Question

Unfortunately, I still don't get this question: A worker of a moving company places a 252-kg trunk on a piece of carpeting and slides it across the floor at constant velocity by exerting a horizontal force of 425 N on the truck.

Answer 1

I have to find the coefficient of kinetic friction and here are the formulas:

Coefficient of Static friction:

|FN| = |Fg| = m|g|

\[\mu\]s = Fs/Fn

Coefficient of kinetic friction:

|FN| = |Fg| = m|g|

\[\mu\]s = FK/FN

Kinetic friction:

FK = \[\mu\]KFN, where |FN| = |Fg| = m|g|

Answer 2

Like we discussed before, the forces of friction equal zero because there is no acceleration. \[F_{fric}=\mu \times F_{nat} \space \space F_{nat}=mg\] \[F_{push} - F_{fric}=0 \rightarrow 425N - \mu F_{nat} \rightarrow 425 - \mu (252kg \times 9.8\frac{m}{s^2}=0\] Now solve for your friction coefficient\[\mu=\frac{425N}{252kg \times 9.8\frac{m}{s^2}} \space \space \mu=0.17\]