A block of mass `m` is connected with a rigid wall by light spring of stiffness `k`. Initially the spring is relaxed.

If the block is pushed with a velocity `v_0`, it oscillates back and forth and stops. Assuming `mu` as the coefficient of kinetic friction between block and ground, find the work done by friction till the block stops.

Explanation

As the block moves, spring force `F_s=(kx)` increases till it nullifies `f_k(=mumg)`. In consequence, the block will stop.
`DeltaK:` When the block will stop,
`DeltaK=-1/2mv_0^2`
`W_(sp):` When the block stops finally, let us assume that the spring is deformed by an amount `x`.
Then the work done by the spring is
`W_(sp)=-(kx^2)/(2)`

The work done by friction `W_f` is given as `W_f+W_(sp)=DeltaK`.
Substituting `DeltaK=-1/2mv_0^2` and `W_(sp)=-(kx^2)/(2)`, we have
`W_f=(kx^2)/(2)-1/2mv_0^2`
where x can be calculated by equating the forces acting on the stationary block.
Since the block is at the verge of sliding, we have
`kx=mumg`
This gives `x=(mumg)/(k)`
Now substituting `x=(mumg)/(k)` in the above expression of work, we have `W_f=(mu^2m^2g^2)/(2k)-1/2mv_0^2`