(A) \(W = c\,d^3\)

(B) \(W = 3 c\,d^2\)

(C) \(W = \frac{1}{4}c\,d^4\)

(D) \(W = \frac{1}{3}c\,d^3\)

(E) \(W = c\,d^4\)

10. Work Done on an Ideal Spring [1 Pt]

An ideal spring (\(F_s = -kx\)) is stretched from \(x = 0\) to \(x = x_1\). The work done by the external stretching agent (triangular area) is:

x_1 ½kx²

(A) \(W = k x_1\)

(B) \(W = -k x_1\)

(C) \(W = \frac{1}{2}k x_1^2\)

(D) \(W = -\frac{1}{2}k x_1^2\)

(E) \(W = \frac{1}{3}k x_1^3\)

11. Work-Energy Theorem Fundamental Relation [1 Pt]

The Work-Energy Theorem states that the net work \(W_{net}\) done on a particle equals:

(A) The change in momentum: \(W_{net} = \Delta p\).

(B) The change in potential energy: \(W_{net} = \Delta U\).

(C) The change in kinetic energy: \(W_{net} = \Delta K = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2\).

(D) The instantaneous rate of power transfer: \(W_{net} = P\).

(E) Zero, whenever forces are non-conservative.

12. Quadratic Speed Stopping Distance [1 Pt]

A car traveling at speed \(v\) skids to rest in distance \(d\). If initial speed is tripled to \(3v\), its new stopping distance will be:

d ∝ v²

(A) \(d\)

(B) \(3d\)

(C) \(6d\)

(D) \(9d\)

(E) \(27d\)

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13. Conservative Force Condition [1 Pt]

A force \(\vec{F}\) is classified as conservative if and only if:

∮F·dr = 0

(A) It causes zero acceleration on any moving particle.

(B) The work done over any closed round-trip path is zero: \(\oint \vec{F}\cdot d\vec{r} = 0\).

(C) It always acts in the direction of the velocity vector.

(D) It dissipates mechanical energy into thermal energy.

(E) Its magnitude increases linearly with velocity.

14. Force from Potential Energy Derivative [1 Pt]

In one dimension, conservative force \(F(x)\) is related to potential energy \(U(x)\) by:

F = -dU/dx

(A) \(F(x) = \frac{dU}{dx}\)

(B) \(F(x) = -\frac{dU}{dx}\)

(C) \(F(x) = \int U(x)\,dx\)

(D) \(F(x) = -\int U(x)\,dx\)

(E) \(F(x) = \frac{d^2U}{dx^2}\)

15. Simple Harmonic Motion Energy Partitioning [1 Pt]

A mass-spring oscillator oscillates with amplitude \(A\). What percentage of total energy is kinetic at \(x = A / 2\)?

K A/2

(A) \(25\%\)

(B) \(50\%\)

(C) \(70.7\%\)

(D) \(75\%\)

(E) \(86.6\%\)

16. Non-Conservative Energy Dissipation [1 Pt]

A block slides down a rough incline where kinetic friction does \(-40\text{ J}\) of work. If the block lost \(100\text{ J}\) of gravitational potential energy, what is the change in its kinetic energy?

(A) \(+40\text{ J}\)

(B) \(+60\text{ J}\)

(C) \(+100\text{ J}\)

(D) \(+140\text{ J}\)

(E) \(-60\text{ J}\)

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17. Stability Classification Criteria [1 Pt]

At an equilibrium coordinate \(x_0\), \(\frac{dU}{dx} = 0\). Which condition proves that \(x_0\) is a point of stable equilibrium?

d²U/dx² > 0

(A) \(\frac{d^2U}{dx^2} < 0\) (local max).

(B) \(\frac{d^2U}{dx^2} > 0\) (local minimum of potential energy).

(C) \(\frac{d^2U}{dx^2} = 0\) (inflection).

(D) \(U(x_0) = 0\).

(E) Total energy \(E < 0\).

18. Potential Energy Curve Turning Points [1 Pt]

A particle with total energy \(E\) moves in a potential well \(U(x)\). What occurs physically at a turning point where \(U(x) = E\)?

E K = 0

(A) Kinetic energy is maximum and acceleration is zero.

(B) Potential energy is zero and speed is maximum.

(C) Kinetic energy drops to zero (\(K = 0\)); particle stops & reverses.

(D) Particle tunnels into classically forbidden zone.

(E) Net force drops to zero.

19. Instantaneous Power Dot Product [1 Pt]

An applied force \(\vec{F}\) acts on an object moving at instantaneous velocity \(\vec{v}\). The instantaneous power delivered by the force is:

v F θ

(A) \(P = \vec{F} \times \vec{v}\)

(B) \(P = \vec{F} \cdot \vec{v} = F v\cos\theta\)

(C) \(P = \frac{1}{2}F v^2\)

(D) \(P = \int \vec{F}\cdot d\vec{v}\)

(E) \(P = \frac{F}{v}\)

20. Incline Vehicle Power Requirement [1 Pt]

A car of mass \(m\) climbs a road inclined at angle \(\alpha\) at constant linear speed \(v\). Total resistive forces equal \(F_R\). The power delivered by the engine must be:

v

(A) \(P = m g v\cos\alpha\)

(B) \(P = F_R\,v\)

(C) \(P = (m g\sin\alpha + F_R)\,v\)

(D) \(P = (m g\tan\alpha + F_R)\,v\)

(E) \(P = (m g + F_R)\,v\)

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Page 2 Answers: 5. (C) • 6. (B) • 7. (C) • 8. (A) Detailed Rationales & AP Exam Pitfalls

5. Sign of Kinetic Friction Work Correct: (C)

Physical Rationale: Kinetic friction \(\vec{f}_k\) by definition opposes relative interfacial motion. For a body sliding on a stationary surface, \(\vec{f}_k\) is directed opposite to displacement (\(\theta = 180.0^\circ\)). Since \(\cos(180^\circ) = -1\), the work is strictly negative: \(W_f = f_k s\cos(180^\circ) = -f_k s\). Negative work indicates mechanical energy is being removed from the object and transformed into internal thermal energy.

• Teacher Note: Clarify that static friction can do positive work (e.g., accelerating a box in the bed of a truck), but kinetic friction on a fixed surface is always negative.

6. Normal Force Modulation Under Upward Pull Correct: (B)

Physical Rationale: Vertical equilibrium requires \(\Sigma F_y = F_N + F\sin\theta - mg = 0 \implies F_N = mg - F\sin\theta\). The kinetic friction force is \(f_k = \mu_k F_N = \mu_k(mg - F\sin\theta)\). Because friction acts at \(180^\circ\) to displacement, the work is \(W_f = -f_k s = -\mu_k(mg - F\sin\theta)s\).

• Teacher Note: Students frequently forget that an angled pull reduces the normal force, erroneously using \(F_N = mg\).

7. Downward Pushing Force vs Upward Pull Correct: (C)

Physical Rationale: When pushing downward at angle \(\theta\), the vertical balance is \(F_N - F\sin\theta - mg = 0 \implies F_N = mg + F\sin\theta\). Because the normal force is increased, the friction force \(f_k = \mu_k(mg + F\sin\theta)\) is larger than in the pulling case (\(F_N = mg - F\sin\theta\)). Consequently, pushing dissipates significantly more mechanical energy over the same distance \(s\).

• Teacher Note: This is an everyday ergonomic principle: pulling a lawnmower or sled is physically easier than pushing it downward.

8. Thermal Dissipation Accounting Correct: (A)

Physical Rationale: By the extended work-energy relation, \(\Delta E_{mech} = W_{nc} = -f_k s = -\mu_k mg s\). The loss in mechanical energy equals the thermal energy generated: \(\Delta E_{thermal} = -\Delta E_{mech} = \mu_k mg s\).

• Teacher Note: Total energy is always conserved; mechanical energy alone decreases by \(\mu_k mgs\).

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Page 3 Answers: 9. (C) • 10. (C) • 11. (C) • 12. (D) Detailed Rationales & AP Exam Pitfalls

9. Definite Integral for 1D Variable Force Correct: (C)

Physical Rationale: Work is the spatial integral of force: \(W = \int_0^d F(x)\,dx = \int_0^d c\,x^3\,dx = c\left[\frac{x^4}{4}\right]_0^d = \frac{1}{4}c\,d^4\).

• Teacher Note: Emphasize that evaluating work by multiplying \(F(d)\cdot d\) yields \(c d^4\), which overestimates the work by a factor of 4! Integration is required.

10. Work Done on an Ideal Spring Correct: (C)

Physical Rationale: To stretch a spring, the external agent must apply \(F_{ext} = +kx\). The work done by this external agent is \(W_{ext} = \int_0^{x_1} kx\,dx = \frac{1}{2}k x_1^2\).

• Teacher Note: Be careful with signs: work done on the spring by the agent is \(+\frac{1}{2}kx_1^2\); work done by the spring on the agent is \(-\frac{1}{2}kx_1^2\).

11. Work-Energy Theorem Fundamental Relation Correct: (C)

Physical Rationale: Integrating Newton's second law along displacement yields \(W_{net} = \int F_{net} dx = \int m v dv = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K\). The theorem applies universally to all forces, whether conservative or non-conservative.

• Teacher Note: Make sure students know \(W_{net}\) is the sum of ALL works: \(W_{net} = W_{cons} + W_{nc}\).

12. Quadratic Speed Stopping Distance Correct: (D)

Physical Rationale: By the Work-Energy Theorem, \(-f_k d = 0 - \frac{1}{2}mv^2 \implies d = \frac{v^2}{2\mu_k g}\). Since stopping distance is proportional to the square of initial speed (\(d \propto v^2\)), tripling speed (\(3v\)) produces a \(3^2 = \mathbf{9}\)-fold increase in distance: \(d' = 9d\).

• Teacher Note: This quadratic dependence is a standard AP C conceptual favorite.

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Page 4 Answers: 13. (B) • 14. (B) • 15. (D) • 16. (B) Detailed Rationales & AP Exam Pitfalls

13. Conservative Force Condition Correct: (B)

Physical Rationale: A force is conservative if and only if the work done along any closed path returning to the initial position is identically zero: \(\oint \vec{F}_{cons}\cdot d\vec{r} = 0\). Equivalently, the work between any two points is completely path-independent.

• Teacher Note: This condition is the prerequisite for defining a scalar potential energy function \(U(\vec{r})\).

14. Force from Potential Energy Derivative Correct: (B)

Physical Rationale: Since \(\Delta U = -W_{cons} = -\int F(x)dx\), applying the Fundamental Theorem of Calculus yields \(F(x) = -\frac{dU}{dx}\). Force is the negative spatial derivative (negative slope) of potential energy.

• Teacher Note: Be vigilant about the negative sign: a positive slope produces a force in the \(-x\) direction.

15. Simple Harmonic Motion Energy Partitioning Correct: (D)

Physical Rationale: Total energy is \(E = \frac{1}{2}kA^2\). At \(x = A/2\), the potential energy is \(U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}\left(\frac{1}{2}kA^2\right) = \frac{1}{4}E = 25\%\). Therefore, kinetic energy is \(K = E - U = \frac{3}{4}E = \mathbf{75\%}\).

• Teacher Note: Many students guess \(50\%\) assuming energy divides linearly with displacement; emphasize \(U \propto x^2\).

16. Non-Conservative Energy Dissipation Correct: (B)

Physical Rationale: \(\Delta E_{mech} = \Delta K + \Delta U = W_{nc}\). Here \(\Delta U = -100\text{ J}\) and \(W_{nc} = -40\text{ J}\). Thus, \(\Delta K + (-100) = -40 \implies \Delta K = -40 + 100 = \mathbf{+60\text{ J}}\).

• Teacher Note: \(100\text{ J}\) of PE was converted: \(60\text{ J}\) into kinetic energy, and \(40\text{ J}\) into thermal heat.

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Page 5 Answers: 17. (B) • 18. (C) • 19. (B) • 20. (C) Detailed Rationales & AP Exam Pitfalls

17. Stability Classification Criteria Correct: (B)

Physical Rationale: At equilibrium, \(\frac{dU}{dx} = 0\). For stable equilibrium, the potential energy curve is concave up (\(\frac{d^2U}{dx^2} > 0\)), corresponding to a local minimum. If displaced to the right (\(+x\)), slope becomes positive \(\implies F = -dU/dx < 0\), creating a restoring force back toward the center.

• Teacher Note: \(\frac{d^2U}{dx^2} < 0\) is unstable (peak); \(\frac{d^2U}{dx^2} > 0\) is stable (bowl).

18. Potential Energy Curve Turning Points Correct: (C)

Physical Rationale: Since \(E = K + U(x)\), when \(U(x) = E\), kinetic energy drops to zero: \(K = E - U = 0\). Because \(K = \frac{1}{2}mv^2 = 0\), the particle's velocity is momentarily zero. The particle cannot penetrate into regions where \(U(x) > E\) because that would require \(K < 0\) (impossible in classical physics). The particle stops and reverses direction.

• Teacher Note: Turning points are the classical boundaries of motion for a given energy level \(E\).

19. Instantaneous Power Dot Product Correct: (B)

Physical Rationale: Differentiating work with respect to time: \(P = \frac{dW}{dt} = \frac{d}{dt}(\vec{F}\cdot\vec{r}) = \vec{F}\cdot\frac{d\vec{r}}{dt} = \vec{F}\cdot\vec{v} = F v\cos\theta\). Power is the scalar dot product of force and instantaneous velocity.

• Teacher Note: Remind students that power is a scalar (measured in Watts) and can be negative if force opposes velocity.

20. Incline Vehicle Power Requirement Correct: (C)

Physical Rationale: At constant speed along the incline (\(a = 0\)), the driving force must balance the downward component of gravity and total resistance: \(\Sigma F = F_{drive} - mg\sin\alpha - F_R = 0 \implies F_{drive} = mg\sin\alpha + F_R\). Power is \(P = F_{drive}\,v = (mg\sin\alpha + F_R)v\).

• Teacher Note: This directly matches Slide 32 of the curriculum slides.

AP Physics C: Mechanics • Chapter 6 Solution Key Page 5 of 5

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