Question 43: Laws of Motion

Question

A block is sliding down on an inclined plane of slope $	heta$ and at an instant $t=0$ this block is given an upward momentum so that it starts moving up on the inclined surface with velocity $u$. The distance $(S)$ travelled by the block before its velocity become zero, is $\_\_\_\_$ .
(g = gravitational acceleration)

Accepted Answer

When the block is moving up the inclined plane, the component of gravity acting down the slope opposes its motion.
Force along the incline, $\mathrm{F}=-\mathrm{mg} \sin 	heta$
Using $\mathrm{F}=$ ma, we get $\mathrm{a}=-\mathrm{g} \sin 	heta$.
The initial velocity $\mathrm{v}_{\mathrm{i}}=\mathrm{u}$
The final velocity $\mathrm{v}_{\mathrm{f}}=0$ (since the block comes to rest)
Acceleration $\mathrm{a}=-\mathrm{g} \sin 	heta$
using the third equation of motion :
$$ \mathrm{v}_{\mathrm{f}}^2=\mathrm{v}_{\mathrm{i}}^2+2 \mathrm{a} S $$
Substitute the values into the equation :
$0^2=\mathrm{u}^2+2(-\mathrm{g} \sin 	heta) \mathrm{S}$
$$\Rightarrow $$ $0=\mathrm{u}^2-2 \mathrm{~g} \sin 	heta (\mathrm{S})$
$$\Rightarrow $$ $$ S=\frac{u^2}{2 g \sin 	heta} $$
Hence, the distance travelled is $\frac{u^2}{2 g \sin 	heta}$. Hence, correct option is (C).

Answer

$\frac{2 u^2}{\mathrm{~g} \cos 	heta}$

Answer

$\frac{u^2}{\sqrt{2} g \cos 	heta}$

Answer

$\frac{u^2}{2 g \sin 	heta}$

Answer

$\frac{u^2}{2 g \cos 	heta}$

Laws of Motion

Choose the correct answer: