Question

A crate of mass 20 kg is being pushed by a person with a horizontal force of 63 N, moving with a constant velocity. Find the coefficient of kinetic friction.

Answer 1

Givens.

• The mass is 20 kg. (m = 20kg)

,

• The applied force is 63 N. (F = 63 N).

First, we have to make a free-body diagram to visualize the problem and its vectors.

Use Newton's Second Law for the horizontal vectors and vertical vectors.

`\begin{gathered} \Sigma F_x=ma=0\to F-F_(\mu)=0 \\ \Sigma F_y=ma=0\to N-W=0 \end{gathered}`

Use the equations for weight and friction force.

`\begin{gathered} W=mg \\ F_(\mu)=\mu N=\mu mg \end{gathered}`
`F-\mu N=0\to F=\mu N\to\mu=(F)/(N)\to\mu=(F)/(mg)`

Where F = 65 N, m = 20kg, and g = 9.8 m/s^2.

Therefore, the kinetic friction coefficient is 0.33.