Power from force and speed

5 A cyclist is riding along a straight horizontal road. The total mass of the cyclist and her bicycle is 70 kg . At an instant when the cyclist's speed is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), her acceleration is \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). There is a constant resistance to motion of magnitude 30 N .

1. Find the power developed by the cyclist.
   The cyclist comes to the top of a hill inclined at \(5 ^ { \circ }\) to the horizontal. The cyclist stops pedalling and freewheels down the hill (so that the cyclist is no longer supplying any power). The magnitude of the resistance force remains at 30 N . Over a distance of \(d \mathrm {~m}\), the speed of the cyclist increases from \(6 \mathrm {~ms} ^ { - 1 }\) to \(12 \mathrm {~ms} ^ { - 1 }\).
2. Find the change in kinetic energy.
3. Use an energy method to find \(d\). \includegraphics[max width=\textwidth, alt={}, center]{4e555003-16f1-4453-ab25-c50929d4b5b3-10_725_785_260_680} Two particles \(P\) and \(Q\), of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley at \(B\) which is attached to two inclined planes. \(P\) lies on a smooth plane \(A B\) which is inclined at \(60 ^ { \circ }\) to the horizontal. \(Q\) lies on a plane \(B C\) which is inclined at \(30 ^ { \circ }\) to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes (see diagram).

4 A car has mass 1600 kg .

1. The car is moving along a straight horizontal road at a constant speed of \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is subject to a constant resistance of magnitude 480 N . Find, in kW , the rate at which the engine of the car is working.
   The car now moves down a hill inclined at an angle of \(\theta\) to the horizontal, where \(\sin \theta = 0.09\). The engine of the car is working at a constant rate of 12 kW . The speed of the car is \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at the top of the hill. Ten seconds later the car has travelled 280 m down the hill and has speed \(32 \mathrm {~ms} ^ { - 1 }\).
2. Given that the resistance is not constant, use an energy method to find the total work done against the resistance during the ten seconds.

A block is pulled along a horizontal floor by a horizontal rope. The tension in the rope is 500 N and the block moves at a constant speed of \(2.75 \text{ m s}^{-1}\). Find the work done by the tension in 40 s and find the power applied by the tension. [4]

A lorry of mass 25 000 kg travels along a straight horizontal road. There is a constant force of 3000 N resisting the motion.

1. Find the power required to maintain a constant speed of 30 m s\(^{-1}\). [2]

The lorry comes to a straight hill inclined at 2° to the horizontal. The driver switches off the engine of the lorry at the point \(A\) which is at the foot of the hill. Point \(B\) is further up the hill. The speeds of the lorry at \(A\) and \(B\) are 30 m s\(^{-1}\) and 25 m s\(^{-1}\) respectively. The resistance force is still 3000 N.

1. Use an energy method to find the height of \(B\) above the level of \(A\). [5]

A particle \(P\) of mass \(3\) kg is moving on a smooth horizontal surface under the influence of a variable horizontal force \(\mathbf{F}\) N. At time \(t\) seconds, where \(t \geqslant 0\), the velocity of \(P\), \(\mathbf{v}\) m s\(^{-1}\), is given by $$\mathbf{v} = (32\sinh(2t))\mathbf{i} + (32\cosh(2t) - 257)\mathbf{j}.$$

1. 1. By considering kinetic energy, determine the work done by \(\mathbf{F}\) over the interval \(0 \leqslant t \leqslant \ln 2\). [5]
   2. Explain the significance of the sign of the answer to part (a)(i). [1]
2. Determine the rate at which \(\mathbf{F}\) is working at the instant when \(P\) is moving parallel to the \(\mathbf{i}\)-direction. [6]