A box slides downwards at a constant velocity on an inclined surface that has a coefficient of friction uK =.58 The angle of the incline, in degrees, is most nearly:

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Physics

Isai Le

15 January, 09:58

A box slides downwards at a constant velocity on an inclined surface that has a coefficient of friction uK =.58 The angle of the incline, in degrees, is most nearly:

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Answers (1)

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April Mathis · Answer 1

Answer: The angle of inclination is nearly 30°

Explanation:

For a body on an inclined plane, the coefficient of friction between the body and the plane is equal to the ratio of the moving force applied to the body to the frictional force acting on the body.

If uK coefficient of friction;

Fm is the moving force

R is the normal reaction on the body

Mathematically uK = Fm/R

Fm = WSin (theta)

R = Wcos (theta)

uK = Wsin (theta) / Wcos (theta)

uK = tan (theta)

theta = arctan (uK)

If uK is 0.58

theta = arctan0.58

theta = 30°

The angle of the inclined will be 30°