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June 2023 p43 q4
3429
A lorry of mass 15,000 kg moves on a straight horizontal road in the direction from A to B. It passes A and B with speeds 20 m/s and 25 m/s respectively. The power of the lorry’s engine is constant and there is a constant resistance to motion of magnitude 6000 N. The acceleration of the lorry at B is 0.5 times the acceleration of the lorry at A.
(a) Show that the power of the lorry’s engine is 200 kW, and hence find the acceleration of the lorry when it is travelling at 20 m/s.
The lorry begins to ascend a straight hill inclined at 1° to the horizontal. It is given that the power of the lorry’s engine and the resistance force do not change.
(b) Find the steady speed up the hill that the lorry could maintain.
Solution
(a) To find the power of the lorry's engine, use the formula for power: \(P = Fv\). At point A, \(P = 20F\) and at point B, \(P = 25F\). Given the resistance force is 6000 N, the net force at A is \(F - 6000\) and at B is \(F - 6000\).
Using Newton's second law, at A: \(\frac{P}{20} - 6000 = 15000a\) and at B: \(\frac{P}{25} - 6000 = 15000 \times 0.5a\).
Solving these equations simultaneously, we find \(P = 200,000 \text{ W} = 200 \text{ kW}\) and \(a = \frac{4}{15} \text{ m/s}^2\).
(b) For the lorry ascending the hill, the equation becomes \(\frac{200,000}{v} - 6000 - 15000g \sin(1^{\circ}) = 0\). Solving for \(v\), we find the steady speed \(v = 23.2 \text{ m/s}\).