A.3 Work, energy and power

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Work done

W = Fs\cos\theta

Kinetic energy

E_{k} = \tfrac{1}{2}mv^{2} = \frac{p^{2}}{2m}

Gravitational PE (near surface)

\Delta E_{p} = mg\Delta h

Elastic potential energy

E_{H} = \tfrac{1}{2}k(\Delta x)^{2}

Power

P = \frac{\Delta W}{\Delta t} = Fv

Efficiency

\eta = \frac{E_{\text{output}}}{E_{\text{input}}} = \frac{P_{\text{output}}}{P_{\text{input}}}

Section A — Multiple choice

A1. A porter carries a heavy suitcase horizontally at constant velocity across a level floor. The work done on the suitcase by the supporting force is:

A Positive
B Negative
C Zero
D Cannot be determined

A2. A pendulum bob is released from a height $h$ above its lowest point. Ignoring air resistance, its speed at the lowest point is:

A $\sqrt{gh}$
B $\sqrt{2gh}$
C $2\sqrt{gh}$
D $gh$

A3. A motor rated at 1.5 kW raises a 60 kg load at a steady $2.0\ \text{m s}^{-1}$ . What is the efficiency of the system?

A 58%
B 66%
C 78%
D 92%

Section B — Short answer

B1. A 65 kg skier starts from rest and descends a slope of vertical height 40 m, reaching the bottom at $22\ \text{m s}^{-1}$ . Determine the energy dissipated by friction and air resistance. [3 marks]

B2. Explain why the work done in stretching a spring by $2x$ is four times, not twice, the work done in stretching it by $x$ . [2 marks]

B3. Sketch a Sankey diagram for a filament lamp that is 10% efficient, and state what each feature represents. [3 marks]

Answer key

For worked explanations, interactive practice and more free resources, visit www.newtonine.com

Section A

A1: Zero — The supporting force is vertical while the displacement is horizontal, so $\theta = 90°$ and $W = Fs\cos 90° = 0$ . Feeling tired is not the physics definition of doing work on the case.

A2: $\sqrt{2gh}$ — Energy conservation: $mgh = \frac{1}{2}mv^2$ , so $v = \sqrt{2gh}$ . Mass cancels — the answer is independent of the bob's mass, which is why energy methods are so quick here.

A3: 78% — Useful power $= Fv = mgv = 60 \times 9.81 \times 2.0 \approx 1177\ \text{W}$ . Efficiency $= 1177/1500 \approx 0.78$ , i.e. 78%. The other 22% is dissipated in the motor and transmission.

Section B

B1: Energy available: $mgh = 65 \times 9.81 \times 40 \approx 25.5\ \text{kJ}$ . Kinetic energy gained: $\frac{1}{2}(65)(22^2) \approx 15.7\ \text{kJ}$ . Energy dissipated $\approx 25.5 - 15.7 = 9.8\ \text{kJ}$ , transferred to the surroundings as thermal energy.

B2: The stored energy is $E_H = \frac{1}{2}k(\Delta x)^2$ , quadratic in extension. Doubling $\Delta x$ quadruples the energy because the force grows with extension — the average force over the second stretch is larger than over the first.

B3: A broad arrow enters representing 100% electrical input; a thin straight arrow (one-tenth the width) continues forward for the useful light output; a large arrow bends away (usually downwards) for the 90% dissipated as thermal energy. Arrow width is proportional to power.