Types of Forces

Example: Your shoes gripping the floor so you don't slip backwards when walking!

KINETIC FRICTION (\(f_k\)) Grinding Past

When Things Are Sliding

• • Once moving, the microscopic peaks snap, grind, and skim over each other.
• • Creates heat and sound while resisting the slide.
• • Universal Law: \(\mu_s > \mu_k\). It is ALWAYS harder to start sliding than to keep sliding!

Example: A heavy sofa breaks free with a huge heave, then glides much easier!

The Roughness Factor (\(\mu\)): Ice has tiny \(\mu \approx 0.05\); rubber on asphalt has huge \(\mu \approx 0.90\)! TEKS 5F

Block 5 Explain • Problem 1

Determining Springiness (GUESS Method)

Hooke's Law

Problem Statement

A soup can of mass \(m = 0.40\text{ kg}\) hangs from a ceiling spring at equilibrium, stretching it by \(\Delta x = 0.08\text{ m}\).

Target Question:

What is the spring stiffness constant (\(k\)) in N/m?

Mass: \(0.40\text{ kg}\)

Stretch: \(0.08\text{ m}\)

Find: \(k = ?\)

G: Given: \(m = 0.40\text{ kg}\), \(\Delta x = 0.08\text{ m}\), \(g = 9.8\text{ m/s}^2\)

U: Unknown: Spring stiffness constant \(k = ?\text{ [N/m]}\)

E: Equation: \(F_{spring} = F_g \implies k\Delta x = mg\)

S: Substitute: \(k(0.08) = (0.40)(9.8) = 3.92\text{ N}\)

S: Solve: \(k = \frac{3.92}{0.08} = \mathbf{49\text{ N/m}}\)

Meaning: It requires 49 Newtons of force to stretch this spring by 1 full meter!

GUESS Step 1 Rule: Always verify that given numbers have matching metric units! Hooke's Law: \(F_s = k\Delta x\)

Block 5 Explain • Problems 2 & 3

Sleds & Wall Bricks (GUESS Breakdown)

GUESS Method

PULLING A SLED \(\theta = 30^\circ\)

G: \(m=20\text{kg}, F_P=50\text{N}, \theta=30^\circ, \mu_k=0.15\)

U: Acceleration \(a = ?\text{ [m/s}^2\text{]}\)

E: \(F_N = mg - F_P\sin\theta\); \(a = \frac{F_P\cos\theta - \mu_k F_N}{m}\)

S: \(F_N = 196 - 25 = 171\text{N}\); \(a = \frac{43.30 - 25.65}{20}\)

S: \(a = \mathbf{0.88\text{ m/s}^2}\)

GUESS Takeaway: Angled pull lifts the sled, dropping \(F_N\) and reducing friction!

STICKING TO WALL Static Equilibrium

G: \(m = 1.5\text{ kg}\), \(\mu_s = 0.40\), \(g = 9.8\text{ m/s}^2\)

U: Minimum push force \(F_{push} = ?\text{ [N]}\)

E: \(f_s = mg \implies \mu_s F_{push} \ge mg \implies F_{push} \ge \frac{mg}{\mu_s}\)

S: \(F_{push} \ge \frac{(1.5)(9.8)}{0.40} = \frac{14.7}{0.40}\)

S: \(F_{push} \ge \mathbf{36.75\text{ N}}\)

GUESS Takeaway: Horizontal push produces vertical friction to support weight!

GUESS guarantees you never miss a step in multi-force vector problems! TEKS 5D

Block 5 Modeling • TE p. 42

The Pinball Launcher: Stored Energy to Speed

Spring Gun Model

How the Spring Launches the Ball

\Delta x (How far you pull back)

More Pull (\(\Delta x\)) = Way More Speed (\(v\))!

Common Student Trap Busted!

Does the spring push with the same force the whole way out? NO! When fully squished, it kicks with maximum force. As it uncoils, the push dies down until it drops to zero right as the ball flies away!

Energy Trade: Squeeze to Speed

You do mechanical work to squish the spring coils. That stored potential energy instantly converts into the kinetic energy of the speeding steel ball!

Assess on the Spot Look-For: Make sure students draw shrinking arrows as the spring stretches back out! TEKS 5F / 5D

Block 6 Wrap-Up • The Grand Connection

The Big Reveal: Everything Is Made of Springs!

Unit Mastery

1. The Secret of Matter

All matter (wood, rubber, steel, bones) is held together by electric atomic bonds that act exactly like microscopic coils.

When you push a wall, you compress those atomic springs. That is all "Normal Force" has ever been!

The Tennis Ball Solved!

2. Friction & Buoyancy

Friction: Microscopic mountain peaks interlocking like Velcro.

Buoyancy: Water shoved aside by an intruder pushing back with the exact weight of the puddle displaced.

Macroscopic Contact Forces

3. Newton's 2nd Law

\[ a = \frac{F_{net}}{m} \]

Add up all the pushes and pulls using GUESS. The leftover net force determines whether an object speeds up, slows down, or stays frozen!

Ready for the Unit Quiz!

You now understand how the physical world pushes back! TEKS 5F & 5D Mastered

TEKS 5D

A book of mass \(m = 1.50\text{ kg}\) is held against a wall. The static coefficient is \(\mu_s = 0.40\). Calculate the minimum horizontal push force (\(F_{push}\)) needed to stop it slipping downward.

Given (G)

Unknown (U)

Equation (E)

Substitute (S)

Solve (S)

Physics Dynamics • Types of Forces Worksheet

Page 2 of 3 • GUESS Calculations (12pt Edition)

Block 6 Evaluation & Model Synthesis

Student Name:

Part 3

Conceptual Mastery Checkpoints

7. A heavy wooden crate sits motionless on a rough floor. What describes the contact forces?

A. The floor cannot push upward because inanimate objects are passive.

B. The normal force balances gravity, while static friction is zero until pushed.

C. Static friction pushes upward on the crate to counteract Earth's gravity.

D. Kinetic friction prevents the crate from accelerating into the ground.

8. A spring compressed by distance \(\Delta x\) is released. What describes the force as it uncoils?

A. The force remains perfectly constant until the ball leaves the coil.

B. The force increases steadily as the spring expands toward equilibrium.

C. The force decreases linearly down to zero at the equilibrium position.

D. The force abruptly switches direction and pulls the ball backward.

Part 4

Math Skill: Absolute Value & Engineering Tolerance (TE p. 44)

TEKS 5D

In physics and engineering, force tolerance measures the maximum allowed difference from a target force regardless of direction: \(|F_{\text{actual}} - F_{\text{target}}| \le \text{Tolerance}\).

A pinball launcher spring must exert a target force of \(50.0\text{ N}\) with a tolerance of \(\pm 4.0\text{ N}\). Write the absolute value inequality and find the acceptable range of forces.

Part 5

Model Synthesis: Pinball Launcher Acceleration (TE p. 42 & 45)

Newtonian Synthesis

When the pinball plunger is pulled back \(3.0\text{ cm}\) and released, does the steel pinball experience constant acceleration or changing acceleration during launch? Use Newton's Second Law (\(a = F_{net}/m\)) and Hooke's Law (\(F_s = k\Delta x\)) to explain your answer.

Physics Dynamics • Types of Forces Worksheet

Page 3 of 3 • Checkpoints & Synthesis (12pt Edition)

Unknown (U)

Spring constant
\(k = \text{? [N/m]}\)

Equation (E)

\(k\Delta x = mg\)
\(k = \frac{mg}{\Delta x}\)

Substitute (S)

\(k = \frac{(0.40)(9.8)}{0.080}\)
\(k = \frac{3.92}{0.080}\)

Solve (S)

\(k = 49\text{ N/m}\)
(Full credit: units required)

Problem 5 Solution Archimedes: Apparent Weight (6 pts)

Answer: \(0.28\text{ N}\)

Weight in air \(F_{g,\text{air}} = 1.00\text{ N}\); apparent submerged weight \(F_{\text{scale}} = 0.72\text{ N}\).

Given (G)

\(F_{g,\text{air}} = 1.00\text{ N}\)
\(F_{\text{sub}} = 0.72\text{ N}\)

Unknown (U)

Buoyant force
\(F_B = \text{? [N]}\)

Equation (E)

\(F_{\text{sub}} = F_{g} - F_B\)
\(F_B = F_{g} - F_{\text{sub}}\)

Substitute (S)

\(F_B = 1.00\text{ N} - 0.72\text{ N}\)

Solve (S)

\(F_B = 0.28\text{ N}\)
Weight disp water = 0.28 N

Problem 6 Solution Wall Static Friction Equilibrium (8 pts)

Answer: \(36.8\text{ N}\)

Vertical: \(f_s = mg\); Horizontal: \(F_N = F_{push}\); Slipping limit: \(f_{s,\max} = \mu_s F_N\).

Given (G)

\(m = 1.50\text{ kg}\)
\(\mu_s = 0.40\)
\(g = 9.8\text{ m/s}^2\)

Unknown (U)

Push force
\(F_{push} = \text{? [N]}\)

Equation (E)

\(mg \le \mu_s F_{push}\)
\(F_{push} = \frac{mg}{\mu_s}\)

Substitute (S)

\(F_{push} = \frac{(1.50)(9.8)}{0.40}\)
\(F_{push} = \frac{14.7}{0.40}\)

Solve (S)

\(F_{push} = 36.75\text{ N}\)
(\(37\text{ N}\) with 2 sig figs)

Physics Dynamics • Force Fundamentals Answer Key

Page 2 of 3 • GUESS Solutions (12pt Edition)

Teacher Edition Block 6 • Synthesis & Evaluation Solutions (15 pts)

Scoring & Misconceptions

Part 3 Checkpoints

Multiple Choice Solutions (4 pts)

Question 7: Correct Answer B

"The normal force balances gravity, while static friction is zero until pushed." Passive objects exert normal forces by atomic compression. Static friction only appears when a tangential applied force is present.

Question 8: Correct Answer C

"The force decreases linearly down to zero at the equilibrium position." By Hooke's Law (\(F_s = k\Delta x\)), as the spring returns to its natural uncompressed length (\(\Delta x \to 0\)), the push force drops to zero.

Part 4 Solution

Absolute Value Force Tolerance (5 pts)

Range: [46.0 N, 54.0 N]

1. Inequality: \(|F_{\text{actual}} - 50.0\text{ N}| \le 4.0\text{ N}\)

2. Expanding double inequality: \(-4.0\text{ N} \le F_{\text{actual}} - 50.0\text{ N} \le +4.0\text{ N}\)

3. Adding \(50.0\text{ N}\) to all sides: \(46.0\text{ N} \le F_{\text{actual}} \le 54.0\text{ N}\)

Full credit: 2 pts for correct absolute value setup, 3 pts for interval bounds.

Part 5 Solution

Pinball Launcher Non-Constant Acceleration (6 pts)

Changing Acceleration

Exemplar Answer: The pinball experiences changing (non-constant) acceleration. By Hooke's Law, the spring force is \(F_s = k\Delta x\). As the plunger pushes the ball forward, the compression distance \(\Delta x\) continuously decreases toward zero. According to Newton's Second Law:

\(a(t) = \frac{F_{net}}{m} = \frac{k\Delta x(t)}{m}\)

Because the restoring force drops continuously as the spring expands, the acceleration of the pinball starts at a maximum and steadily decreases to zero when the ball disconnects from the plunger.

Physics Dynamics • Force Fundamentals Answer Key

Page 3 of 3 • Evaluation & Synthesis (12pt Edition)

Station 1 Question: Why does the foam pad make the bounce quieter and lower than the hard table?

Station 2

Hooke's Law: Spring Stretch (\(F_s = k\Delta x\))

\(F_g = mg\)

Hang increasing masses on the spring. Measure initial position \(x_0\) and stretched position \(x\) to find stretch \(\Delta x = x - x_0\).

Station 2 Question: When you doubled the hanging weight from \(0.100\text{ kg}\) to \(0.200\text{ kg}\), did the stretch also double? What does this tell you about the spring?

Physics Dynamics • Contact Forces Lab Guide

Page 2 of 4 • Stations 1 & 2 (12pt Edition)

Page 3 Station 3 & Station 4 Data Collection

Student Name:

Station 3

Surface Friction: Starting vs. Sliding

Spring scale pull

Pull the wooden block across different surfaces. Record the peak force needed to start moving (Static Friction \(f_s\)) and the steady force to keep it sliding (Kinetic Friction \(f_k\)).

Station 3 Question: Why does it take more force to break the object free from a standstill than to keep it sliding?

Station 4

Buoyancy: The Water Push (Archimedes)

\(F_B = F_{\text{air}} - F_{\text{sub}}\)

Weigh the metal block in air with a spring scale. Submerge it into an overflow cup of water, measure the new scale reading, and weigh the overflow water that spilled out.

Station 4 Question: How does the upward push of the water (\(F_B\)) compare to the weight of the water that spilled out?

Physics Dynamics • Contact Forces Lab Guide

Page 3 of 4 • Stations 3 & 4 (12pt Edition)

Page 4 Real-World Engineering Synthesis

Student Name:

Synthesis

Connecting Lab Results to Real-World Designs

1. Car Crash Crumple Zones: Old vintage cars had rock-hard steel bumpers, while modern cars have front ends designed to crumple and accordion during a crash. Based on what you observed at Station 1 with hard vs. soft surfaces, why are crumple zones much safer for the passengers inside?

2. Athletic Running Shoes: Basketball and running shoes have thick, squishy foam soles. Using the concept of normal force and contact time, explain how squishy shoe soles protect a runner's knees and bones when landing from a jump.

3. Pinball Launch Springs: In a pinball machine, you pull back a spring plunger to launch a heavy steel ball. Using Hooke's Law from Station 2, explain what happens to the launch force and the ball's acceleration if you pull the plunger back \(4\text{ cm}\) instead of \(2\text{ cm}\).

Big Physics Takeaway

All contact forces (normal force, friction, springs, and collisions) arise from atoms pushing, pulling, or sliding against each other. When an interaction allows for more squish or deformation, the contact time increases and the peak impact force drops!

Physics Dynamics • Contact Forces Lab Guide

Page 4 of 4 • Synthesis (12pt Edition)

Station 1 Exemplar Answer:

The foam pad deforms macroscopically and absorbs a significant portion of the ball's kinetic energy, converting it into internal friction and thermal energy within the open-cell foam. Because less elastic potential energy is stored and returned to the ball, the rebound height is much lower and the sound is muffled.

Station 2 Data Hooke's Law: Spring Stretch (8 pts)

\(k \approx 24.5\text{ N/m}\)

Station 2 Exemplar Answer:

Yes, doubling the hanging weight from \(0.98\text{ N}\) to \(1.96\text{ N}\) doubles the stretch from \(0.040\text{ m}\) to \(0.080\text{ m}\). This confirms Hooke's Law: restoring force is directly proportional to displacement (\(F \propto \Delta x\)) with a constant spring stiffness \(k\).

Physics Dynamics • Contact Forces Lab Answer Key

Page 2 of 4 • Stations 1 & 2 Solutions (12pt Edition)

Teacher Edition Stations 3 & 4 Sample Data & Solutions

Expected Values

Station 3 Data Friction: Starting vs. Sliding (8 pts)

\(f_s > f_k\) Always

Station 3 Exemplar Answer:

When stationary, surface microscopic ridges interlock deeply and form temporary electrostatic bonds. Breaking these interlocking ridges requires a large initial force (\(f_s\)). Once in motion, the surfaces skate across the tops of the ridges before settling, creating less resistance (\(f_k\)).

Station 4 Data Buoyancy & Displaced Water (8 pts)

Archimedes Principle

Station 4 Exemplar Answer:

The upward buoyant force (\(F_B = 1.00\text{ N} - 0.72\text{ N} = 0.28\text{ N}\)) exactly equals the weight of the spilled water (\(0.28\text{ N}\)). This confirms Archimedes' principle: any submerged body experiences an upward buoyant push equal to the weight of the fluid it displaces.

Physics Dynamics • Contact Forces Lab Answer Key

Page 3 of 4 • Stations 3 & 4 Solutions (12pt Edition)

Teacher Edition Synthesis Solutions & 50-Point Rubric

Final Evaluation

1. Car Crash Crumple Zones (6 pts)

Exemplar Answer: Old rigid steel cars acted like the hard tabletop: stopping immediately in milliseconds, resulting in massive, lethal peak impact forces transferred to passengers. Modern crumple zones act like the soft foam pad: they accordion and crush over several feet. This extends the collision duration (\(\Delta t\)), spreading out the change in momentum and dramatically reducing the peak impact force on the cabin.

2. Running Shoes & Knee Protection (6 pts)

Exemplar Answer: The foam midsole compresses macroscopically during foot strike. This extends the deceleration contact time from \(\approx 0.01\text{ s}\) to \(\approx 0.03\text{ s}\). By \(F_{avg} = \Delta p / \Delta t\), tripling the stopping time cuts the peak normal impact force transmitted up the tibia and knee joints by two-thirds.

3. Pinball Plunger Hooke's Law Launch (6 pts)

Exemplar Answer: By Hooke's Law (\(F_s = k\Delta x\)), pulling the plunger \(4\text{ cm}\) instead of \(2\text{ cm}\) doubles the compression distance, thereby doubling the launch restoring force. By Newton's Second Law (\(a = F/m\)), doubling the net force doubles the initial acceleration of the steel ball.

Grading Rubric Summary (50 Points Total)

Pre-Lab CFUs

10 pts

2 questions @ 5 pts

Station 1 Data

8 pts

Drops & sound notes

Station 2 Data

8 pts

Hooke's \(k\) calculation

Stations 3 & 4

16 pts

Friction & Buoyancy

Synthesis

8 pts

Safety engineering

Physics Dynamics • Contact Forces Lab Answer Key

Page 4 of 4 • Synthesis Solutions (12pt Edition)

Pregunta de la Estación 1: ¿Por qué la espuma hace que el rebote sea más bajo y silencioso que en la mesa dura?

Estación 2

Ley de Hooke: Estiramiento de Resorte (\(F_s = k\Delta x\))

\(g = 9.8\text{ m/s}^2\)

Cuelga masas crecientes del resorte. Mide la posición inicial \(x_0\) y estirada \(x\) para calcular la elongación \(\Delta x = x - x_0\).

Pregunta de la Estación 2: Al duplicar el peso colgado de \(0.100\text{ kg}\) a \(0.200\text{ kg}\), ¿se duplicó también el estiramiento? ¿Qué nos indica esto sobre el resorte?

Física Dinámica • Laboratorio de Fuerzas de Contacto

Página 2 de 4 • Estaciones 1 y 2 (12pt)

Página 3 Datos: Estación 3 y Estación 4

Nombre del Estudiante:

Estación 3

Fricción: Iniciar el Movimiento vs. Deslizar

Tirón con dinamómetro

Tira del bloque de madera en distintas superficies. Registra la fuerza máxima para empezar a moverlo (fricción estática \(f_s\)) y la fuerza constante para mantenerlo deslizándose (\(f_k\)).

Pregunta de la Estación 3: ¿Por qué se requiere más fuerza para poner el bloque en movimiento que para mantenerlo deslizándose?

Estación 4

Flotabilidad: El Empuje del Agua (Arquímedes)

\(F_B = F_{\text{aire}} - F_{\text{sum}}\)

Pesa el bloque de metal en el aire. Sumérgelo en el vaso de desbordamiento, anota el nuevo peso en el agua y pesa el agua desbordada que cae en el vaso receptor.

Pregunta de la Estación 4: ¿Cómo se compara el empuje hacia arriba del agua (\(F_B\)) con el peso del agua que se desbordó?

Física Dinámica • Laboratorio de Fuerzas de Contacto

Página 3 de 4 • Estaciones 3 y 4 (12pt)

Página 4 Síntesis y Aplicaciones en el Mundo Real

Nombre del Estudiante:

Síntesis

Conexión de los Resultados con el Mundo Real

1. Zonas de Deformación en Autos: Los autos antiguos tenían parachoques de acero rígido, mientras que los modernos tienen frentes que se arrugan y colapsan en un choque. Con base en lo observado en la Estación 1 (superficies duras vs. blandas), ¿por qué las zonas de deformación protegen mucho mejor a los pasajeros?

2. Zapatillas Deportivas y Rodillas: Las zapatillas de correr tienen suelas gruesas de espuma flexible. Usando los conceptos de fuerza normal y tiempo de contacto, explica cómo la suela protege las articulaciones y los huesos al saltar o correr sobre pavimento.

3. Lanzador de Pinball: En una máquina de pinball se jala un resorte para lanzar una bola de acero. Usando la Ley de Hooke de la Estación 2, explica qué sucede con la fuerza de empuje y la aceleración de la bola si jalas el resorte \(4\text{ cm}\) en lugar de \(2\text{ cm}\).

Gran Conclusión de la Física

Todas las fuerzas de contacto (fuerza normal, fricción, resortes y choques) provienen de átomos que se empujan, jalan o deslizan entre sí. Cuando una superficie puede deformarse o amortiguar, el tiempo de contacto aumenta y la fuerza máxima de impacto disminuye notablemente.

Física Dinámica • Laboratorio de Fuerzas de Contacto

Página 4 de 4 • Síntesis (12pt)

Datos (G)

Incógnita (U)

Ecuación (E)

Sustituir (S)

Resolver (S)

Problema 6 Equilibrio con Fricción Estática Vertical

TEKS 5D

Un libro de física con masa \(m = 1.50\text{ kg}\) se sostiene contra una pared vertical. El coeficiente estático es \(\mu_s = 0.40\). Calcula la fuerza de empuje horizontal mínima (\(F_{empuje}\)) para evitar que resbale hacia abajo.

Datos (G)

Incógnita (U)

Ecuación (E)

Sustituir (S)

Resolver (S)

Física Dinámica • Hoja de Trabajo de Fuerzas Fundamentales

Página 2 de 3 • Cálculos con GUESS (12pt)

Bloque 6 Evaluación y Síntesis de Modelos

Nombre del Estudiante:

Parte 3

Puntos de Verificación Conceptual

7. Una caja de madera descansa sin moverse en el suelo. ¿Qué describe las fuerzas de contacto?

A. El suelo no puede empujar hacia arriba porque es inanimado.

B. La fuerza normal equilibra el peso; la fricción estática es cero hasta empujar.

C. La fricción estática empuja hacia arriba contra la gravedad.

D. La fricción cinética evita que la caja se hunda en el suelo.

8. Se suelta un resorte comprimido una distancia \(\Delta x\). ¿Qué describe la fuerza mientras se expande?

A. La fuerza permanece constante hasta que la bola se separa.

B. La fuerza aumenta continuamente conforme se descomprime.

C. La fuerza disminuye linealmente hasta cero en el equilibrio.

D. La fuerza invierte su sentido y hala la bola hacia atrás.

Parte 4

Matemáticas: Valor Absoluto y Tolerancia de Fuerza

TEKS 5D

En ingeniería, la tolerancia mide la desviación máxima permitida respecto a una fuerza objetivo: \(|F_{\text{real}} - F_{\text{objetivo}}| \le \text{Tolerancia}\).

Un resorte de pinball debe ejercer una fuerza de \(50.0\text{ N}\) con una tolerancia de \(\pm 4.0\text{ N}\). Escribe la desigualdad con valor absoluto y calcula el rango aceptable de fuerzas.

Parte 5

Síntesis: Aceleración en el Lanzador de Pinball

Síntesis Newtoniana

Cuando el émbolo del pinball se comprime \(3.0\text{ cm}\) y se suelta, ¿experimenta la bola una aceleración constante o aceleración cambiante durante el lanzamiento? Usa la Segunda Ley de Newton (\(a = F_{net}/m\)) y la Ley de Hooke (\(F_s = k\Delta x\)) para fundamentar tu respuesta.

Física Dinámica • Hoja de Trabajo de Fuerzas Fundamentales

Página 3 de 3 • Verificación y Síntesis (12pt)

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Path: 1 (Bottom Left) ➔ Zigzags Upward ➔ 100 (Top Left)

PAGE 2 OF 7 • COMPLETE 10×10 BOARD

OVERSIZE BOARD • PART 1

UPPER HALF: SQUARES 51 — 100

Tape to Page 4 to make Giant Floor Board!

100WIN!

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OVERLAP SEAM: TAPE THIS BOTTOM MARGIN OVER TOP OF PAGE 4 CONNECTS TILE 51 TO TILE 50

ALIGN BOTTOM OF PAGE 3 PRECISELY ALONG THIS LINE TILE 50 MEETS TILE 51 ABOVE

OVERSIZE BOARD • PART 2

LOWER HALF: SQUARES 1 — 50

Start at Square 1 (Bottom Left)

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🚀 Roll die to advance • Star squares trigger Forces & Motion Task Cards! PAGE 4 OF 7 • OVERSIZE PART 2

Illustrated Task Cards Cards 01 — 12 (Set A)

FORCES & MOTION CHALLENGE CARDS

Draw when landing on a ⭐ Star Tile or defending against a snake!

CARD 01 Net Force

5 N ⬅

BOX

➡ 15 N

What is the net force \(F_{net}\) on the box above?

A) 20 N right

B) 10 N right

C) 10 N left

Correct: +2 squares boost!

CARD 02 Equilibrium

10 N ⬅

\(\Sigma F = 0\)

➡ 10 N

If \(\Sigma F = 0\), what is true about the object's motion?

A) It must be accelerating

B) At rest or constant velocity

C) It instantly halts

Correct: Roll again!

CARD 03 Newton 1st

BRAKE! 🛑 ➡ Inertia

Why do passengers lurch forward during sudden braking?

A) Inertia resists velocity change

B) Friction pushes forward

C) Gravity increases ahead

Correct: +2 squares boost!

CARD 04 \(F = ma\)

m = 4 kg

➡ \(a = 3\text{ m/s}^2\) \(F_{net} = ?\)

What net force produces this acceleration?

A) 7 N

B) 12 N

C) 1.33 N

Correct: +3 squares boost!

CARD 05 Mass vs Weight

Earth

➔

Moon

When moving Earth to Moon, which stays unchanged?

A) Weight (\(F_g\))

B) Mass (\(m\))

C) Gravity pull

Correct: Roll again!

CARD 06 Newton 3rd

A

80 N ⬅ ➡ ? N

B

Swimmer pushes water back 80 N. Reaction force forward?

A) Exactly 80 N forward

B) 40 N forward

C) 0 N

Correct: +2 squares boost!

CARD 07 Normal Force

⬆ \(F_N\) (\(90^\circ\))

Direction of normal force relative to the contact surface?

A) Parallel

B) Perpendicular (\(90^\circ\))

C) Downward

Correct: Roll again!

CARD 08 Friction

⬅ \(F_f\) Static

REST

➡ Push

Friction that must be overcome to start moving a rest box?

A) Kinetic friction

B) Static friction

C) Rolling friction

Correct: +2 squares boost!

CARD 09 Tension (\(T\))

⬆ T = ?

100 N

At Rest

A 100 N light hangs motionless. What is cord tension?

A) 0 N

B) 100 N upward

C) 200 N

Correct: Roll again!

CARD 10 Hooke's Law

〰〰〰

k = 200 N/m ➡ \(\Delta x = 0.1\text{ m}\)

What restoring force \(F = k\Delta x\) is produced?

A) 20 N

B) 2000 N

C) 2 N

Correct: +3 squares boost!

CARD 11 FBD Diagram

⬆ \(F_N\)

m

⬇ \(F_g\)

A book rests quietly on a desk. Vertical forces?

A) Gravity down & Normal up

B) Friction left & Push right

C) Tension up & Drag down

Correct: Roll again!

CARD 12 Action Pairs

\(F_{A\text{ on }B}\) vs \(F_{B\text{ on }A}\)

Why do action and reaction pairs NEVER cancel?

A) Act on different bodies

B) One force is stronger

C) Same direction

Correct: +2 squares boost!

✂ Cut along dashed lines to make individual playing cards PAGE 5 OF 7 • TASK CARDS SET A

Illustrated Task Cards Cards 13 — 24 (Set B)

FORCES & MOTION CHALLENGE CARDS

Draw when landing on a ⭐ Star Tile or defending against a snake!

CARD 13 Centripetal

⬇ \(F_c\)

In circular motion, what direction is the net force?

A) Outward

B) Inward toward center

C) Forward tangent

Correct: +2 squares boost!

CARD 14 Projectiles

v_x = const

Ignoring air drag, what happens to horizontal velocity \(v_x\)?

A) Remains constant

B) Increases

C) Drops to zero

Correct: Roll again!

CARD 15 Gravity Flight

⬇ \(g = 9.8\text{ m/s}^2\) Down

Vertical acceleration on a projectile throughout flight?

A) \(0\text{ m/s}^2\)

B) \(9.8\text{ m/s}^2\) downward

C) \(9.8\text{ m/s}^2\) upward

Correct: +2 squares boost!

CARD 16 Displacement

A 400 m loop ➔ A \(\Delta x = ?\)

Runner finishes 400 m loop at the start line. \(\Delta x =\)?

A) 400 m

B) 0 m

C) 200 m

Correct: Roll again!

CARD 17 Acceleration

Turning Corner 🔄

Which describes an object undergoing acceleration?

A) Straight line 50 mph

B) Bike turning at 10 mph

C) Book at rest

Correct: +2 squares boost!

CARD 18 Weight \(F_g\)

5 kg

⬇ \(F_g = (5)(9.8)\)

Weight of 5 kg ball on Earth (\(g = 9.8\text{ m/s}^2\))?

A) 49 N

B) 5 N

C) 2 N

Correct: +3 squares boost!

CARD 19 Unbalanced

3 N ⬅

BOX

➡ 12 N

When unbalanced forces act, what must happen?

A) Constant speed

B) Accelerates (\(a \neq 0\))

C) Mass shrinks

Correct: Roll again!

CARD 20 Friction Vector

⬅ \(F_f\) ?

➡ Sliding East

Puck slides east across ice. Friction direction?

A) East

B) West (opposes slide)

C) Downward

Correct: +2 squares boost!

CARD 21 Buoyancy

⬆ \(F_B\) (Upthrust)

⬇ \(F_g\) (Weight)

An object floats at rest. Buoyant force vs weight?

A) Buoyant is greater

B) Buoyant equals weight

C) Buoyant is zero

Correct: Roll again!

CARD 22 Solve for a

10 kg

➡ F = 50 N

50 N net force pushes a 10 kg sled. Acceleration?

A) \(500\text{ m/s}^2\)

B) \(5\text{ m/s}^2\)

C) \(0.2\text{ m/s}^2\)

Correct: +3 squares boost!

CARD 23 Tension Force

Taut String = Pull Only

Can an ideal flexible rope exert a pushing force?

A) Yes, if pushed fast

B) No, tension only pulls

C) Yes, if heavy

Correct: Roll again!

CARD 24 Equilibrium

ELEV

⬆ steady 2 m/s

Elevator moves up at steady \(2\text{ m/s}\). Net force \(\Sigma F =\)?

A) Upward & positive

B) Exactly 0 N

C) Downward

Correct: +2 squares boost!

✂ Cut along dashed lines to make individual playing cards PAGE 6 OF 7 • TASK CARDS SET B

Referee & Teacher Guide Complete Solutions

FORCES & MOTION MASTER ANSWER KEY

Quick reference for student referees and teachers during classroom game sessions

24 Challenge Cards

Full Mathematical Explanations

Set A Solutions (Cards 1–12) Answer & Rationale

01. [B] 10 N right: \(\Sigma F = 15\text{ N} - 5\text{ N} = 10\text{ N}\) in the direction of the larger force.

02. [B] At rest or constant velocity: \(\Sigma F = 0\) implies zero acceleration (\(a = 0\)).

03. [A] Inertia: Newton's 1st Law states mass resists any change in its existing state of motion.

04. [B] 12 N: \(F_{net} = ma = (4\text{ kg})(3\text{ m/s}^2) = 12\text{ N}\).

05. [B] Mass (\(m\)): Mass is the amount of matter; weight varies with local gravitational field \(g\).

06. [A] Exactly 80 N forward: Newton's 3rd Law equal and opposite reaction force from water.

07. [B] Perpendicular (\(90^\circ\)): "Normal" in physics and mathematics strictly means perpendicular to surface.

08. [B] Static friction: Static friction holds resting surfaces in place; coefficient \(\mu_s > \mu_k\).

09. [B] 100 N upward: In vertical equilibrium, \(\Sigma F_y = T - F_g = 0\), so \(T = 100\text{ N}\).

10. [A] 20 N: Hooke's Law magnitude \(F = k\Delta x = (200\text{ N/m})(0.1\text{ m}) = 20\text{ N}\).

11. [A] Gravity down & Normal up: Gravitational weight balanced by table upward support.

12. [A] Act on different bodies: Force \(F_{A\text{ on }B}\) and \(F_{B\text{ on }A}\) never act on the same object.

Set B Solutions (Cards 13–24) Answer & Rationale

13. [B] Inward toward center: Centripetal acceleration always points toward the center of curvature.

14. [A] Remains constant: With zero horizontal acceleration (\(a_x = 0\)), \(v_x\) never changes.

15. [B] \(9.8\text{ m/s}^2\) downward: Gravity pulls downward at constant \(g\) throughout entire flight.

16. [B] 0 m: Displacement is the straight-line vector from start to finish; here \(\Delta x = 0\).

17. [B] Bicycle turning a corner: Changing direction changes the velocity vector, producing acceleration.

18. [A] 49 N: \(F_g = mg = (5\text{ kg})(9.8\text{ m/s}^2) = 49\text{ N}\).

19. [B] Accelerates: A non-zero net force always produces acceleration according to \(a = \Sigma F / m\).

20. [B] West (opposing sliding): Kinetic friction acts opposite to the direction of relative surface motion.

21. [B] Buoyant force equals weight: For vertical floating equilibrium, buoyant upthrust \(F_B = F_g\).

22. [B] \(5\text{ m/s}^2\): \(a = F/m = 50\text{ N} / 10\text{ kg} = 5\text{ m/s}^2\).

23. [B] No, tension only pulls: Flexible cords cannot transmit compressive loads without buckling.

24. [B] Exactly 0 N: Steady velocity means acceleration \(a = 0\), which requires \(\Sigma F = 0\).

Game Facilitation & Tournament Tips

Peer Verification:

Appoint one student in each group as the "Physics Chief" who holds this answer key to referee cards.

Fast-Play Mode:

Roll 2 dice or grant an automatic +5 tile boost on every correct card to complete games in under 20 minutes.

Snake Defense Challenge:

Students who land on a snake get 1 chance to answer a card: correct answer allows them to ignore the snake!

High School Physics Instructional Resource • Types of Forces Unit PAGE 7 OF 7 • MASTER ANSWER KEY