2.3 Work done and energy transfer — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

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

A force of 60 N pushes a crate 4.0 m along the line of the force. Calculate the work done on the crate.
A force of 35 N pulls a sledge 12 m along the line of the force. Calculate the work done on the sledge.
A machine does 900 J of work as it moves a load 12 m along the line of the force. Calculate the force.
A force of 90 N does 720 J of work. Calculate the distance moved along the line of the force.
A box of mass 8.0 kg is lifted 3.0 m onto a shelf. Take $g = 9.8\,\text{N/kg}$ . Calculate the work done against gravity. Use the force needed to lift the box.
A friction force of 45 N acts on a sledge that is pulled 22 m. Calculate the work done against friction.
A car of mass 1200 kg travels at 20 m/s. Calculate the energy in its kinetic store. This is the work done by the brakes when the car stops.
The brakes of the car transfer $2.4 \times 10^5\,\text{J}$ of energy over a braking distance of 50 m. Calculate the average braking force.
A pulley does 350 N m of work. Convert this value to joules.
A motor does 3000 J of work in 20 s. Calculate the power of the motor.
A bag of mass 25 kg is lifted 1.2 m onto a wall. Take $g = 9.8\,\text{N/kg}$ . Say which force does the work, then calculate the work done.
A box slides across a floor and stops. Describe the energy transfer that happens as the box slows, and name the store that receives the energy.

Answers and concise solutions

1. Work done on the crate

Answer: $W = Fs = 60 \times 4.0 = 240\,\text{J}$ .

2. Work done on the sledge

Answer: $W = Fs = 35 \times 12 = 420\,\text{J}$ .

3. Force exerted by the machine

Answer: $F = \dfrac{W}{s} = \dfrac{900}{12} = 75\,\text{N}$ .

4. Distance moved

Answer: $s = \dfrac{W}{F} = \dfrac{720}{90} = 8.0\,\text{m}$ .

5. Work done against gravity

Answer: The lifting force equals the weight: $F = mg = 8.0 \times 9.8 = 78.4\,\text{N}$ . Then $W = Fs = 78.4 \times 3.0 = 235.2\,\text{J}$ , so $240\,\text{J}$ to 2 significant figures.

6. Work done against friction

Answer: $W = Fs = 45 \times 22 = 990\,\text{J}$ . This energy is transferred to the thermal store of the sledge and the snow.

7. Work done by the brakes

Answer: $E_k = \tfrac{1}{2}mv^2 = 0.5 \times 1200 \times 20^2 = 240000\,\text{J} = 2.4 \times 10^5\,\text{J}$ .

8. Average braking force

Answer: $F = \dfrac{W}{s} = \dfrac{240000}{50} = 4800\,\text{N} = 4.8 \times 10^3\,\text{N}$ .

9. Work done in joules

Answer: One newton-metre is one joule, so $350\,\text{N m} = 350\,\text{J}$ .

10. Power of the motor

Answer: $P = \dfrac{W}{t} = \dfrac{3000}{20} = 150\,\text{W}$ .

11. Work done lifting the bag

Answer: The lifting force equals the weight, $F = mg = 25 \times 9.8 = 245\,\text{N}$ . Then $W = Fs = 245 \times 1.2 = 294\,\text{J}$ , so $290\,\text{J}$ to 2 significant figures.

12. The energy transfer

The kinetic store of the box decreases as the box slows down.
The friction force does work on the box as it moves, and the work done is $W = Fs$ .
The energy is transferred to the thermal store of the floor and the box, so their temperature rises.