Question

A box when dropped from a certain height reaches the ground with a speed v. When it skids from rest from the same height down a rough inclined plane inclined at in angle ${45}^o$ to the horizontal, it reaches the ground with a speed $v/3$. The coefficient of sliding friction between the box and the plane is(acceleration due to gravity $10m{s}^{-2}$).

• A. $\frac{8}{9}$
• B. $\frac{1}{9}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

$\frac{8}{9}$

When box of mass m is dropped from a height h, then speed at ground is v, therefore using mechanical energy conservation
$\frac{1}{2}m{v}^2=mgh$
In inclined plane let coefficient of sliding friction be $\mu$
Now net acceleration $a=(g\sin 45-\mu g\cos 45)$
$v^{\prime 2}=2a\frac{h}{\sin 45}=2g(\sin 45-\mu \cos 45)\frac{h}{\sin 45}=\frac{v^2}{9}=\frac{2gh}{9}$
$\mu =\frac{8}{9}$