A 755 N box rests on a floor with a coefficient of kinetic friction of 0.600 and a coefficient of static friction of 0.800. What horizontal force is required to start the box moving?
A) 755 N
B) 453 N
C) 254 N
D) 653 N
E) 604 N

Question

Quizplus · Accepted Answer

To start the box moving, we need to overcome static friction. The force of static friction is given by $F_{\text{static}} = \mu_{\text{static}} \times N$, where $\mu_{\text{static}}$ is the coefficient of static friction and $N$ is the normal force. The normal force here is equal to the weight of the box, which is $755 \, \text{N}$ (since weight = mass $\times$ gravity, and the mass times gravity gives the force in newtons directly here). Thus, $F_{\text{static}} = 0.800 \times 755 \, \text{N} = 604 \, \text{N}$. Therefore, a force of $604 \, \text{N}$ is required to start the box moving.

A 755 N Box Rests on a Floor with a Coefficient