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15.0 N object is pulled up an inclined plane at constant velocity. If the inclined plane makes an

angle with the horizontal of 35.0°, and the coefficient of friction is 0.300, what is the force of friction?​

1 Answer

6 votes

Answer:

3.69 N

Step-by-step explanation:

Draw a free body diagram. There are four forces acting on the object:

Weight force W pulling straight down.

Normal force N pushing perpendicular to the inclined plane.

Friction force f pushing down the inclined plane.

Applied force P pushing up the inclined plane.

Sum of forces in the perpendicular direction:

∑F = ma

N − W cos θ = 0

N = W cos θ

Friction force is normal force times the coefficient of friction:

f = Nμ

f = Wμ cos θ

Plug in values:

f = (15.0 N) (0.300) cos 35.0°

f = 3.69 N

15.0 N object is pulled up an inclined plane at constant velocity. If the inclined-example-1
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