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 f is the acceleration due to gravity, the work required to pull the hanging part on the table is
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The work required to pull the hanging part on the table is MgL / 18.
Explanation:
Weight of chain of length "x" =(M/Lx)g
- Amount of work done in moving dx to the top
dW = F→. dx→
=F.dx = (MLx)gdx
- The total amount of work done in moving the one third length of the hanging chain on the table will be
W=∫L/3 - 0 MLxgdx = M / Lg
∫L / 3 - 0 xdx
= M/ Lg[x^2 / 2] L/3 - 0
=MgL / 18
Hence the work required to pull the hanging part on the table is MgL / 18.
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Calculate the work done by a person in lifting a load of 20 kg from the ground and placing it on a 1 m high table ?
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