# Arrangement of two block system is as shown in the figure . mass of block A is  and mass of B is  A constant force F = 100 N applied on upper block A . Friction between A and B is  . and b/w B and ground surface is smooth , then find work done by frictional force on block B in t = 2 sec  if system starts from rest

friction force on block A, $f_{k} = 0.5\times 5\times 10 = 25N$

An equal and opposite force will be applied on block B

$m_{b}a_{b}=f_{k}$

$10a_{b}=25$

$\Rightarrow a_{b}=2.5 m/s^{2}$

$s = \frac{1}{2}a_{b}t^{2}$

$s = \frac{1}{2}2.5\times 2^{2} = 5\: m$

Work done = $25\times 5 = 125\: J$

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