A 2kg Block slides on a horizontal floor with speed of 4 m/s.   It strikes a uncompressed spring and compresses it till the block is motionless.  The kinetic friction force is 15N and spring constant is 10,000 N/m.   The spring compressed by

Let the spring compresses by x

Initial K.E. of the mass  =   P.E. of the spring + work done due to friction
$\mathrm{KE} \mathrm{of} \mathrm{the} \mathrm{mass}=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{2}{\mathrm{mv}}^2; \mathrm{PE} \mathrm{of} \mathrm{spring}=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{2}{\mathrm{kx}}^2;$

$\mathrm{work} \mathrm{done} \mathrm{by} \mathrm{friction}=\mathrm{friction} \mathrm{force} . \mathrm{displacement}$

$$\begin{gathered} \mathrm{substitute} \\ \frac{1}{2}{\mathrm{mv}}^2=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{2}{\mathrm{kx}}^2+\mathrm{f}.\mathrm{x} \end{gathered}$$

$\mathrm{substitute} \mathrm{the} \mathrm{given} \mathrm{values}$

$\therefore \frac{1}{2}\times 2\times 4^2=\frac{1}{2}\times 10000\times {\mathrm{x}}^2+15\mathrm{x}$

$$\begin{gathered} \mathrm{on} \mathrm{solving} \mathrm{the} \mathrm{quadratic} \mathrm{equation}, \mathrm{we} \mathrm{get} \\ \mathrm{x}=0.055 \mathrm{m} \end{gathered}$$