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

Answer:

$\large \boxed{2.04}$

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

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

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.

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