Showing posts with label Work Energy and Power. Show all posts
Showing posts with label Work Energy and Power. Show all posts

Monday, November 12, 2018

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

Sunday, November 11, 2018

9702/Oct Nov/12/2017/Q22

When sound travels through air, the air particles vibrate. A graph of displacement against time for
a single air particle is shown.


Which graph best shows how the kinetic energy of the air particle varies with time?


Solution:
Answer: D

Kinetic energy = ½ mv2

A graph of displacement against time for a single air particle is shown.  The gradient of the displacement-time graph gives the velocity of the air particle at that point in time. This is done by calculating the gradient of the tangent at that point.

The gradient (and hence, velocity) is found to be zero at the maximum displacement (the tangent is horizontal) and maximum when the displacement is zero (the tangent is steepest).

Thus, at time = 0, T and 2T the velocity is zero and hence kinetic energy is zero. [A and C incorrect]

But, between time = 0 and T or between time = T and 2T the displacement is zero (in case) twice. So, the velocity (and kinetic energy) reaches its maximum value 2 times in each of the 2 intervals.[B is incorrect]

Reference: PYQ - Oct/Nov 2017 Paper 12 Q22

Tuesday, November 6, 2018

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 1st 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 2nd 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