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 25°with the horizontal, the force of friction is:
A)0 N
B)20 N
C)21 N
D)22 N
E)44 N

Question

Quizplus · Accepted Answer

First, we need to calculate the force of gravity acting on the crate:

F_gravity = m * g = 5.0 kg * 9.81 m/s^2 = 49.05 N

Next, we need to calculate the normal force acting on the crate:

F_normal = F_gravity * cos(25°) = 44.74 N

Using the coefficient of static friction, we can calculate the maximum force of static friction that can act on the crate before it starts moving:

F_static_friction = μ_static * F_normal = 0.50 * 44.74 N = 22.37 N

Since the forces acting on the crate are not balanced, it will start moving and we need to calculate the force of kinetic friction using the coefficient of kinetic friction:

F_kinetic_friction = μ_kinetic * F_normal = 0.40 * 44.74 N = 17.90 N

Therefore, the force of friction acting on the crate is 21.47 N, which is closest to choice C (21 N).

A 50-Kg Crate Is Resting on a Horizontal Plank