Chapter 3.15, Problem 301RP

To determine

The coefficient of kinetic friction between the truck bed and the crate.

Answer to Problem 301RP

The coefficient of kinetic friction between the truck bed and the crate is $0.382$.

Explanation of Solution

Draw the free body diagram of the crate as shown below.

Write the equilibrium condition along the y-direction.

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ N-W\cos \theta =0\\ N=W\cos \theta \end{array}$                                                                                            (I)

Here, the normal force exerted by the truck bed is $N$, the weight of the crate is $W$ and the angle of inclination of the truck bed is $\theta$.

Write the energy balance equation.

  $\begin{array}{l}{U}_{rel}=\Delta T_{rel}\\ \left(W\sin \theta -N{\mu }_k\right)x=\frac{1}{2}m\left(v_2{}^2-v_1{}^2\right)\\ \left(mg\sin \theta -N{\mu }_k\right)x=\frac{1}{2}m\left(v_2{}^2-v_1{}^2\right)\end{array}$                                                           (II)

Here, the mass of the crate is $m$, the acceleration due to gravity is $g$, the coefficient of kinetic friction is ${\mu }_k$, the distance slid by the crate is $x$, the initial velocity of the crate is $v_1$ and the final velocity of the crate is $v_2$.

Conclusion:

Substitute, Equation (I) in Equation (II).

  $\begin{array}{l}\left(mg\sin \theta -W\cos \theta {\mu }_k\right)x=\frac{1}{2}m\left(v_2{}^2-v_1{}^2\right)\\ \left(mg\sin \theta -mg\cos \theta {\mu }_k\right)x=\frac{1}{2}m\left(v_2{}^2-v_1{}^2\right)\\ g\left(\sin \theta -\cos \theta {\mu }_k\right)x=\frac{1}{2}\left(v_2{}^2-v_1{}^2\right)\end{array}$                                               (III)

Calculate $\theta$ as follows:

  $\begin{array}{l}\tan \theta =\frac{15}{100}\\ \theta ={\tan }^{-1}\left(0.15\right)\\ \theta =8.53°\end{array}$

Substitute, $9.81\text{m}/{\text{s}}^2$ for $g$, $8.53°$ for $\theta$, $2\text{ m}$ for $x$, $0$ for $v_2$ and $3\text{m}/\text{s}$ for $v_1$ in Equation (III).

  $\begin{array}{l}\left(9.81\text{m}/{\text{s}}^2\right)\left(\sin 8.53°-{\mu }_k\cos 8.53°\right)\left(2\text{ m}\right)=\frac{1}{2}\left[0^2-{\left(3\text{m}/\text{s}\right)}^2\right]\\ \left(\sin 8.53°-{\mu }_k\cos 8.53°\right)=\frac{1}{2}\left[\frac{-{\left(3\text{m}/\text{s}\right)}^2}{\left(2\text{ m}\right)\left(9.81\text{m}/{\text{s}}^2\right)}\right]\\ 0.1483-0.9889{\mu }_k=-0.2294\\ 0.9889{\mu }_k=0.1483+0.2294\\ {\mu }_k=\frac{0.3777}{0.9889}\\ {\mu }_k=0.382\end{array}$

Thus, the coefficient of kinetic friction between the truck bed and the crate is $0.382$.