A 20 kg block attached to a spring of spring constant 5 N/m is released from rest at A. The spring at this instantis having an elongation of 1m. The blockis allowed to move in a smooth horizontal slot with the help of constant force of 50 N in the rope as shown. Find the velocity of block as it reaches B. [Assume the rope to be light ].

Answer

2

Explanation

Consider the rope and the block as the system. From work - energy theorem.
`DeltaK = W_("net")`
Here only two forces do non-zero work on the sytem one is the spring force and the order is the constant force of 50 N acting on cable. Let v be the speed of block when it reaches B, then

`(mv^(2))/(2) - 0 = -[(k xx 5^(2))/(2) - (k xx J^(2))/(2) ] + ` work done by 50 N force.
Horizontal displacement of the rope over pulley as the block moves from A to B is,
`x = sqrt(4^(2) + 3^(2)) - 3 = 2` m, this is same as displacement of point of application of 50 N force.
So, `(mv^(2))/(2) =- [ (5^(2) xx 5^(2))/(2) - (5 xx 1^(2))/(2)] + 50 xx 2 rArr (20 xx v^(2))/(2) = 40 rArr v = 2m//s`