A uniform chain having length L and mass M overhangs a horizontal table such that two-thirds of its length rests on the table. The friction coefficient between the table and the chain is . Determine the work done by friction during the period the chain slips off the table.

Step-by-step solution

Let x be the length of the chain on the table at any instant. The mass of the chain on the table is m = M L x . The normal reaction on the chain is N = mg = M L xg . The frictional force acting on the chain is f = N = M L xg . When the chain slips by a small distance ds , the length of the chain on the table changes by dx = -ds . The work done by friction during this small displacement is dW = -f ds = f dx . Total work done by friction as the chain slips completely off the table is obtained by integrating from x =