Work and Kinetic Energy

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

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⚡ 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=FdcosθW = Fd\cos\theta

where:

  • WW = work (J, joules)
  • FF = magnitude of force (N)
  • dd = magnitude of displacement (m)
  • θ\theta = angle between force and displacement vectors

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

Key Points

  1. Force must cause displacement - No displacement = no work
  2. Only the component of force parallel to displacement does work
  3. 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>0W > 0)

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

Negative Work (W<0W < 0)

  • Force has component opposite to motion (90°<θ180°90° < \theta \leq 180°)
  • Energy is removed from the system
  • Example: Friction on a sliding box (friction opposes motion)

Zero Work (W=0W = 0)

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

Special Cases

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

W=Fdcos(0°)=FdW = Fd\cos(0°) = Fd

Maximum positive work.

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

W=Fdcos(180°)=FdW = Fd\cos(180°) = -Fd

Maximum negative work.

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

W=Fdcos(90°)=0W = Fd\cos(90°) = 0

No work done.


Kinetic Energy

Kinetic energy is energy of motion:

KE=12mv2KE = \frac{1}{2}mv^2

where:

  • KEKE = kinetic energy (J)
  • mm = mass (kg)
  • vv = speed (m/s)

Properties of Kinetic Energy

  1. Always positive (or zero) - v2v^2 is always positive
  2. Scalar quantity - no direction
  3. Depends on speed squared - Double the speed = 4× the kinetic energy
  4. Frame-dependent - KE depends on reference frame

The Work-Energy Theorem

The work-energy theorem connects work and kinetic energy:

Wnet=ΔKE=KEfKEiW_{net} = \Delta KE = KE_f - KE_i

Wnet=12mvf212mvi2W_{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 Wnet>0W_{net} > 0: object speeds up (gains KE)
  • If Wnet<0W_{net} < 0: object slows down (loses KE)
  • If Wnet=0W_{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

Wnet=W1+W2+W3+...W_{net} = W_1 + W_2 + W_3 + ...

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

Method 2: Find net force first

Wnet=FnetdcosθW_{net} = F_{net} \cdot d \cdot \cos\theta

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

Both methods give the same result!


Variable Forces and Work

For a variable force (force changes during motion):

W=xixfF(x)dxW = \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=FdW = Fd always ✅ Right: W=FdcosθW = 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

  1. Identify all forces acting on the object
  2. Draw a free body diagram
  3. For each force, determine:
    • Magnitude FF
    • Angle θ\theta relative to displacement
    • Sign of work (positive/negative/zero)
  4. Calculate work for each force: W=FdcosθW = Fd\cos\theta
  5. Sum to find net work
  6. 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: Wfriction=12mv2W_{friction} = -\frac{1}{2}mv^2

Doubling speed requires 4 times the stopping distance!

Accelerating Vehicles

Engine does positive work to increase kinetic energy: Wengine=12m(vf2vi2)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=FdcosθW = Fd\cos\theta | J (joules) | | Kinetic Energy | KE=12mv2KE = \frac{1}{2}mv^2 | J | | Work-Energy Theorem | Wnet=ΔKEW_{net} = \Delta KE | J | | Power (rate of work) | P=WtP = \frac{W}{t} | W (watts) |

📚 Practice Problems

1Problem 1easy

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

Find: Final speed vfv_f


Step 1: Calculate work done by applied force

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

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


Step 2: Apply work-energy theorem

Wnet=ΔKEW_{net} = \Delta KE

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

200=12mvf212mvi2200 = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

200=12(5)vf20200 = \frac{1}{2}(5)v_f^2 - 0

200=2.5vf2200 = 2.5v_f^2

vf2=80v_f^2 = 80

vf=80=458.94 m/sv_f = \sqrt{80} = 4\sqrt{5} \approx 8.94 \text{ m/s}


Answer: The final speed is approximately 8.94 m/s or 454\sqrt{5} m/s.

2Problem 2medium

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

Find: Work done by gravity


Step 1: Identify the gravitational force

Fg=mg=(2)(9.8)=19.6 NF_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: θ=90°30°=60°\theta = 90° - 30° = 60°

Step 3: Calculate work

Wg=FgdcosθW_g = F_g d \cos\theta

Wg=(19.6)(5)cos(60°)W_g = (19.6)(5)\cos(60°)

Wg=(19.6)(5)(0.5)W_g = (19.6)(5)(0.5)

Wg=49 JW_g = 49 \text{ J}


Alternative Method: Use vertical displacement

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

Work by gravity = mgh=(2)(9.8)(2.5)=49mgh = (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!

3Problem 3hard

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

(a) Find work done by friction


Step 1: Calculate initial kinetic energy

KEi=12mvi2=12(1200)(25)2KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(1200)(25)^2

KEi=600(625)=375,000 JKE_i = 600(625) = 375,000 \text{ J}


Step 2: Calculate final kinetic energy

KEf=12mvf2=12(1200)(0)2=0 JKE_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(1200)(0)^2 = 0 \text{ J}


Step 3: Apply work-energy theorem

Wfriction=ΔKE=KEfKEiW_{friction} = \Delta KE = KE_f - KE_i

Wfriction=0375,000=375,000 JW_{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=FdcosθW = Fd\cos\theta

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

375,000=F(50)cos(180°)-375,000 = F(50)\cos(180°)

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

375,000=50F-375,000 = -50F

F=7,500 NF = 7,500 \text{ N}


Alternative approach:

W=Fd|W| = Fd

375,000=F(50)375,000 = F(50)

F=7,500 NF = 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.