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A chain of mass M is placed on a smooth table with 1/n of its length L hanging over the edge.The work done in pulling the hanging portion of the chain back to the surface of the table is
- a)MgL/n
- b)MgL/2n
- c)MgL/n2
- d)MgL/2n2
Correct answer is option 'D'. Can you explain this answer?
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A chain of mass M is placed on a smooth table with 1/n of its length L...
W = change in PE of COM of hanging part
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A chain of mass M is placed on a smooth table with 1/n of its length L...
Given: Mass of chain = M, Length of chain = L, Portion of chain hanging over the edge = 1/n*L
To find: Work done in pulling the hanging portion of the chain back to the surface of the table
Assumptions: The table is smooth, i.e., there is no friction between the chain and the table.
Solution:
1. Finding the center of mass of the chain:
- The center of mass of the chain will be at a distance of 1/2*L from the end of the chain that is resting on the table.
- Therefore, the distance between the center of mass of the chain and the point where the hanging portion of the chain touches the table is (1/2 - 1/n)*L.
2. Finding the potential energy of the hanging portion of the chain:
- The potential energy of the hanging portion of the chain is given by mgh, where m is the mass of the hanging portion, g is the acceleration due to gravity, and h is the height to which the hanging portion is lifted.
- Here, m = M*(1/n), g = g, and h = (1/n)*L.
- Therefore, the potential energy of the hanging portion of the chain is (MgL/n^2)/2.
3. Finding the work done in pulling the hanging portion of the chain back to the surface of the table:
- The work done in pulling the hanging portion of the chain back to the surface of the table is equal to the change in potential energy of the hanging portion of the chain.
- Therefore, the work done is (MgL/n^2)/2.
Hence, the correct answer is option 'D' - MgL/2n^2.
To find: Work done in pulling the hanging portion of the chain back to the surface of the table
Assumptions: The table is smooth, i.e., there is no friction between the chain and the table.
Solution:
1. Finding the center of mass of the chain:
- The center of mass of the chain will be at a distance of 1/2*L from the end of the chain that is resting on the table.
- Therefore, the distance between the center of mass of the chain and the point where the hanging portion of the chain touches the table is (1/2 - 1/n)*L.
2. Finding the potential energy of the hanging portion of the chain:
- The potential energy of the hanging portion of the chain is given by mgh, where m is the mass of the hanging portion, g is the acceleration due to gravity, and h is the height to which the hanging portion is lifted.
- Here, m = M*(1/n), g = g, and h = (1/n)*L.
- Therefore, the potential energy of the hanging portion of the chain is (MgL/n^2)/2.
3. Finding the work done in pulling the hanging portion of the chain back to the surface of the table:
- The work done in pulling the hanging portion of the chain back to the surface of the table is equal to the change in potential energy of the hanging portion of the chain.
- Therefore, the work done is (MgL/n^2)/2.
Hence, the correct answer is option 'D' - MgL/2n^2.
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A chain of mass M is placed on a smooth table with 1/n of its length L hanging over the edge.The work done in pulling the hanging portion of the chain back to the surface of the table isa)MgL/nb)MgL/2nc)MgL/n2d)MgL/2n2Correct answer is option 'D'. Can you explain this answer?
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