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Two blocks asre connected by a string as...

Two blocks asre connected by a string as shown in figure. They are released from rest. Show that after they moved a distance L, their common speed is given by `sqrt((2(m_(2)-mu m_(1))gL)/((m_(1)+m_(2))))` where `mu` is the coefficient of friction.

Text Solution

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In moving a distance L by `m_(1) and m_(2)` the PE will be lost by `m_(2)` only and will be `m_(2)gl`, while work done by friction will be only on `m_(1)` and will be `mu m_(1)gL`. Now as the string is inextensible `v_(1) = v_(2) = v ` and so, the gain in KE wil be = `(1//2) (m_(1) + m_(2)) v^(2)`. By conservation of energy.
Loss in PE = Grain in KE + WD against friction
i.e., `m_(2)gL =(1)/(2)(m_(1)+m_(2))v^(2)+mu m_(1)gL`
`v=sqrt((2(m_(2)-mum_(1))gL)/((m_(1)+m_(2))))`
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