9702/Oct Nov/13/2013/Q19

An electrical generator is started at time zero. The total electrical energy generated during the
first 5 seconds is shown in the graph.

What is the maximum electrical power generated at any instant during these first 5 seconds?

A 10 W
B 13 W
C 30 W
D 50 W

Solution:

Answer: C

Power = Energy / time

From the energy-time graph, the power generated is given by the gradient. Maximum electrical power generated at any instant is given by the gradient at that instant.

The steeper the graph, the greater the value of gradient and thus the greater the power generated. The graph is steepest between times t = 2s and t = 3s.

Consider the points: (2, 10) and (3, 40)

Maximum power = gradient = (40 – 10) / (3 – 2) = 30 W

Reference: PYQ - Oct/Nov 2013 Paper 13 Q19

9702/May Jun/13/2015/Q19

When a horizontal force F is applied to a frictionless trolley over a distance s, the kinetic energy
of the trolley changes from 4.0 J to 8.0 J.

If a force of 2F is applied to the trolley over a distance of 2s, what will the original kinetic energy
of 4.0 J become?

A 16 J
B 20 J
C 32 J
D 64 J

Solution:

Answer: B.

Initial kinetic energy of trolley = 4J

Consider the 1^st case.

We need to consider the initial 4J here.

Final kinetic energy = 8J

Work done by force F (= Fs) = 8 – 4 = 4J

So, Fs = 4J

Consider the 2^nd case.

Work done by force 2F = (2F) (2s) = 4(Fs) = 4(4) = 16J since Fs = 4J

Total kinetic energy = 4 + 16 = 20J

Reference: PYQ - May/Jun 2015 Paper 13 Q19