A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from.......?

Question

A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

1. 7.2 J

2. 1200 J

3. 120 J

4. 3.6 J

Answer 1

Mass of 2m = 4 kg

Mass of 60 cm i.e. 0.6 m = (4/2) * 0.6 kg = 1.2 kg

When the chain is pulled up then the gravitational potential energy of the part of the chain, which is already on the table does not change.

But the gravitational potential energy of the part initially hanging changes.

Mass of the part hanging m = 1.2 kg

Initially its centre of gravity is at its mid point i.e. 0.6/2 = 0.3 m

So, when it is pulled up, then the centre of gravity moves up by height h = 0.3 m

Work done = change in gravitational potential energy = mgh

= 1.2 * 10 * 0.3

= 3.6 J

Ans: (4)

I took g as 10 m/s^2. On taking 9.8 m/s^2, the answer comes as 3.52 J. It looks like the question is only expecting you to find approximate value.

I have assumed that there is no friction.

BOND,

You assumed that the tension is constant. But that is wrong. When any part of the chain is pulled up, then there is smaller part which is hanging. Therefore, the mass of the part hanging becomes smaller. This makes tension smaller.

Tension changes. Therefore, force required changes. Therefore, you should not multiply initial tension with total distance to find the work. But this is what you have done.

I have used work energy theorem to find the answer.

If you want to use tension, then the following method should be used.

Let α = linear density

Then α = mass/length = 4 kg/2m = 2 kg/m

When length x is hanging, then the mass of hanging part = αx

Therefore, tension = αxg

Or force required = αxg

Work done in pulling by an infinitesimal height dx is

dW = αxgdx

Work required in pulling the entire chain W = integral of dW

Or, W = integral of αxgdx

Or, W = integral of αg * xdx

Or, W = αg * x^2/2

The integral is to be taken from 0 to the length of the hanging part (i.e. 0.6 m)

So,

W = αg * 0.6^2/2 = 2*10*0.36/2 = 3.6 J

As you can see, from this method also, you get 3.6 J

But I find work energy theorem as simpler approach.

Answer 2

l and m are length and mass of part of chain hanging down .

M and L are total lenght and mass of the chain.

M = 4 kg

L = 2 m

l = 0.6 m

m =  l / L * 4 = (0.6 / 2) * 4 = 1.2 kg

Tension in chain hanging down =  mg = 12 N    (g=10 m/s^2)

Force required = Tension

Displacement = 0.6 m        (part of chain hanging comes up)

W= Force * Displacement = 12 * 0.6 = 7.2 J

WORK DONE = 7.2 J.

OPTION (A) IS CORRECT.