A uniform chain of length l and mass m overhangs a horizontal table with its two third part on the table. the friction coefficient between the table and the chain is μ.find the work done by the friction during the period the chain slips off the table?
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Force of friction is against the movement (sliding) of the chain on the table.
dF due to friction = μ x dm x g
= μ g (m/L) dx
when the chain slides the element gets displaced by the value x
so the work done = dF . x= μ g (m/L) dx .x
total work by applying limit = 2/9μ mgL
dF due to friction = μ x dm x g
= μ g (m/L) dx
when the chain slides the element gets displaced by the value x
so the work done = dF . x= μ g (m/L) dx .x
total work by applying limit = 2/9μ mgL
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