Question

Block sliding down... moving up... distance S before stopping is _____.

• A. $\frac{u^2}{2g \cos \theta}$
• B. $\frac{u^2}{\sqrt{2}g \cos \theta}$
• C. $\frac{u^2}{4g \sin \theta}$
• D. $\frac{2u^2}{g \cos \theta}$

Hint:

Retardation up an incline with friction is $g(\sin\theta + \mu\cos\theta)$.

Answer 1

The Correct Option is C

Sliding down constant velocity implies $\mu = \tan\theta$.
Moving up: $a = g\sin\theta + \mu g\cos\theta = 2g\sin\theta$.
$S = u^2 / 2a = u^2 / 4g\sin\theta$.