Question

A uniform chain of mass m and length l is kept on a horizontal table with half of its length hanging from the edge of the table work done in pulling a chain on the table so that only one by 5 of its length now hangs from the edge

Answer 1

Given:  Uniform chain of mass m and length l.

To find: work done in pulling a chain on the table

Solution:

• Now let us assume that the potential energy at any point in this line is 0.

• As we have given that this is an uniform chain, so we can conclude that it's mass is proportional to it's length.

• If mass of given length 'l' is 'm', then mass of length (l/2) is (m/2) respectively.
• As, innitial length l/2 is below the reference line, so we can say that:

                  v(innitial) = 0 - mgh

                                  = - (m/2) x g x (l/2)

                                  = - (mgl/4)

• Now we have given that the l/5 length of the uniform chain is below the reference line.
• So:

                   v(final) = 0 - mgh

                               = - (m/5) x g x (l/5)

                               = - (mgl/25)

• So now, we can find the work done:

                  W = v (final) - v (innitial)

                      = 21 mgl/10

Answer:

                Work done in pulling a chain on the table is  21 mgl/10