Q10 of 101 Page 132

A block of mass m is kept over another block of mass M and the system rests on a horizontal surface. A constant horizontal force F acting on the lower block produces an acceleration in the systems the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal Surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system.

(b)


(c)



(a) Weight of the combined system = (m+M)g = N Frictional force = Given that acceleration a = Now by equating the forces along the surface we have the





(b) Let f be the frictional force acting on the smaller block, along the movement only this force is on the block whose acceleration is a =


Hence this force should be equal to mass× acceleration =


(c) Now during the displacement d, velocity of the block can be evaluated by using uniform motion equation


Now substituting for the values, we get v= , u=0


Now initial K.E =0 since u=0 and


final K.E =


Therefore, the change in K.E is =


So, work done by the frictional force on the smaller block by the larger block = change in kinetic energy ( from work energy theorem) =


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