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 ($W > 0$)

• 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

1. Always positive (or zero) - $v^2$ 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:

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

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

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

1. Identify all forces acting on the object
2. Draw a free body diagram
3. For each force, determine:
   • Magnitude $F$
   • Angle $\theta$ relative to displacement
   • Sign of work (positive/negative/zero)
4. Calculate work for each force: $W = Fd\cos\theta$
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: $W_{friction} = -\frac{1}{2}mv^2$

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

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.