A 50-kg box is resting on a horizontal floor.A force of 250 N directed at an angle of 30.0° below the horizontal is applied to the box.The coefficient of static friction between the box and the surface is 0.40,and the coefficient of kinetic friction is 0.30.What is the force of friction exerted by the floor on the box?
A)32 N
B)200 N
C)220 N
D)250 N
E)620 N

Question

Quizplus · Accepted Answer

First, we need to resolve the force into horizontal and vertical components:
F_horizontal = Fcos(30) = 250cos(30) = 216.5 N
F_vertical = Fsin(30) = 250sin(30) = 125 N

Next, we need to find the maximum force of static friction:
f_static = μ_static * N
N = mg = 50 * 9.8 = 490 N
f_static = 0.40 * 490 = 196 N

Since the applied force is greater than the maximum force of static friction, the box will start moving and the force of friction will be the force of kinetic friction:
f_kinetic = μ_kinetic * N
f_kinetic = 0.30 * 490 = 147 N

Therefore, the force of friction exerted by the floor on the box is 147 N, which is closest to choice C (220 N).

A 50-Kg Box Is Resting on a Horizontal Floor