Question

Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if `mu` is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?

The question is understood as below:
Block of mass `m` has initial velocity `u` having direction along `+x` axis. The block stops after covering distance `S` causing similar extension in the spring of spring constant `K` holding the block. if μ is the coefficient of kinetic friction between the block and surface on which it was moving, the distance `S` is given by the equation

Answer 1

I got a different answer than the published result.
Did I make a mistake!

Explanation:

Assuming that initially the spring is in its equilibrium position.

Initial energy of block is its kinetic energy `=1/2mv^2`

Using Law of Conservation of energy:
When the block stops it initial kinetic energy is converted in to mechanical potential energy of spring which gets stretched by distance `S` and remaining energy is spent doing work against force of friction during its movement.

PE of the spring`=1/2KS^2`

Force of friction `=mumg`

Work done against force of friction `="Force"xx"distance"`
`=mumgxxS`

Equating the initial and final energies we have

`1/2mv^2=1/2KS^2+mumgS`
`=>KS^2+2mumgS-mv^2=0`

Solving the quadratic in `S` we get

`S=(-2mumg+-sqrt((2mumg)^2 -4xxK(-mv^2)))/(2K)`
`S=1/K(-mumg+-sqrt((mumg)^2 +Kmv^2))`

Ignoring the `-ve` root as the movement is in `+x` direction we get
`S=1/K(sqrt((mumg)^2 +Kmv^2)-mumg)`