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:

  1. (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{. } $