Q. A block is moving up an inclined plane of inclination $60{^\circ}$ with velocity of 20 $ms^{- 1}$ and stops after 2.00 s. If $g = 10ms^{- 2}$ then the approximate value of coefficient of friction is
A) 3
B) 3.3
C) 0.27
D) 0.33
Explanation
(c) Retardation = $g(\sin{60{^\circ} + \mu\cos{60{^\circ}}})$
$= 10\left( \frac{\sqrt{3}}{2} + \mu\frac{1}{2} \right) = 5(\sqrt{3} + \mu)$
$v = u - at$
$0 = 20 - 5(\sqrt{3} + \mu) \times 2$
$\mu = 2 - 1.732 \approx 0.27$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.