Laws of Motion - Result Question 49
49. A block has been placed on an inclined plane with the slope angle $\theta$, block slides down the plane at constant speed. The coefficient of kinetic friction is equal to
[1993]
(a) $\sin \theta$
(b) $\cos \theta$
(c) $g$
(d) $\tan \theta$
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Answer:
Correct Answer: 49. (d)
Solution:
- (d) When the block slides down the plane with a constant speed, then the inclination of the plane is equal to angle of repose $(\theta)$.
Coeff. of friction $=\tan$ of the angle of repose
$ =\tan \theta \text{. } $
