Spring Foundations Slides Springing Into Work
Applications of Integration in Physics
Lesson 01: Variable Force
The Stretch Factor
Think about these two tasks:
Task A: Lifting a 10kg Weight
The weight feels the same at inch 1, inch 10, and inch 20.
Task B: Pulling a Rubber Band
The further you pull, the harder it pushes back.
"If the force is constantly changing, how do we calculate the total energy used?"
Work: The Physics Baseline
Constant Force Formula
\[ W = F \cdot d \]
1
Force (\(F\))
Units: Newtons (N) or Pounds (lb)
2
Distance (\(d\))
Units: Meters (m) or Feet (ft)
Result: Work (\(W\)) is in Joules (J) or ft-lb
When Force varies with Position...
If force is a function of position, \(F(x)\), we can't just multiply. We must accumulate .
The Calculus Definition:
\[ W = \int_{a}^{b} F(x) \, dx \]
Work is the area under the force-position curve .
\(a\)
\(b\)
Force \(F(x)\)
Distance \(x\)
Hooke's Law
Compressing
Pushing a spring from its natural length requires more force as it gets shorter.
Stretching
Pulling a spring requires more force as it gets longer.
The Model
\[ F(x) = kx \]
\(k\) is the spring constant.
Crucial Insight: In these problems, \(x\) represents the displacement from the natural length of the spring.
Live Example 01
Spring Work Calculation
A force of 40 N is required to hold a spring that has been stretched 0.1 m from its natural length.
How much work is done stretching the spring from 0.1 m to 0.3 m?
1
Find \(k\): \(40 = k(0.1) \implies k = 400\) N/m
2
Set up integral: \(W = \int_{0.1}^{0.3} 400x \, dx\)
3
Evaluate: \([200x^2]_{0.1}^{0.3} = 200(0.09 - 0.01) = 200(0.08)\)
Final Work
16 J
Calculated in Joules (N·m)
Your Turn!
If a spring has a natural length of 10 cm and a force of 12 N is required to maintain it at a length of 16 cm...
What is the displacement \(x\) used in Hooke's Law?
A) 10 cm
B) 16 cm
C) 6 cm
Spring Power Worksheet Spring Power
Calculus II: Applications of Integration
NAME:
DATE:
Hooke's Law
\( F(x) = kx \)
\(x\) = displacement from natural length
Work Definition
\( W = \int_{a}^{b} F(x) \, dx \)
Integral of force over displacement
1
Calibration
A spring has a natural length of 20 cm. A force of 40 N is required to stretch the spring to a total length of 25 cm.
A) Find the spring constant \(k\).
B) How much work is done stretching the spring from 20 cm to 30 cm?
2
Compression Dynamics
Suppose that 2 Joules of work is needed to compress a spring from its natural length of 30 cm to a length of 20 cm.
How much work is required to compress the spring an additional 5 cm (to a total length of 15 cm)?
FINAL ANSWER:
3
The Constant vs. The Variable
A heavy metal block is being pushed against a spring with spring constant \(k = 500 \text{ N/m}\).
Compare the work required to stretch the spring from 0 to 10 cm versus 10 to 20 cm. Use calculus to justify why the latter is greater, even though the distance traveled (10 cm) is the same.
Critical Thinking
In Problem 1, why would it be incorrect to use \(x = 25\) in the formula \(F = kx\)? Explain the role of "natural length" in setting up your bounds of integration.
Variable Force Teacher Guide Variable Force
Teacher Facilitation Guide
Lesson 1 of 5
Lesson Objectives
Define work as the integral of force with respect to displacement.
Apply Hooke’s Law (\(F = kx\)) to determine spring constants from given physical data.
Calculate the work required to stretch or compress springs between specific intervals.
Pacing
10 min
The Hook & Review
20 min
Direct Instruction (Slides)
25 min
Guided Worksheet
Instructional Scripting & Prompts
The Hook: stretching a rubber band vs. lifting a weight
"Ask: Does a rubber band feel harder to pull at the beginning or at the end? Why? Now, if I lift a 5lb dumbbell, does it feel harder to lift it the first foot or the second foot? Why the difference?"
Common Misconception: Units and Displacement
Students often use the total length (e.g., 25 cm) as the \(x\) value. Remind them: \(x\) is the change from natural length . If natural length is 20 cm, then 25 cm means \(x = 5\). Also, always convert cm to meters to keep units in Newtons and Joules.
Worksheet Answer Key (Spring Power)
1. Calibration
A) \(F = kx \implies 40 = k(0.05) \implies k = 800 \text{ N/m}\)
B) \(W = \int_{0}^{0.1} 800x \, dx = [400x^2]_{0}^{0.1} = 400(0.01) = 4 \text{ Joules}\)
2. Compression Dynamics
First, find \(k\): \(W = \int_{0}^{0.1} kx \, dx = 2 \implies [\frac{1}{2}kx^2]_{0}^{0.1} = 2 \implies 0.005k = 2 \implies k = 400 \text{ N/m}\)
Work for additional 5 cm: \(W = \int_{0.1}^{0.15} 400x \, dx = [200x^2]_{0.1}^{0.15} = 200(0.0225 - 0.01) = 200(0.0125) = 2.5 \text{ Joules}\)
3. Constant vs. Variable
Interval 1: \( \int_{0}^{0.1} 500x \, dx = 2.5 \text{ J}\) | Interval 2: \( \int_{0.1}^{0.2} 500x \, dx = 7.5 \text{ J}\)
Reasoning: The force \(F(x) = 500x\) is larger at every point in the second interval, so the area under the curve (the work) must be larger.
Pumping Power Slides Pumping Power
Fluid Displacement & Integration
Lesson 02: Tank Problems
The Civil Challenge
You've been hired to empty a flooded subterranean reservoir.
"Why can't we just multiply the weight of the water by the depth of the tank?"
Because every "slice" of water travels a different distance to reach the top!
Lifting Liquid
[PROJECT: RESERVOIR_01]
The Slice Method
Step 1: Slice
Imagine a thin horizontal disk of liquid at height \(y\).
Step 2: Area
Find the area of that slice, \(A(y)\), based on the tank's shape.
Step 3: Force
Weight of slice = \(\text{Density} \times A(y) \times \Delta y\).
Step 4: Distance
Distance slice travels = (Top of Tank - \(y\)).
Master Formula
\[ W = \int_{a}^{b} \rho g \cdot A(y) \cdot D(y) \, dy \]
\(\rho g\): Weight Density
\(D(y)\): Distance to Top
The Geometry Challenge: Cones
In a cone, the radius of our slice changes as we move up the tank.
Similar Triangles
\[ \frac{r}{y} = \frac{R}{H} \implies r = \frac{R}{H}y \]
This means \(A(y) = \pi r^2\) becomes \(A(y) = \pi \left(\frac{R}{H}y\right)^2\).
r(y)
HEIGHT (y)
Pumping Above the Tank
The "Spout" Effect
If the water must be pumped to a height H meters above the tank, your distance term changes.
\(D(y) = (H_{tank} + H_{spout}) - y\)
Every extra foot of pipe
The Pumping Checklist
1
Geometry
Draw the tank. Define your origin (bottom is easiest).
2
Weight
Density of water: \(1000 \, \text{kg/m}^3\) or \(62.5 \, \text{lb/ft}^3\).
3
Distance
Distance from slice \(y\) to top of the outlet.
4
Bounds
Integrate from the current liquid level to the bottom of the tank.
Tank Troubleshooting Worksheet Tank Troubleshooting
Fluid Dynamics & Pumping Work
NAME:
DATE:
General Integral
\( W = \int \rho g \cdot A(y) \cdot D(y) \, dy \)
Water Density (Metric)
\( 9800 \text{ N/m}^3 \)
Water Density (Imperial)
\( 62.5 \text{ lb/ft}^3 \)
1
The Standard Well
A vertical cylindrical tank has a height of 10 feet and a radius of 4 feet. It is currently half-full of water.
How much work is required to pump all of the water over the top rim of the tank?
Work Area: Sketch and Integral Setup
2
The Grain Silo Spill
An inverted conical tank (point down) has a height of 6 meters and a radius of 2 meters at the top. It is full of a liquid with weight density \( \rho g = 9000 \text{ N/m}^3 \).
Calculate the work required to pump all of the liquid to a level 2 meters above the top of the tank.
Work Area: Similar Triangles and Integral
3
The Spherical Basin
A hemispherical bowl with a radius of 5 feet is filled with water.
A) Determine the radius of a horizontal slice \(r(y)\) at height \(y\). (Hint: \(y=0\) is the bottom of the bowl).
B) Set up the integral to find the work done pumping the top 2 feet of water over the rim.
Final Calculation (Problem 3)
Show final integration steps here...
The Efficiency Check
Compare Problems 1 and 2. In which tank does the work increase most rapidly as the liquid level drops? Why? Use your distance functions to justify your answer.
Tank Pumping Teacher Guide Pumping Power
Teacher Facilitation Guide
Lesson 2 of 5
Facilitation Strategies
The Coordinate Choice
Encourage students to define \(y=0\) at the bottom of the tank. This makes \(D(y) = \text{Top} - y\). If they define \(y=0\) at the top, \(D(y) = y\), which simplifies the distance but can confuse the area function. Stick to one method for consistency.
Imperial Units Warning
Remind students that \(62.5 \text{ lb/ft}^3\) is weight density (\(\rho g\)), not mass density. They do not need to multiply by \(g\) (32.2) when working in pounds and feet.
Tank Troubleshooting Key
1. The Standard Well (Cylinder)
Radius \(r = 4\), Area \(A(y) = 16\pi\). Distance \(D(y) = 10 - y\).
2. The Grain Silo Spill (Cone)
Similar Triangles: \(\frac{r}{y} = \frac{2}{6} \implies r = \frac{y}{3}\). Area \(A(y) = \pi(\frac{y}{3})^2 = \frac{\pi y^2}{9}\).
3. The Spherical Basin (Hemisphere)
A) Radius of slice at height \(y\): Center of sphere is at \((0, 5)\). Equation: \(x^2 + (y-5)^2 = 25 \implies r^2 = 25 - (y-5)^2 = 10y - y^2\).
Heavy Lifting Slides Heavy Lifting
Variable Mass & Linear Density
Lesson 03: Chains & Cables
The Anchor Problem
Imagine pulling a heavy 100ft chain up a cliffside.
The Paradox of Lifting:
At the start, you are pulling the entire weight of the chain.
By the end, most of the chain is already at the top. You are pulling almost nothing .
Linear Density (\(\lambda\))
How much weight does one foot of cable have?
Metric
kg/m
(Need to multiply by \(g\))
Imperial
lb/ft
(Already a force!)
Variable Force Function:
\[ F(x) = \text{Density} \times \text{Length Hanging} \]
Where is \(x\)?
Most cable problems are easiest if you let \(x\) be the amount of cable already pulled up .
If total length is \(L\), then the amount still hanging is \((L - x)\).
The force at position \(x\) is \(F(x) = \lambda(L - x)\).
The Integral Setup
\[ W = \int_{0}^{L} \lambda(L - x) \, dx \]
"Integrating from 0 (nothing pulled) to L (everything pulled)."
Lifting a Payload
The Payload Rule
If there is a constant weight (like a bucket or anchor) at the end:
\[ W_{total} = W_{cable} + W_{payload} \]
Shortcut:
You can also just add the payload weight into your force function:
[CONST_MASS: 500lb]
Quick Challenge
A 50ft cable weighing 2 lb/ft is used to lift a 100lb motor. What is the force required to hold the motor when 10ft of cable has been pulled up?
A) 100 lb
B) 180 lb
C) 200 lb
Reasoning: 40ft of cable is still hanging (40 * 2 = 80) plus the 100lb motor.
Cable and Chain Calculus Worksheet Heavy Lifting
Variable Mass Systems & Linear Density
NAME:
DATE:
Force Function (Chain)
\( F(x) = \lambda \cdot (L - x) \)
Work Integral
\( W = \int_{0}^{L} F(x) \, dx \)
1
The Simple Pull
A 40-meter heavy chain lies on the ground. Its linear density is 5 kg/m. How much work is required to lift the chain so that it is hanging vertically from its end? (Use \( g = 9.8 \text{ m/s}^2 \)).
Work Area: Setup and Integration
2
The Salvage Mission
A 200-lb anchor is attached to a 50-ft chain that weighs 3 lb/ft. The anchor is currently at the bottom of the ocean floor.
How much work is required to pull the anchor and the chain all the way up to the deck of the salvage ship?
Work Area: Force function \( F(x) \) should include both components
3
The Leaking Payload
A bucket weighing 10 lbs is lifted 20 feet using a cable of negligible weight. The bucket starts with 40 lbs of water but has a hole and leaks at a constant rate such that it is empty just as it reaches the top.
A) Express the total weight of the bucket and water as a function of the distance lifted \(x\).
B) Use integration to find the total work done.
Conceptual Extension
Consider a 100ft chain. Why is the work required to pull it up the first 50 feet significantly more than the work required to pull it up the last 50 feet? Use the geometry of the integral to explain.
Lifting Logic Teacher Guide Heavy Lifting
Teacher Facilitation Guide
Lesson 3 of 5
Instructional Focus
The "Force" Confusion
Students will consistently forget to multiply mass (kg) by gravity (9.8) to get force (Newtons). In US Customary units, pounds (lb) ARE a unit of force. Emphasize this early!
Decomposition Method
Encourage students to split work into pieces: \( W = W_{chain} + W_{payload} \). This is often less intimidating than trying to write one complex force function for everything.
Lifting Logic Key
1. The Simple Pull (Chain)
Density \(\lambda = 5 \text{ kg/m} \cdot 9.8 = 49 \text{ N/m}\).
2. The Salvage Mission (Payload)
Chain: \( \int_{0}^{50} 3(50 - x) \, dx = 3 [50x - \frac{1}{2}x^2]_0^{50} = 3(1250) = 3,750 \text{ ft-lb} \).
3. The Leaking Payload (Bucket)
A) Water weight starts at 40 and goes to 0 over 20 feet. \( \text{Water}(x) = 40 - 2x \).
Fluid Force Slides Under Pressure
Hydrostatic Force & Submerged Planes
Lesson 04: Fluid Mechanics
Why are dams thick at the bottom?
"The water at the surface doesn't push as hard as the water 500 feet down."
Pressure is depth-dependent.
To find the total force on a vertical wall, we must account for the changing pressure from top to bottom.
Pressure
PRESSURE!
Pressure vs. Force
Hydrostatic Pressure
\[ P = \rho g d \]
"Pressure at depth \(d\)"
Fluid Force
\[ F = \int P \, dA \]
"Summing pressure over the area"
The Integrated Approach
Master Formula
To find force on a vertical plate from depth \(a\) to \(b\):
\[ F = \int_{a}^{b} \rho g \cdot \text{depth} \cdot \text{width} \, dy \]
Depth = \(y\) (or distance from surface)
Width = \(w(y)\) (depends on plate shape)
Plate Geometry
Rectangle
Width = constant
Triangle
Width = linear \(f(y)\)
Circle
Width = Pythagorean
Remember!
1. Identify the depth function (distance from surface).
2. Identify the width function (geometry of the plate).
3. Integrate from the top of the plate to the bottom of the plate.
Total Force = Accumulation of Depth-Dependent Pressure
Deep Sea Dams Worksheet Deep Sea Dams
Hydrostatic Pressure & Fluid Force
NAME:
DATE:
Pressure at Depth \(d\)
\( P = \rho g d \)
Total Fluid Force
\( F = \int_{a}^{b} \rho g y \cdot w(y) \, dy \)
1
The Canal Lock
A rectangular plate 3 meters wide and 4 meters high is submerged vertically in water such that the top of the plate is 2 meters below the surface.
Find the total fluid force exerted by the water on one side of the plate. (Use \( \rho g = 9800 \text{ N/m}^3 \)).
Work Area: Define your variable \(y\) as depth from the surface
2
The Triangular Sluice
A vertical dam has the shape of an inverted isosceles triangle with a height of 10 feet and a base (at the top) of 12 feet. The water level is exactly at the top of the dam.
A) Find a width function \(w(y)\) in terms of the depth \(y\) from the surface.
B) Set up the integral for the total force. (Use \( \rho g = 62.5 \text{ lb/ft}^3 \)).
Complete Integration and Final Calculation here...
3
The Submarine Observation Window
A circular porthole on a submarine has a radius of 1 foot. The center of the porthole is 20 feet below the surface.
Set up an integral that represents the total fluid force on the porthole. (Hint: Let \(y = 0\) at the center of the porthole. How does depth relate to \(y\)? How does width relate to \(y\)?).
Why integrate?
If the plate was horizontal (lying flat at the bottom), would we need calculus to find the total force? Why or why not?
Pressure Points Answer Key Under Pressure
Teacher Facilitation Guide
Lesson 4 of 5
Depth vs. Height
The Coordinate Origin
Unlike tank problems where we often put the origin at the bottom, fluid force problems are usually easiest if you put the origin at the water surface . This makes depth \(d = y\), simplifying the pressure term \(\rho g y\).
Horizontal Force
Remind students that hydrostatic pressure acts in all directions equally (Pascal's Principle). Even though the water is pushing "sideways" against a vertical plate, the magnitude is still determined by the vertical depth.
Deep Sea Dams Key
1. The Canal Lock (Rectangle)
Depth \(y\) goes from 2 to 6. Width \(w = 3\) (constant).
2. The Triangular Sluice (Triangle)
A) Similar Triangles: At depth \(y\), width \(w\). Base is at depth \(y=0\), vertex is at \(y=10\).
3. Submarine Window (Circle)
Let \(y=0\) be center of circle. Width \(w(y) = 2\sqrt{1 - y^2}\). Depth \(d(y) = 20 - y\).
Engineering Challenge Assessment Official Assessment
Engineering Challenge
Physics Applications of Integration
EXAMINEE:
ID NUMBER:
Testing Protocol
Complete the following five engineering problems. Show all integral setups, coordinate systems used, and final numerical answers with appropriate units. Partial credit will be awarded for correct geometric modeling.
01
Spring Tension Analysis
[15 PTS]
A spring has a natural length of 0.5 meters. A force of 20 N is required to stretch it to a length of 0.8 meters. Calculate the work done in stretching the spring from 0.8 meters to 1.1 meters.
02
High-Rise Cable Dynamics
[20 PTS]
A construction crane lifts a 500-kg steel beam to the top of a 100-meter building. The cable used has a mass of 2 kg per meter. Calculate the total work done by the crane. (Use \( g = 9.8 \text{ m/s}^2 \)).
03
Hydro-Storage Extraction
[25 PTS]
A hemispherical tank with a radius of 10 feet is full of water (\( 62.5 \text{ lb/ft}^3 \)). The water must be pumped out through a spout that is 2 feet above the top of the tank. Set up the integral and calculate the work required to empty the top half of the tank (the first 5 feet of depth).
04
Structural Integrity Check
[20 PTS]
A vertical plate in the shape of a trapezoid is submerged in water. The top edge is 8 meters wide and lies at the water surface. The bottom edge is 4 meters wide and lies at a depth of 6 meters. Calculate the total fluid force on one side of the plate.
05
The Leaking Bucket Redux
[20 PTS]
A 5-lb bucket containing 20 lbs of sand is being lifted 30 feet. The sand is leaking out at a constant rate such that only 5 lbs of sand remain when the bucket reaches the top. If the rope used weighs 0.5 lb/ft, find the total work done.
End of Engineering Assessment // Force Accumulation Division
Engineering Challenge Key Mastery Challenge Solutions
Assessment Master Key & Rubric
Sequence Finale
Universal Scoring Rubric (Per Problem)
40%
Modeling
Correct coordinate system and \(F(x)\) setup.
30%
Setup
Correct limits and integral formulation.
20%
Calculus
Correct antiderivative and evaluation.
10%
Precision
Final answer with correct units.
Problem 01: Springs
\( F = kx \implies 20 = k(0.3) \implies k = \frac{200}{3} \approx 66.67 \text{ N/m} \).
\( W = \int_{0.3}^{0.6} \frac{200}{3}x \, dx = [\frac{100}{3}x^2]_{0.3}^{0.6} \).
\( W = \frac{100}{3} (0.36 - 0.09) = \frac{100}{3}(0.27) = \mathbf{9 \text{ Joules}} \).
Problem 02: Lifting
\( F_{beam} = 500 \cdot 9.8 = 4900 \text{ N} \). \( F_{cable}(x) = 2 \cdot 9.8 \cdot (100 - x) = 19.6(100 - x) \).
\( W = \int_0^{100} [4900 + 19.6(100 - x)] \, dx \).
\( W = [4900x + 1960x - 9.8x^2]_0^{100} = 490,000 + 196,000 - 98,000 \).
\( W = \mathbf{588,000 \text{ Joules}} \).
Problem 03: Pumping
\( A(y) = \pi(100 - (y-10)^2) = \pi(20y - y^2) \). Distance \( D(y) = 12 - y \).
\( W = \int_{5}^{10} (62.5)\pi(20y - y^2)(12 - y) \, dy \).
\( W = 62.5\pi \int_{5}^{10} (240y - 32y^2 + y^3) \, dy \).
\( W = 62.5\pi [120y^2 - \frac{32}{3}y^3 + \frac{1}{4}y^4]_5^{10} \approx \mathbf{1,085,270 \text{ ft-lb}} \).
Problem 04: Fluid Force
Width \( w(y) \) at depth \( y \): Line from \((0, 8)\) to \((6, 4)\). \( w - 8 = \frac{4-8}{6-0}(y - 0) \implies w(y) = 8 - \frac{2}{3}y \).
\( F = \int_0^6 9800 \cdot y \cdot (8 - \frac{2}{3}y) \, dy = 9800 [4y^2 - \frac{2}{9}y^3]_0^6 \).
\( F = 9800 (144 - 48) = 9800(96) = \mathbf{940,800 \text{ Newtons}} \).
Problem 05: Leaking Bucket
\( F_{bucket} = 5 \). \( F_{sand}(x) = 20 - \frac{15}{30}x = 20 - 0.5x \). \( F_{rope}(x) = 0.5(30 - x) \).
\( W = \int_0^{30} [5 + 20 - 0.5x + 15 - 0.5x] \, dx = \int_0^{30} (40 - x) \, dx \).
\( W = [40x - 0.5x^2]_0^{30} = 1200 - 450 = \mathbf{750 \text{ ft-lb}} \).
