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Nov 2018 p43 q6
3425
A van of mass 3200 kg travels along a horizontal road. The power of the van’s engine is constant and equal to 36 kW, and there is a constant resistance to motion acting on the van.
When the speed of the van is 20 m s-1, its acceleration is 0.2 m s-2. Find the resistance force.
When the van is travelling at 30 m s-1, it begins to ascend a hill inclined at 1.5° to the horizontal. The power is increased and the resistance force is still equal to the value found in part (i). Find the power required to maintain this speed of 30 m s-1.
The engine is now stopped, with the van still travelling at 30 m s-1, and the van decelerates to rest. Find the distance the van moves up the hill from the point at which the engine is stopped until it comes to rest.
Solution
(i) The driving force is given by the power equation: \(Fv = 36000\). Therefore, \(F = \frac{36000}{20} = 1800 \text{ N}\). Using Newton's Second Law: \(1800 - R = 3200 \times 0.2\). Solving for \(R\), we get \(R = 1160 \text{ N}\).
(ii) The driving force required to maintain the speed on the incline is \(F = 3200g\sin(1.5^{\circ}) + 1160\). The power required is \(P = (3200g\sin(1.5^{\circ}) + 1160) \times 30\). Calculating gives \(P = 59900 \text{ W}\).
(iii) Using Newton's Second Law: \(- (3200g\sin(1.5^{\circ}) + 1160) = 3200a\). Solving for \(a\), we find \(a = -0.62426 \text{ m/s}^2\). Using the equation \(0^2 = 30^2 + 2as\), solve for \(s\) to get \(s = 721 \text{ m}\).