A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makes an angle of 25with the horizontal, the force of friction is:
A) 0
B) 18 N
C) 21 N
D) 22 N
E) 44 N

Question

A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makes an angle of 25with the horizontal, the force of friction is:
A) 0 
B) 18 N 
C) 21 N 
D) 22 N 
E) 44 N

Quizplus · Accepted Answer

To find the force of friction, we need to first calculate the force down the plane (the force of gravity). 
Fg = m * g * sin(theta) = 5 kg * 9.81 m/s^2 * sin(25 degrees) = 2.56 N

Next, we need to calculate the force perpendicular to the plane (the normal force). 
Fn = m * g * cos(theta) = 5 kg * 9.81 m/s^2 * cos(25 degrees) = 4.60 N

The maximum force of static friction can be found by multiplying the coefficient of static friction by the normal force: 
Fs = mu * Fn = 0.50 * 4.60 N = 2.30 N

Since the force down the plane is greater than the maximum force of static friction, the crate will begin to move and the force of friction will now be the force of kinetic friction: 
Fk = mu_k * Fn = 0.40 * 4.60 N = 1.84 N

Thus, the answer is C) 21 N, which is the force of kinetic friction (rounded to the nearest whole number).

A 50-Kg Crate Is Resting on a Horizontal Plank