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 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 applied force into horizontal and vertical components. The vertical component does not affect the motion of the box since it is perpendicular to the surface.

The horizontal component is 250N*cos(30°) = 216.5 N.

Since the box is not moving, the force of static friction must be equal in magnitude and opposite in direction to the horizontal component of the applied force. Therefore, the force of static friction is also 216.5 N.

The question asks for the force of friction, which would be the force of kinetic friction if the box were moving. Since the box is not moving, we use the coefficient of static friction and the force of static friction to find the force of friction:

Force of friction = coefficient of static friction * normal force 
Normal force = weight of box = 50 kg * 9.81 m/s^2 = 490.5 N 
Force of friction = 0.40 * 490.5 N = 196.2 N

Therefore, the force of friction on the box is 196.2 N, which is closest to choice C.

A 50-Kg Box Is Resting on a Horizontal Floor