When box is dropped from height $h$, its speed when it reaches the ground is
$v=\sqrt{2 g h}$
When block slides down the inclined plane $\theta=45^{\circ}$,
Net downward acceleration of block is
$a=\frac{m g \sin \theta-f}{m}$
where, $f=$ friction force.
$\Rightarrow a=\frac{m g}{m} \sin \theta-\mu m g \cos \theta$
$=g(\sin \theta-\mu \cos \theta)$
$=\frac{g}{\sqrt{2}}(1-\mu)$
${[\because \sin \theta}\left.=\cos \theta=\frac{1}{\sqrt{2}}, \text { when } \theta=45^{\circ}\right]$
Velocity of block when it reaches bottom of inclined plane is
$v^{\prime}=\sqrt{2 a s}$
where, $s=$ slope length of inclined plane.
$\Rightarrow \quad v^{\prime}=\sqrt{2 a h / \sin \theta}$
$=\sqrt{\left(2 \frac{g h}{\sqrt{2}}(1-\mu) \times \sqrt{2}\right)}$
$=\sqrt{2 g h(1-\mu)}$
$=\frac{v}{3} \text { (given) }$
So, $\sqrt{2 g h(1-\mu)}=\frac{1}{3} \sqrt{2 g h}$
$\Rightarrow \quad 1-\mu=\frac{1}{9}$
$\Rightarrow \quad \mu=1-\frac{1}{9}=\frac{8}{9}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.