## Thursday, December 31, 2009

### Two Questions (MCQ) involving Kinetic Energy and Potential Energy

The following question appeared in Kerala Engineering Entrance 2007 question paper:

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and the length of the pendulum, the gain in kinetic energy at B is

(a) mgl/2

(b) mgl/√2

(c) (mgl√3)/2

(d) 2mgl/√3

(e) mgl From position A, the bob of the pendulum has fallen through a distance l cos 30º (fig.). Therefore, the loss of potential energy is mg l cos 30º = (mgl√3)/2.

Therefore, the gain in kinetic energy by the bob (by the law of conservation of energy) is (mgl√3)/2.

Here is another question which will be useful in the present context:

The bob of a simple pendulum has mass 0.2 kg. The bob is drawn aside so that the string is horizontal and is released. When it swings through 60º its kinetic energy is 0.5 J. The kinetic energy when the sting becomes vertical will be

(a) √3 J

(b) 1/√3 J

(c) √3/2 J

(d) 2/√3 J

(e) 1 J

You can use the figure used with the previous question for working out the present question as well since the angle turned is 60º. The loss of potential energy by the bob on swinging through 60º is mg l cos30º itself. [Or, you may take it as mg l sin60º].

Since the loss of potential energy is equal to the gain of kinetic energy, we have

mg l cos30º = 0.5 J

Or, (mgl√3)/2 = 0.5 J.

The kinetic energy when the string becomes vertical will be mgl since the bob falls through a distance l and loses its entire initial potential energy mgl.

From the above equation mgl = 1/√3 J [Option (b)].

[Note that the mass of the bob given in the question just serves as a distraction].

You will find some useful questions in this section here.