A `0.500 kg` block is attached to a spring with length `0.60m` and force constant `k=40.0 N//m`. The mass of the spring is negligible. You pull the block to the right along the surface with a constant horizontal force `F=20.0 N`. (a) What is the block's speed when the block reaches point B,which is `0.25 m` to the right of right of point A? (b) What the block reaches point B, you let go off the block. In the subsequent motion, how close does the block get to the wall where the lift end of the spring is attached? Neglect size of block and friction.
.

Options

A. `3.87ms^(-1) , 0.1m`
B. `4.87ms^(-1) , 0.1m`
C. `1.87ms^(-1) , 0.5m`
D. `3.87ms^(-1) , 0.6m`

Answer

A, B, C

Explanation

(a) From work energy theorem,
Work done by all forces = change in kinetic energy
`:. Fx-(1)/(2)kx_(2) =1/2mv^(2)`
`:. vsqrt((2Fx-kx^(2))/m)`
Substituting the values we have,
`v=sqrt((2xx20xx0.25-40xx25xx0.25)/0.5)`
`= sqrt(15) m//s = 3.87 m//s`
(b) From conservation of mechanical energy,
`E_(i) =E(f)`
or `1/2mv_(i)^(2)+1/2kx_(i)^(2) =1/2kx_(f)^(2)`
or `x_(f) =sqrt((mv_(i)^(2))/k+x_(i)^(2))`
`=sqrt((0.5xx15)/(40) +(0.25)^(2))`
`0.5m` (compression
`:.` Distance of block from the wall
`=(0.6-0.5) m`
`=0.1 m`.