Question

A block is given a certain velocity up an inclined plane to reach just top of it. If the length of the inclined plane is $l$ and its angle of incline is equal to angle of repose, then
a. the retardation is $2g$ $sin\theta$
b. the retardation is $2g$ $tan\theta$
c. the time of ascent is $\sqrt{\frac{l}{gsin\theta }}$
d. the time of ascent is $\sqrt{\frac{l}{2gsin\theta }}$

• A. a & c are correct
• B. a & d are correct
• C. b & c are correct
• D. b & d are correct

Answer 1

Answer: (A)

a & c are correct

$\mu =\tan \theta$
$-a=g(\sin \theta +\mu \cos \theta )$

Negative sign shows that motion against gravity.

But,
$\therefore a=gsin\theta +tan\theta gcos\theta$
$\therefore a=2gsin\theta$
Time of ascent,
$T=\sqrt{\frac{2l}{a}}=\sqrt{\frac{2l}{2gsin\theta }}=\sqrt{\frac{l}{gsin\theta }}$