Work and Kinetic Energy

Definition of work, kinetic energy, and the work-energy theorem

⚡ Work and Kinetic Energy

What is Work?

In physics, work has a very specific meaning: work is done when a force causes a displacement.

Definition of Work

$W = Fd\cos\theta$

where:

$W$ = work (J, joules)
$F$ = magnitude of force (N)
$d$ = magnitude of displacement (m)
$\theta$ = angle between force and displacement vectors

Units: 1 joule (J) = 1 N·m = 1 kg·m²/s²

Key Points

Force must cause displacement - No displacement = no work
Only the component of force parallel to displacement does work
Work can be positive, negative, or zero

💡 Important: Work is a scalar quantity (not a vector), even though it's calculated from two vectors.

Sign of Work

Positive Work ( $W > 0$ )

Force has component in direction of motion ( $0° \leq \theta < 90°$ )
Energy is added to the system
Example: Pushing a box forward while it moves forward

Negative Work ( $W < 0$ )

Force has component opposite to motion ( $90° < \theta \leq 180°$ )
Energy is removed from the system
Example: Friction on a sliding box (friction opposes motion)

Zero Work ( $W = 0$ )

Force perpendicular to motion ( $\theta = 90°$ )
OR no displacement ( $d = 0$ )
Example: Carrying a box horizontally (weight is perpendicular to motion)

Special Cases

Case 1: Force Parallel to Displacement ( $\theta = 0°$ )

$W = Fd\cos(0°) = Fd$

Maximum positive work.

Case 2: Force Opposite to Displacement ( $\theta = 180°$ )

$W = Fd\cos(180°) = -Fd$

Maximum negative work.

Case 3: Force Perpendicular to Displacement ( $\theta = 90°$ )

$W = Fd\cos(90°) = 0$

No work done.

Kinetic Energy

Kinetic energy is energy of motion:

$KE = \frac{1}{2}mv^2$

where:

$KE$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = speed (m/s)

Properties of Kinetic Energy

Always positive (or zero) - $v^2$ is always positive
Scalar quantity - no direction
Depends on speed squared - Double the speed = 4× the kinetic energy
Frame-dependent - KE depends on reference frame

The Work-Energy Theorem

The work-energy theorem connects work and kinetic energy:

$W_{net} = \Delta KE = KE_f - KE_i$

$W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Interpretation

Net work done on an object equals the change in its kinetic energy
If $W_{net} > 0$ : object speeds up (gains KE)
If $W_{net} < 0$ : object slows down (loses KE)
If $W_{net} = 0$ : speed unchanged (constant KE)

Calculating Work by Multiple Forces

When multiple forces act on an object:

Method 1: Sum the work done by each force

$W_{net} = W_1 + W_2 + W_3 + ...$

Each force can do positive, negative, or zero work.

Method 2: Find net force first

$W_{net} = F_{net} \cdot d \cdot \cos\theta$

where $\theta$ is angle between $\vec{F}_{net}$ and $\vec{d}$ .

Both methods give the same result!

Variable Forces and Work

For a variable force (force changes during motion):

$W = \int_{x_i}^{x_f} F(x)\,dx$

The work equals the area under the force vs. position graph.

For AP Physics 1, you usually just need to recognize this graphically.

⚠️ Common Misconceptions

Misconception 1: Confusing Force and Work

❌ Wrong: "I'm doing work just by holding this box" ✅ Right: Holding stationary = no displacement = no work (even though it's tiring!)

Misconception 2: Forgetting the Angle

❌ Wrong: $W = Fd$ always ✅ Right: $W = Fd\cos\theta$ - angle matters!

Misconception 3: Thinking Work is a Vector

❌ Wrong: Work has direction ✅ Right: Work is a scalar (can be positive or negative, but that's not direction)

Misconception 4: Normal Force Always Does Zero Work

Usually true (when surface is level), but NOT always! On an incline or accelerating elevator, normal force can do work.

Problem-Solving Strategy

Identify all forces acting on the object
Draw a free body diagram
For each force, determine:
- Magnitude $F$
- Angle $\theta$ relative to displacement
- Sign of work (positive/negative/zero)
Calculate work for each force: $W = Fd\cos\theta$
Sum to find net work
Apply work-energy theorem if finding speed change

Real-World Applications

Stopping Distance

When brakes are applied, friction does negative work to remove kinetic energy: $W_{friction} = -\frac{1}{2}mv^2$

Doubling speed requires 4 times the stopping distance!

Accelerating Vehicles

Engine does positive work to increase kinetic energy: $W_{engine} = \frac{1}{2}m(v_f^2 - v_i^2)$

Sports

Baseball: Work done by bat on ball increases ball's KE
Braking in cars: Friction converts KE to thermal energy

Key Formulas Summary

| Concept | Formula | Units | |---------|---------|-------| | Work | $W = Fd\cos\theta$ | J (joules) | | Kinetic Energy | $KE = \frac{1}{2}mv^2$ | J | | Work-Energy Theorem | $W_{net} = \Delta KE$ | J | | Power (rate of work) | $P = \frac{W}{t}$ | W (watts) |

📚 Practice Problems

1Problem 1medium

❓ Question:

A 2.0 kg block is pushed 5.0 m across a floor by a constant 15 N force parallel to the floor. (a) How much work is done by the applied force? (b) If the block starts from rest, what is its final kinetic energy? (c) What is its final speed?

💡 Show Solution

Solution:

Given: m = 2.0 kg, d = 5.0 m, F = 15 N, v₀ = 0

(a) Work done: W = Fd cos θ (θ = 0° since force is parallel to displacement) W = 15 × 5.0 × cos 0° W = 15 × 5.0 × 1 W = 75 J

(b) Final kinetic energy: Assuming no friction, by Work-Energy Theorem: W_net = ΔKE = KE_f - KE_i 75 = KE_f - 0 KE_f = 75 J

2Problem 2easy

❓ Question:

A 5 kg box is pushed 10 m across a floor by a horizontal force of 20 N. If the box starts from rest, what is its final speed? (Assume no friction.)

💡 Show Solution

Given Information:

Mass: $m = 5$ kg
Displacement: $d = 10$ m
Applied force: $F = 20$ N (horizontal)
Initial velocity: $v_i = 0$ m/s (starts from rest)
No friction

Find: Final speed $v_f$

Step 1: Calculate work done by applied force

Since force is horizontal and displacement is horizontal, $\theta = 0°$ :

$W = Fd\cos\theta = (20)(10)\cos(0°) = 200 \text{ J}$

Step 2: Apply work-energy theorem

$W_{net} = \Delta KE$

Since there's no friction, $W_{net} = W = 200$ J

$200 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$200 = \frac{1}{2}(5)v_f^2 - 0$

$200 = 2.5v_f^2$

$v_f^2 = 80$

$v_f = \sqrt{80} = 4\sqrt{5} \approx 8.94 \text{ m/s}$

Answer: The final speed is approximately 8.94 m/s or $4\sqrt{5}$ m/s.

3Problem 3medium

❓ Question:

💡 Show Solution

Solution:

Given: m = 2.0 kg, d = 5.0 m, F = 15 N, v₀ = 0

(a) Work done: W = Fd cos θ (θ = 0° since force is parallel to displacement) W = 15 × 5.0 × cos 0° W = 15 × 5.0 × 1 W = 75 J

(b) Final kinetic energy: Assuming no friction, by Work-Energy Theorem: W_net = ΔKE = KE_f - KE_i 75 = KE_f - 0 KE_f = 75 J

4Problem 4hard

❓ Question:

A 3.0 kg box is pulled 8.0 m up a frictionless 30° incline by a force of 25 N parallel to the incline. (a) Find the work done by the applied force. (b) Find the work done by gravity. (c) Find the net work. (d) If the box starts from rest, what is its final speed?

💡 Show Solution

Solution:

Given: m = 3.0 kg, d = 8.0 m, F = 25 N, θ = 30°, g = 10 m/s²

(a) Work by applied force: W_F = Fd cos 0° = 25 × 8.0 = 200 J

(b) Work by gravity: Vertical height: h = d sin 30° = 8.0 × 0.5 = 4.0 m Weight: mg = 3.0 × 10 = 30 N (downward) Angle between F_g and displacement = 180° - 30° = 150°

W_g = mgd cos 150° = 30 × 8.0 × (-cos 30°) W_g = 30 × 8.0 × (-0.866) = -208 J

Or: W_g = -mgh = -30 × 4.0 = -120 J ✓ (simpler method)

(d) Final speed: W_net = ΔKE = ½mv² - 0 80 = ½(3.0)v² v² = 160/3 = 53.3 v = 7.3 m/s

5Problem 5hard

❓ Question:

💡 Show Solution

Solution:

Given: m = 3.0 kg, d = 8.0 m, F = 25 N, θ = 30°, g = 10 m/s²

(a) Work by applied force: W_F = Fd cos 0° = 25 × 8.0 = 200 J

(b) Work by gravity: Vertical height: h = d sin 30° = 8.0 × 0.5 = 4.0 m Weight: mg = 3.0 × 10 = 30 N (downward) Angle between F_g and displacement = 180° - 30° = 150°

W_g = mgd cos 150° = 30 × 8.0 × (-cos 30°) W_g = 30 × 8.0 × (-0.866) = -208 J

Or: W_g = -mgh = -30 × 4.0 = -120 J ✓ (simpler method)

(d) Final speed: W_net = ΔKE = ½mv² - 0 80 = ½(3.0)v² v² = 160/3 = 53.3 v = 7.3 m/s

6Problem 6medium

❓ Question:

A 2 kg block slides 5 m down a frictionless incline that makes a 30° angle with the horizontal. What is the work done by gravity?

💡 Show Solution

Given Information:

Mass: $m = 2$ kg
Displacement along incline: $d = 5$ m
Angle of incline: $30°$
Frictionless (no friction)

Find: Work done by gravity

Step 1: Identify the gravitational force

$F_g = mg = (2)(9.8) = 19.6 \text{ N}$

This force points vertically downward.

Step 2: Find the angle between force and displacement

Gravity points straight down (vertical)
Displacement is along the incline (30° below horizontal)
Angle between them: $\theta = 90° - 30° = 60°$

Step 3: Calculate work

$W_g = F_g d \cos\theta$

$W_g = (19.6)(5)\cos(60°)$

$W_g = (19.6)(5)(0.5)$

$W_g = 49 \text{ J}$

Alternative Method: Use vertical displacement

Vertical drop: $h = d\sin(30°) = 5(0.5) = 2.5$ m

Work by gravity = $mgh = (2)(9.8)(2.5) = 49$ J ✓

Answer: Gravity does 49 J of work (positive because it has a component along the motion).

Note: Even though the block moves 5 m along the incline, only the vertical component of that displacement matters for gravitational work!

7Problem 7hard

❓ Question:

A 1200 kg car traveling at 25 m/s applies its brakes and comes to a stop in 50 m. (a) What is the work done by friction? (b) What is the magnitude of the average frictional force?

💡 Show Solution

Given Information:

Mass: $m = 1200$ kg
Initial velocity: $v_i = 25$ m/s
Final velocity: $v_f = 0$ m/s (stops)
Displacement: $d = 50$ m

(a) Find work done by friction

Step 1: Calculate initial kinetic energy

$KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(1200)(25)^2$

$KE_i = 600(625) = 375,000 \text{ J}$

Step 2: Calculate final kinetic energy

$KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(1200)(0)^2 = 0 \text{ J}$

Step 3: Apply work-energy theorem

$W_{friction} = \Delta KE = KE_f - KE_i$

$W_{friction} = 0 - 375,000 = -375,000 \text{ J}$

The negative sign indicates friction removes energy from the car.

(b) Find magnitude of frictional force

Step 4: Use work formula

$W = Fd\cos\theta$

Since friction opposes motion, $\theta = 180°$ :

$-375,000 = F(50)\cos(180°)$

$-375,000 = F(50)(-1)$

$-375,000 = -50F$

$F = 7,500 \text{ N}$

Alternative approach:

$|W| = Fd$

$375,000 = F(50)$

$F = 7,500 \text{ N}$

Answers:

(a) Work done by friction: -375,000 J (or -375 kJ)
(b) Magnitude of friction force: 7,500 N

Interpretation: The friction force of 7,500 N acting over 50 m removes all the car's kinetic energy, bringing it to rest.

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Work and Kinetic Energy

⚡ Work and Kinetic Energy

What is Work?

Definition of Work

Key Points

Sign of Work

Positive Work (W>0W > 0W>0)

Negative Work (W<0W < 0W<0)

Zero Work (W=0W = 0W=0)

Special Cases

Case 1: Force Parallel to Displacement (θ=0°\theta = 0°θ=0°)

Case 2: Force Opposite to Displacement (θ=180°\theta = 180°θ=180°)

Case 3: Force Perpendicular to Displacement (θ=90°\theta = 90°θ=90°)

Kinetic Energy

Properties of Kinetic Energy

The Work-Energy Theorem

Interpretation

Calculating Work by Multiple Forces

Method 1: Sum the work done by each force

Method 2: Find net force first

Variable Forces and Work

⚠️ Common Misconceptions

Misconception 1: Confusing Force and Work

Misconception 2: Forgetting the Angle

Misconception 3: Thinking Work is a Vector

Misconception 4: Normal Force Always Does Zero Work

Problem-Solving Strategy

Real-World Applications

Stopping Distance

Accelerating Vehicles

Sports

Key Formulas Summary

📚 Practice Problems

1Problem 1medium

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4hard

❓ Question:

5Problem 5hard

❓ Question:

6Problem 6medium

❓ Question:

7Problem 7hard

❓ Question:

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Positive Work ( $W > 0$ )

Negative Work ( $W < 0$ )

Zero Work ( $W = 0$ )

Case 1: Force Parallel to Displacement ( $\theta = 0°$ )

Case 2: Force Opposite to Displacement ( $\theta = 180°$ )

Case 3: Force Perpendicular to Displacement ( $\theta = 90°$ )