A 20-kg block attached to a spring of spring constant 5 N m-1 is released from rest at A. The spring at this instant is having an elongation of 1 m. The block is allowed to move in smooth horizontal slot with the help of a constant force 50 N in the rope as shown. The velocity of the block as it reaches B is (assume the rope to be light)

Consider the rope and the block as a system.From work-energy theorem, ΔK=Wnet Here only two forces do non- zero work on the system : one is the spring force and the other is the constant force of 50 N acting on the cable.Let v be the speed of the block when it reaches B, then                  mv22−0=−k×522−k×122+ work done by 50N force. Horizontal displacement of the rope over pulley as the block moves from A to B is x=42+32−3=2m; this is same as displacement of point of application of 50 N force.So,  mv22=−5×522−5×122+50×2                  =20×v22=40⇒v=2ms−1