Question

You slide a 325N trunk up a 20.0° inclined plane with a constant velocity by exerting a force of 211N parallel to the inclined plane. a. What is the component of the trunks weight parallel to the plane? b. What is the sum of your applied force, friction and the parallel component of the trunks weight? Why? What is the size and direction of the frictional force? What is the coefficient of friction? c. d.
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Answer #1

a)Wx =w*sin theta

=325 N*sin 20 deg

=111.1565466 N

Wx =111 N

b)

since the object is traveling at a constant velocity,the sum of the forces are zero.

c)

Since the kinetic friction force needs to balance the component of the weight, the

magnitude is equal to 111 N, the direction is opposite i.e -111 N

d)

normal force N =w*cos theta

coefficient of friction =111N/(325 N*cos 20 deg)

=0.36

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