a chain of mass M and length L is placed on a table with one sixth of it hanging freely from the table as amount of the work done to pull the chain on the table
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Supposing that the table is frictionless,
and the external agent brings the chain up slowly on the table.
There will be no change in kinetic energy and hence,
Work external + Work gravity = 0
Work external + ( -M/6*g*L/12) = 0
Therefore,
Work external = MgL/72
and the external agent brings the chain up slowly on the table.
There will be no change in kinetic energy and hence,
Work external + Work gravity = 0
Work external + ( -M/6*g*L/12) = 0
Therefore,
Work external = MgL/72
chayansoni:
this is right
ok cool
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