IB — physicsMedium3 min read

Work, Energy, and Power

W = Fs cos θ; KE, GPE, EPE; conservation of energy; P = Fv; efficiency

Tutor AI Team2026-02-25T05:34:12.852Z

# Work, Energy, and Power — IB Physics

The concepts of work, energy, and power provide a powerful alternative to Newton's laws for solving mechanics problems, particularly where forces vary or where we don't need to know about individual forces.

1. Work Done

$\boxed{W = Fs\cos\theta}$

$W$ = work (J)
$F$ = force (N)
$s$ = displacement (m)
$\theta$ = angle between $F$ and $s$

Work is a scalar. 1 J = 1 N·m.

If $\theta = 90°$ : $W = 0$ (force perpendicular to motion).

2. Kinetic Energy

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

Work-Energy Theorem

$W_{\text{net}} = \Delta KE$

The net work done on an object equals its change in kinetic energy.

3. Gravitational Potential Energy

$\Delta GPE = mg\Delta h$

4. Elastic Potential Energy

$EPE = \frac{1}{2}ke^2$

Where $k$ = spring constant, $e$ = extension/compression.

5. Conservation of Energy

Total energy is conserved in an isolated system.

With only conservative forces: $KE_i + PE_i = KE_f + PE_f$

With friction/drag: $KE_i + PE_i = KE_f + PE_f + W_{\text{friction}}$

6. Power

$P = \frac{W}{t} = \frac{\Delta E}{t}$

For constant force and velocity: $P = Fv$

Units: watts (W).

7. Efficiency

$\eta = \frac{\text{useful output energy/power}}{\text{total input energy/power}} \times 100\%$

Worked Example: Example 1

Problem

A 60 kg skier descends 200 m vertically. Starting from rest, find speed at bottom if friction does 30 kJ of work.

$mgh = \frac{1}{2}mv^2 + W_f$ $60(9.81)(200) = \frac{1}{2}(60)v^2 + 30000$ $117720 = 30v^2 + 30000$ $v = \sqrt{87720/30} = \sqrt{2924} = 54.1$ m/s

Solution

Worked Example: Example 2

Problem

A 1500 kg car at 30 m/s: engine power 50 kW, drag force 1200 N.

Driving force: $F = P/v = 50000/30 = 1667$ N Net force: $1667 - 1200 = 467$ N Acceleration: $a = 467/1500 = 0.311$ m/s²

Max speed (when $F_{\text{drive}} = F_{\text{drag}}$ ): $v_{\max} = P/F_{\text{drag}} = 50000/1200 = 41.7$ m/s

Solution

Worked Example: Example 3

Problem

A motor lifts a 500 kg load 20 m in 30 s. Efficiency is 75%. Find input power.

Useful output: $W = mgh = 500(9.81)(20) = 98100$ J Output power: $P_{\text{out}} = 98100/30 = 3270$ W Input power: $P_{\text{in}} = 3270/0.75 = 4360$ W

Solution

9. Practice Questions

1. A 50 N force pulls a box 10 m at 30° to the horizontal. Calculate the work done. (2 marks)
1. A 2 kg mass falls 5 m from rest. Find its speed just before hitting the ground (ignore air resistance). (2 marks)
1. A pump raises 200 kg of water per minute through 15 m. Calculate the minimum power. (3 marks)
1. A car engine has 40% efficiency and burns fuel at 0.004 kg/s (fuel energy = 45 MJ/kg). Find the useful power. (3 marks)
Answers
1. $W = 50(10)\cos30° = 433$ J.

Want to check your answers and get step-by-step solutions?

Summary

$W = Fs\cos\theta$ ; work-energy theorem: $W_{\text{net}} = \Delta KE$
Conservation: $KE_i + PE_i = KE_f + PE_f + W_{\text{friction}}$
$P = W/t = Fv$
Efficiency = useful output / total input × 100%

Tags:

#ib#physics#mechanics#work#energy#power#conservation

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# Work, Energy, and Power — IB Physics

1. Work Done

2. Kinetic Energy

Work-Energy Theorem

3. Gravitational Potential Energy

4. Elastic Potential Energy

5. Conservation of Energy

6. Power

7. Efficiency

Worked Example: Example 1

Worked Example: Example 2

Worked Example: Example 3

9. Practice Questions

Answers

Summary

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