First slide

Potential energy

Question

A uniform chain of length I 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 acceleration due to gravity, the work required to pull the hanging part on to the table is

Moderate

A

M g L
B

M g L / 3
C

M g L / 9
D

M g L / 18

Solution

The situation is shown in fig. (9).
Given that the length of hanging part is L/3, so its mass would be M/3.Hence the weight (Mg/3) will act at the center of gravity G of hanging part. It will act at a distance (L/6) ) below the surface of table. Work done in pulling the hanging part on the table = increase in P.E.
$= m \times g \times h = (\frac{M}{3}) \times g \times \frac{L}{6} = \frac{mgL}{18}$

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