Question

A block slides down a rough inclined plane of slope angle $\theta$ with a constant velocity. It is then projected up the same plane with an initial velocity v. The distance travelled by the block up the plane before coming to rest is

• A. $\frac{v^2}{4gsin\theta }$
• B. $\frac{v^2}{2gsin\theta }$
• C. $\frac{v^2}{gsin\theta }$
• D. $\frac{4g{v}^2}{sin\theta }$

Answer 1

The correct option is A $\frac{v^2}{4gsin\theta }$

Since block is moving with constant velocity.

So, $\mu mgcos\theta =mgsin\theta \Rightarrow \mu =tan\theta$,

When it is projected upward, retardation $a=\mu gcos\theta +gsin\theta =2gsin\theta$

So, we know that:

$v^2=2as$

$s=\frac{v^2}{4gsin\theta }$