Question

A block slides down an inclined plane of slope angle $\theta$ with constant velocity. If it is then projected up the same plane with an initial speed $v_0$, the distance in which it will come to rest is:

• A. $\frac{v_0^2}{gtan\theta }$
• B. $\frac{v_0^2}{4gsin\theta }$
• C. $\frac{v_0^2}{2g}$
• D. $\frac{v_0^2}{2gsin\theta }$

Answer 1

The correct option is B $\frac{v_0^2}{4gsin\theta }$

From the free body diagram.
If the body move with constant velocity $mgsin\theta =\mu N$
$N-mgcos\theta =0$
So, $\mu =tan\theta$
Now in case (ii) if it is projected with velocity $v_0$ upward then net retardation against it is
$a=\frac{mgsin\theta +\mu mgcos\theta }{m}=gsin\theta +\mu gcos\theta$
$=gsin\theta +gsin\theta$
$=2gsin\theta$
So distance travelled=$\frac{v_0^2}{2a}=\frac{v_0^2}{4gsin\theta }$
So option (B) is correct.