3.3 Work, energy and power — Practise

Work through these on your own. Start each energy question by naming the store that loses energy and the store that gains it.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A force of 120 N moves a trolley 6.0 m along the direction of the force. Find the work done.
A force of 300 N acts at 25° to the horizontal and moves a crate 10 m along the ground. Find the work done by the force.
A person carries a box along a level floor at a constant height. Explain why the person does no work on the box against gravity.
A car of mass 1200 kg travels at $15\,\text{m}\,\text{s}^{-1}$ . Find its kinetic energy.
A mass of 5.0 kg is lifted through 8.0 m. Find the gravitational potential energy gained. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
An object is dropped from rest through 20 m. Use conservation of energy to find its speed just before it lands. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A car of mass 800 kg speeds up from $5.0\,\text{m}\,\text{s}^{-1}$ to $15\,\text{m}\,\text{s}^{-1}$ . Find the work done by the resultant force.
A machine does 3000 J of useful work in 6.0 s. Find its useful power.
A car travels at a constant $12\,\text{m}\,\text{s}^{-1}$ against a resistive force of 500 N. Find the useful power developed by the engine.
A block slides down a slope. It loses 50 J of gravitational potential energy and gains 30 J of kinetic energy. Find the energy transferred to thermal stores.
A motor is supplied with 5000 J of energy and uses it to raise a load of mass 40 kg through 10 m. Find the useful energy transferred and the efficiency. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A ball is dropped from a height of 3.0 m and you must find its speed at the ground. Say which method you would use and why, and name the assumption that makes the answer simple.

Answers and concise solutions

1. The work done

Answer: The force is along the displacement, so $W = Fs = 120 \times 6.0 = 720\,\text{J}$ .

2. The work done

Answer: Only the horizontal component does work: $W = Fs\cos 25^\circ = 300 \times 10 \times 0.906 = 2.7 \times 10^3\,\text{J}$ .

3. Why the work against gravity is zero

The upward force on the box is perpendicular to the horizontal displacement, and cos 90 degrees is zero.
The box does not change height, so its gravitational potential energy does not change and no work is done against gravity.

Answer: The lifting force is perpendicular to the horizontal displacement, so it does no work. The height does not change, so the gravitational potential energy does not change either.

4. The kinetic energy

Answer: $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 1200 \times 15^2 = 1.35 \times 10^5\,\text{J} = 135\,\text{kJ}$ .

5. The gravitational potential energy gained

Answer: $\Delta E_p = mg\Delta h = 5.0 \times 9.81 \times 8.0 = 392\,\text{J}$ .

6. The speed just before landing

Answer: $mg\Delta h = \tfrac{1}{2}mv^2$ , so $v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.81 \times 20} = 19.8\,\text{m}\,\text{s}^{-1}$ . The mass cancels, so the speed does not depend on it.

7. The work done

Answer: The work done equals the gain in kinetic energy: $W = \tfrac{1}{2} \times 800 \times (15^2 - 5.0^2) = 400 \times 200 = 80000\,\text{J} = 80\,\text{kJ}$ .

8. The useful power

Answer: $P = \dfrac{W}{t} = \dfrac{3000}{6.0} = 500\,\text{W}$ .

9. The useful power

Answer: At a constant speed the driving force balances the resistance, so $P = Fv = 500 \times 12 = 6000\,\text{W} = 6.0\,\text{kW}$ .

10. The thermal energy transferred

Answer: Energy is conserved, so the missing energy has been transferred to thermal stores: $50 - 30 = 20\,\text{J}$ .

11. The useful energy and the efficiency

Answer: The useful energy is $mg\Delta h = 40 \times 9.81 \times 10 = 3924\,\text{J}$ . The efficiency is $\dfrac{3924}{5000} = 0.78$ , or 78%. The rest is transferred to thermal stores by friction in the motor.

12. The method and its assumption

Use conservation of energy: the potential energy lost equals the kinetic energy gained, so v = sqrt(2gh).
Assume air resistance is negligible, so no energy is transferred to thermal stores.
The energy method is shorter than the equations of motion here, because it does not need the time and the mass cancels.

Answer: Use conservation of energy, $mg\Delta h = \tfrac{1}{2}mv^2$ , giving $v = \sqrt{2gh}$ , and assume air resistance is negligible.