A force of F - 0.5 ~N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, .g=10 ~m / s ^2 )

Step-by-step solution

Maximum acceleration of 1 kg block may be a_ = g=1 ~m / s ^2 Common acceleration without relative motion between two blocks may be, a= 0.5 3 ~m / s ^2 Since, a a_ There will be no relative motion and blocks will move with acceleration 0.5 3 ~m / s ^2 . Force of friction by lower block on upper block, aligned f & =m a=(1) ( 0.5 3 )= 1 6 ~N ( towards right ) W & =f s & = 1 6 3=0.5 ~J aligned