A classical mechanics problem by Md Zuhair

By applying law of conservation of energy,

$\large \frac{mv^{2}}{2}\ = \frac{kx^{2}}{2}\ + μmgx$

Here,

$\large \frac{mv^{2}}{2}$ - Initial kinetic energy of the block

$\large \frac{kx^{2}}{2}$ - Elastic potential energy stored in the spring at the final position

$\large μmgx$ - Work done by friction against the motion of the block

$\large x$ - Displacement of the block

After substituting the given values, we get,

$\large v$ = $0.4$

$\large \boxed{\color{#20A900}{10v} = \color{#20A900}{4}}$

Are you at fiitjee ? @Md Zuhair