You slide a 325-N trunk up a 20.0° inclined plane with a constant velocity by exerting a
force of 211 N parallel to the inclined plane.
a. What is the component of the trunk's
weight parallel to the plane?
b. What is the sum of your applied force, fric-
tion, and the parallel component of the
trunk's weight? Why?
C. What is the size and direction of the friction
force?
d. What is the coefficient of friction?

A) 111 N

B) F_net = 0 N

C) 100 N, in the opposite direction of the motion.

D) μ = 0.3274

Explanation:

A) The component of the trunks weight acts parallel to the plane but in the vertical direction and so;

W = Fsin θ

W = 325 sin 20

W = 325 × 0.3420

W ≈ 111 N

B) We are told the trunk is moving with a constant velocity and for this reason the sun of the forces will equal to zero.

Thus; F_net = 0 N

C) Since the kinetic friction force needs to balance the component of the weight, the magnitude is;

F_f = 211 - 111 = 100 N

The direction will be in the opposite direction since it's against the motion.

D) From friction formula, we know that;

F_f = μW cos θ

100 = μ × 325 × cos 20

μ = 100/(325 × cos 20)

μ = 0.3274

it is easily magnetized.