Question

A uniform cable of mass m and length l is placed on a horizontal surface such that its th n 1 part is hanging below the edge of the surface. to lift the hanging part of the cable upto the surface, the work done should be -

Answer 1

The work done to lift the hanging part of the cable up to the surface is MgL/2n²

Given-

• Mass of uniform cable = M
• Length of cable = L
• Length of cable hanging below the edge of the surface is = 1/n th part
• Mass of the hanging part = M/n

Center of mass of h = L/2n

So work done should be-

W = mgh = (M/n) g (L/2n)

W = MgL/2n²

Refer to the attached figure for more details.