-7mgl/18 J
=4.66 J
a chain of mass m=0.8 kg and length l=1.5 m rests on a rough table so that one of its ends hangs over the edge.The chain starts sliding off the table all by itself provided the hanging part is n=1/3 of chain's length.What will be the total work done by the frictional forces acting on the chain by the moment it completely slides off the table?
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7 Answers
Sahil Jain
·2014-01-22 08:15:45
P.e (i)= -Mgl/18
Pe (f) = -Mgl/2
According to law of energy con
Work done is= pe(i) - pe(f)
=-8mgl/18=-5.333