A worker pulls a 350-N packing crate at constant velocity across a rough floor by exerting a force F = 200.0 N at an angle of 30.0° above the horizontal.What is the coefficient of kinetic friction of the floor?
A)0.793
B)0.693
C)0.593
D)1.386
E)0.400

Question

A worker pulls a 350-N packing crate at constant velocity across a rough floor by exerting a force F = 200.0 N at an angle of 30.0° above the horizontal.What is the coefficient of kinetic friction of the floor?
A)0.793
B)0.693
C)​0.593
D)​1.386
E)​0.400

Quizplus · Accepted Answer

The coefficient of kinetic friction ($\mu_k$) can be found using the equation $F_{friction} = \mu_k F_{normal}$. Since the crate moves at constant velocity, the net force is zero, meaning the horizontal component of the applied force equals the frictional force, and the vertical component of the applied force plus the normal force equals the weight of the crate. The horizontal component of the applied force is $F \cos(\theta) = 200 \cos(30^\circ) = 173.2 \, \text{N}$, and the vertical component is $F \sin(\theta) = 200 \sin(30^\circ) = 100 \, \text{N}$. The normal force is the weight of the crate minus the vertical component of the applied force, $350 \, \text{N} - 100 \, \text{N} = 250 \, \text{N}$. Thus, $\mu_k = \frac{F_{friction}}{F_{normal}} = \frac{173.2}{250} = 0.693$.

A Worker Pulls a 350-N Packing Crate at Constant Velocity