Two identical blocks A and B, each of mass `m=2kg` are connected to the ends of an ideal spring having force constant `K=1000Nm^-1`. System of these blocks and spring is placed on a rough floor. Coefficient of friction between blocks and floor is `m=0.5`. Block B is passed towards left so that spring gets compressed.

Calculate initial minimum compression `x_0` of spring such that block A leaves contact with the wall when system is released.

Explanation

Since, spring is initially compressed, therefore, on releasing the system, block B experiences a resultant force ( to the right) and system, block B experiences a resultant force (to the right) and starts moving to the right.
Block A leaves contact with the wall when rightward force on it applied by spring becomes just equal to the force of friction between this block and floor. Let at that instant, elongation of spring (from its natural length) be `x^'`, then tension in spring is `T=Kx^'=1000x^'` newton. Considering free body diagram of A at that instant.
`V_1=mg=20N`
`1000.x^'=mu.V_1=0.5xx20`
`x^'=0.01m`
Since spring was initially compressed by `x_0` and it has an elongation of `x^'`, therefore, displacement of block B (upto this instant)`=x_0=x^'`.
According to the law of conservation of energy,
Decrease in energy stored in the spring = Work done by block B against friction
`[1/2Kx_0^2-1/2K(x^')^2]=mumg(x_0+x^')`
Substituting `x^'=0.01m`, `K=1000Nm^-1`, `mu=0.5`, `n=2kg` and `x_0=0.03cm`.
If initial compression of spring is `x=2 x_0=6cm`, initial energy stored in spring is
`U_0=1/2K(2x_0)^2=1.80J`
When block A loses contanct with the wall, at that instant, elongation of spring (from natural length) is `x^'=1cm`.
At that instant energy stored in the spring is ,
`U=1/2K(x^')^2=0.05J`
Decrease in energy stored in the spring is,
`(U_0-U)=1.75J`
This energy is used in two ways.
To over come frictional loss during motion of block. But displacement of block B is `(2x_0+x^')=7cm`. Hence, energy lost against friction is
`mumg=(2x_0+x^')=0.7J`