A-level · Work, energy & power

PowerExam-style questions

Power as the rate of doing work, average power = work ÷ time, and the key working relationship P = Fv for a driving force — used to find engine power at constant speed and, by combining P = Fv with Newton's second law, the instantaneous acceleration of a vehicle on a level road or a hill.

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22 questions · A Level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 1 Level 2 Level 3 Level 4 Level 5

1
A tractor of mass $3500$ kg is used on a farm.

The tractor is moving along a straight horizontal track with its engine working at a constant rate of $24$ kW. The resistance to the motion of the tractor is modelled as a constant force of magnitude $R$ newtons. When the tractor is moving at $6\ \text{m s}^{-1}$ , its acceleration is $0.8\ \text{m s}^{-2}$
(a) Using the model, show that $R = 1200$

(4 marks)

The tractor is then driven along a different track, from a point $A$ in the farmyard to a point $B$ at the gate of a field. The track from $A$ to $B$ is $450$ m long and $B$ is $18$ m higher than $A$ . The tractor passes $A$ with speed $2\ \text{m s}^{-1}$ and passes $B$ with speed $5\ \text{m s}^{-1}$

The resistance to the motion of the tractor from non-gravitational forces is modelled as a constant force of magnitude $1200$ N.
(b) Find the total work done by the engine of the tractor as the tractor moves from $A$ to $B$ .

(5 marks)

Later, the tractor climbs a straight hill which is inclined to the horizontal at an angle $\beta$ , where $\sin\beta = \frac{1}{12}$ . The engine is working at a constant rate of $15$ kW and the tractor moves up the hill at a constant speed of $V\ \text{m s}^{-1}$ . The resistance to the motion of the tractor from non-gravitational forces is again modelled as a constant force of magnitude $1200$ N.
(c) Find the value of $V$ .

(4 marks)
Level 5 · 13 marksTry it
2
A lorry of mass $6000$ kg is moving up a straight hill which is inclined to the horizontal at an angle $\alpha$ , where $\sin\alpha = \frac{1}{35}$ . The engine of the lorry is working at a constant rate of $36$ kW and the lorry is moving at a constant speed of $15\ \text{m s}^{-1}$ . The resistance to the motion of the lorry from non-gravitational forces is modelled as a constant force of magnitude $R$ newtons.
(a) Show that $R = 720$

(4 marks)

On another day, the lorry is towing a trailer of mass $2000$ kg along a straight horizontal road, as shown in Figure 1.

Figure 1Diagram not accurately drawn.

The trailer is attached to the lorry by a towbar which is parallel to the road. The towbar is modelled as a light rod. The resistance to the motion of the lorry is $720$ N and the resistance to the motion of the trailer is $400$ N. The engine of the lorry is again working at a constant rate of $36$ kW.
(b) Find the acceleration of the lorry at the instant when its speed is $10\ \text{m s}^{-1}$

(4 marks)
(c) Find the tension in the towbar at this instant.

(3 marks)

Later, the lorry tows the trailer up the hill described in part (a). At the instant when the lorry and the trailer are moving up the hill at $9\ \text{m s}^{-1}$ , the towbar breaks. The resistance to the motion of the trailer from non-gravitational forces is still $400$ N.
(d) Use the work–energy principle to find how far the trailer travels up the hill between the towbar breaking and the trailer coming to instantaneous rest.

(4 marks)
Level 5 · 15 marksTry it
3
A lorry of mass $5000\text{ kg}$ moves along a level road. Its engine works at a constant rate of $60\text{ kW}$ , and the total resistance to motion is constant at $1500\text{ N}$ .
(a) Find the maximum speed of the lorry on the level road. (3 marks)
(b) Find the acceleration of the lorry at the instant when its speed is $20\text{ m s}^{-1}$ . (3 marks)
Level 5 · 6 marksTry it
4
A car of mass $800\text{ kg}$ descends a hill inclined at angle $\theta$ , where $\sin\theta = \dfrac{1}{40}$ , with its engine working at $12\text{ kW}$ . The non-gravitational resistance is a constant $1000\text{ N}$ . Take $g = 10\text{ m s}^{-2}$ .

Find the car's maximum speed on the descent. (5 marks)
Level 5 · 5 marksTry it