Solveeit Logo
Login

Question

Physics Question on work

A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, the work required to pull the hanging part on to the table is:

A

MgL18\frac{MgL}{18}

B

MgL9\frac{MgL}{9}

C

MgL3\frac{MgL}{3}

D

MgLMgL

Answer

MgL18\frac{MgL}{18}

Explanation

Solution

There will be the change in position of centre of gravity, which was L/6L\text{/}6 down now is shifted to the top of table hence, Work done =M3×g×L6=MgL18=\frac{M}{3}\times g\times \frac{L}{6}=\frac{MgL}{18}