Question

A block is released from the top of a smooth inclined plane of inclination $\theta$ as shown in figure. Let $v$ be the speed of the particle after travelling a distance $s$ down the pane. Then which of the following will remain constant

• A. $v^2+2gs\sin \theta$
• B. $v^2-2gs\sin \theta$
• C. $v-\sqrt{2gs}\sin \theta$
• D. $v+\sqrt{2gs}\sin \theta$

Answer 1

The correct option is C $v-\sqrt{2gs}\sin \theta$

If H is the vertical height of the block, then $h=ssin\theta$ .....................(1)

$\frac{1}{2}m{v}^2+mg(H-h)=const.$

$\frac{1}{2}m{v}^2-mgh=const.$

$v^2-2gh=const.$

Using eqn. 1,

we get,

$v^2-2gssin\theta$