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)
2 ms−1
Consider the rope and the blocks as a system.From work-energy theorem, △k=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=−[5×522−5×122]+50×2
= 20×v22=40
⇒v=2ms−1