Question

A 155.0 kg crate is pushed horizontally with a force of 650.0 N. If the coefficient of friction is 0.42, calculate the net force of the crate.

Answer 1

`12.02N`

Step-by-step explanation

Parameters given:

Mass of crate, m = 155.0 kg

Force of push, Fp = 650.0 N

Coefficient of friction, μ = 0.42

The net force of the crate is the sum of all forces acting on the crate. The forces acting on the crate are the force of push and friction force.

Therefore, we have that:

`\begin{gathered} F=F_p+(-F_r) \\ F=F_p-F_r \end{gathered}`

The friction force is given by:

`F_r=\mu N=\mu mg`

where g = acceleration due to gravity; N = normal force

Hence, the net force is:

`\begin{gathered} F=F_p-\mu mg \\ F=650-(0.42\cdot155\cdot9.8) \\ F=650-637.98 \\ F=12.02N \end{gathered}`

That is the net force of the crate.