Question

what is the coefficient of friction of a 20.0-kg box that requires 400 N of force to move across the table at constant velocity?

Answer 1

`\large \boxed{2.04}`

Step-by-step explanation:

The coefficient of friction (µ) is the ratio of the frictional force (F) to the normal force (N).

µ =F/N

1. Calculate the normal force

`\begin{array}{rcl}F& = & ma\\& = & \text{20.0 kg $*$ 9.807 m$\cdot$s}^(-2)\\& = & \text{196 N}\\\end{array}\\\text{The normal force is 196 N.}`

2. Calculate the coefficient of friction

The box is moving at a constant velocity, so the frictional force is equal and opposite to the applied force.

`\begin{array}{rcl}\mu & = & (F)/(N)\\\\& = & \frac{\text{400 N}}{\text{196 N}}\\\\& = & \mathbf{2.04}\\\end{array}\\\text{The coefficient of friction is $\large \boxed{\mathbf{2.04}}$}`