A uniform chain of mass 'm' and length 'l' is lying on a smooth
horizontal table with one-third of its length hanging from the edge
of the table. Work done in pulling the chain on to the table so that 5/
6th of the chain is on the table is
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Answer:
The weight of hanging part (
3
L
) of chain is (
3
1
Mg). This weight acts at centre of gravity of the hanging part, which is at a distance of (
6
L
) from the table.
As work done =force×distance
∴W=
3
Mg
×
6
L
=
18
MgL
.
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