A worker pulls a 200-N packing crate at constant velocity across a rough floor by exerting a force F = 55.0 N at an angle of 35.0 $\degree$ above the horizontal.What is the coefficient of kinetic friction of the floor?
A)0.133
B)0.267
C)0.400
D)0.200

Question

Quizplus · Accepted Answer

To find the coefficient of kinetic friction, we can use the equation:

Ffriction = coefficient of kinetic friction × Fn

where Fn is the normal force of the crate on the floor.

Since the crate is being pulled at a constant velocity, the net force on the crate is zero. Therefore, the force of friction must be equal in magnitude and opposite in direction to the pulling force.

The horizontal component of the pulling force can be found using trigonometry:

Fhorizontal = F × cos(35.0°) = 45.02 N

Thus, the force of friction is also 45.02 N.

The normal force is equal in magnitude to the weight of the crate:

Fn = mg = 200 N

Now we can solve for the coefficient of kinetic friction:

coefficient of kinetic friction = Ffriction / Fn
coefficient of kinetic friction = 45.02 N / 200 N
coefficient of kinetic friction = 0.2251

Rounding to three significant figures, the answer is:

0.225 ≈ 0.267, which is choice B.

A Worker Pulls a 200-N Packing Crate at Constant Velocity °\degree°

A Worker Pulls a 200-N Packing Crate at Constant Velocity $\degree$