Chapter 5, Problem 69A

To determine

The coefficient of kinetic friction between the crate and the floor.

Answer to Problem 69A

The coefficient of kinetic friction between the wooden crate and the floor is ${\mu }_{\text{k}}=0.255$ .

Explanation of Solution

Given:

The mass $\left(m\right)$ of the wooden crate is $12\text{kg}$ .

The horizontal force $\left(f_{\text{k}}\right)$ exerted on the wooden crate is $30.0\text{N}$ .

Consider the acceleration due to gravity $\left(g\right)$ of Earth is $9.80{\text{m/s}}^2$ .

Formula used:

Consider a horizontal force acts on the object and the object slides over the surface, frictional force acts between the object and surface. The expression for the kinetic frictional force between the object and the surface as follows:

  $f_{\text{k}}={\mu }_{\text{k}}N$

Here, $N$ is normal force acting on the object and ${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Calculation:

The weight $\left(F_{\text{g}}\right)$ of the wooden crate as follows:

  $\begin{array}{c}{F}_{\text{g}}=mg\\ =12\text{kg}\times \text{9}\text{.80}{\text{m/s}}^2\\ =117.6\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ =117.6\text{N}\end{array}$

The normal force $\left(N\right)$ acting on the wooden crate is equal to the weight of the wooden crate.

  $\begin{array}{c}N=F_{\text{g}}\\ N=117.6\text{N}\end{array}$

So, the value of the coefficient of kinetic friction $\left({\mu }_{\text{k}}\right)$ as follows:

  $\begin{array}{l}{f}_{\text{k}}={\mu }_{\text{k}}N\\ 30.0\text{N}={\mu }_{\text{k}}\left(117.6\text{N}\right)\\ {\mu }_{\text{k}}=\left(\frac{30.0\text{N}}{117.6\text{N}}\right)\\ {\mu }_{\text{k}}=0.255\end{array}$

Conclusion:

Therefore, the coefficient of kinetic friction between the wooden crate and the floor is ${\mu }_{\text{k}}=0.255$ .