Question

A uniform chain of length L and mass M overhungs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is μ. The work done by the friction during the period, the chain slips off the table is

(a) -2/9 μMgL

(b) -6/7 μMgL

(c) -1/4 μMgL

(d) -4/9 μMgL

Answer 1

Correct option (a) -2/9 μMgL   

Explanation:

Total mass of the chain = M 

Total length of the chain = L

Linear mass density = M/L

dW=-F.dx

Let dW be the small work done by friction when chain of length l slides a small distance dl

Frictional force applied during this period

Length of the chain on the table = 2L/3

Now, total work done