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

Options

A. (a) `4ms^-1`
B. (b) `2ms^-1`
C. (c) `1ms^-1`
D. (d) `3ms^-1`

Answer

B

Explanation

Consider the rope and the block as a system. From work-energy theorem, `DeltaK=W_(n et)`
Here only two forces do not-zero work on the system: one is the spring force and the other is the constant force of `50N` acting on the cable. Let v be the speed of the block when it reaches B, then
`(mv^2)/(2)-0=-[(kxx5^2)/(2)-(kxx1^2)/(2)]`+work done by `50N` force.
Horizontal displacement of the rope over pulley as the block moves from A to B is `x=sqrt(4^2+3^2)-3=2m`, this is same as displacement of point of application of `50N` force.
So, `(mv^2)/(2)=-[(5xx5^2)/(2)-(5xx1^2)/(2)]+50xx2`
`=(20xxv^2)/(2)=40impliesv=2ms^-1`