Question

A body which is moving with $10\, ms ^{-1}$ 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

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