Question

A body which is moving with $10m{s}^{-1}$ is sliding up on a rough inclined plane having inclination of $30^{\circ }$. Find the height upto which it can go, if coefficient of friction of the inclined surface is $0.1$.

• A. 4.25 m
• B. 5.25 m
• C. 6.25 m
• D. 7.25 m

Answer 1

Answer: (A)

According to work-energy theorem
Kinetic energy $=$ Work against gravity $+$ Work against friction
Thus, using the figure given below, we have

$\frac{1}{2}m{v}^2=mgh+\mu R\cdot AB$
$\Rightarrow \frac{1}{2}m{v}^2=mgh+\mu mg\cos 30^{\circ }\cdot \frac{h}{\sin 30^{\circ }}$
$\left[\because AB=\frac{h}{\sin 30^{\circ }}\right]$
$\Rightarrow \frac{1}{2}\times m\times (10{)}^2=mgh+\mu mg\cot 30^{\circ }\times h$
$\Rightarrow 50=h[10+0.1\times 10\times \sqrt{3}]=h[10+1.732]$
$\Rightarrow h=\frac{50}{11.732}=4.25m$