Two Multiple Choice Questions on Work and Energy

The following MCQ which appeared in KEAM (Engineering) 2007 question paper is worth noting:

A particle is released from a height S. At a certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

(a) S/4, 3gS/2 (b) S/4, [√(3gS)]/2 (c) S/2, [√(3gS)]/2

(d) S/4, √(3gS/2) (e) S/3, √(3gS/2)

Suppose the given conditions occur at the instant when the particle is at height ‘h’.

Potential energy at the height h = mgh.

Since the initial energy is completely potential and is equal to mgS, the kinetic energy at the height ‘h’ is equal to the loss of potential energy and is equal to mg(S–h).

Since the K.E. is three times the P.E., we have

mg(S–h) = 3mgh, from which h = S/4.

But, kinetic energy = ½ mv².

Therefore, ½ mv² = mg(S–h) = mg[S– (S/4)] = 3 mgS/4 so that

v =√(3gS/2)

Now, consider the following MCQ which is simple. But simple qiestions may cheat you some times. Here is the question:

A uniform chain of mass M and length L is placed on a table with 20% of its length overhanging from the edge of the table. The minimum work that has to be done to place the hanging portion on the table is

(a) MgL/5 (b) MgL/10 (c) MgL/20

(d) MgL/25 (e) MgL/50

The hanging portion has mass M/5 and therefore weight Mg/5. The length of the hanging portion is L/5 and hence the centre of gravity of the hanging portion is at a depth L/10 from the surface of the table. The entire weight of the hanging portion can be imagined to act through its centre of gravity. When you pull the hanging portion up to keep it on the table, you are moving the point of application of the force (equal to the magnitude of the the weight Mg/5) through a distance L/10 along the direction of the force.

Therefore, the work done is (Mg/5)×(L/10) = MgL/50