Chapter 4, Problem 47P

To determine

The coefficient of kinetic friction between the sled and the plane.

Answer to Problem 47P

The coefficient of kinetic friction between the sled and the plane is $\underset{\_}{0.12}$.

Explanation of Solution

Figure 1 represents the free body diagram of the sled.

Figure 2 represents the free body diagram of the sled with added mass.

Write the expression for net force acting on the sled along $x$ axis.

    ${\displaystyle \sum F_x=m_{\text{sled}}{a}_{\text{sled}}}$

Use the above condition in Figure 1.

    $F_{\text{applied}}-F_{\text{f}}=m_{\text{sled}}{a}_{\text{sled}}$

Since the frictional force $F_{\text{f}}$ is given by ${\mu }_{\text{k}}{m}_{\text{sled}}g$, the above equation is written as

    $\begin{array}{c}{F}_{\text{applied}}-{\mu }_{\text{k}}{m}_{\text{sled}}g=m_{\text{sled}}{a}_{\text{sled}}\\ F_{\text{applied}}=m_{\text{sled}}{a}_{\text{sled}}+{\mu }_{\text{k}}{m}_{\text{sled}}g\\ m_{\text{sled}}=\frac{F_{\text{applied}}}{a_{\text{sled}}+{\mu }_{\text{k}}g}\end{array}$        (I)

Write the expression for net force acting on the sled with mass along $x$ axis.

    ${\displaystyle \sum F_x=0}$

Use the above condition in Figure 2.

    $F_{\text{applied}}-{F^{\prime }}_{\text{f}}=0$

Since the frictional force $F_{\text{f}}{}^{\prime }$ is given by ${\mu }_{\text{k}}\left(m_{\text{sled}}+m\right)g$, the above equation is written as

    $\begin{array}{c}{F}_{\text{applied}}-{\mu }_{\text{k}}\left(m_{\text{sled}}+m\right)g=0\\ {\mu }_{\text{k}}=\frac{F_{\text{applied}}}{\left(m_{\text{sled}}+m\right)g}\end{array}$        (II)

Use equation (I) in (II).

    $\begin{array}{l}{\mu }_{\text{k}}=\frac{F_{\text{applied}}}{\left(\frac{F_{\text{applied}}}{a_{\text{sled}}+{\mu }_{\text{k}}g}+m\right)g}\\ {\mu }_{\text{K}}=\frac{F_{\text{applied}}}{\frac{g{F}_{\text{applied}}}{a_{\text{sled}}+{\mu }_{\text{K}}g}+\frac{mg(a_{\text{sled}}+{\mu }_{\text{K}}g)}{a_{\text{sled}}+{\mu }_{\text{K}}g}}\\ {\mu }_{\text{K}}=\frac{F_{\text{applied}}(a_{\text{sled}}+{\mu }_{\text{K}}g)}{g{F}_{\text{applied}}+mg(a_{\text{sled}}+{\mu }_{\text{K}}g)}\end{array}$

Rearrange the above equation.

    $\begin{array}{c}\left(g{F}_{\text{applied}}+mg{a}_{\text{sled}}+m{\mu }_{\text{K}}{g}^2\right){\mu }_{\text{K}}=F_{\text{applied}}{a}_{\text{sled}}+F_{\text{applied}}{\mu }_{\text{K}}g\\ \left(m{g}^2\right){\mu }_{\text{K}}^2+\left(g{F}_{\text{applied}}+mg{a}_{\text{sled}}-F_{\text{applied}}g\right){\mu }_{\text{K}}-F_{\text{applied}}{a}_{\text{sled}}=0\\ \left(m{g}^2\right){\mu }_{\text{K}}^2+\left(mg{a}_{\text{sled}}\right){\mu }_{\text{K}}-F_{\text{applied}}{a}_{\text{sled}}=0\\ {\mu }_{\text{K}}=\frac{-mg{a}_{\text{sled}}\pm \sqrt{{\left(mg{a}_{\text{sled}}\right)}^2-4\left(m{g}^2\right)\left(-F_{\text{applied}}{a}_{\text{sled}}\right)}}{2\left(m{g}^2\right)}\end{array}$        (III)

Conclusion:

Substitute $20\text{N}$ for $F_{\text{applied}}$, $9.8\text{m}/{\text{s}}^2$ for $g$, $0.40\text{m}/{\text{s}}^{\text{2}}$ for $a_{\text{sled}}$, and $4.5\text{kg}$ for $m$ in equation (III), to find ${\mu }_k$.

    $\begin{array}{c}{\mu }_{\text{K}}=\frac{-\left(4.5\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(0.40\text{m}/{\text{s}}^{\text{2}}\right)\pm \sqrt{{\left(\left(4.5\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(0.40\text{m}/{\text{s}}^{\text{2}}\right)\right)}^2-4\left(\left(4.5\text{kg}\right){\left(9.8\text{m}/{\text{s}}^2\right)}^2\right)\left(-\left(20\text{N}\right)\left(0.40\text{m}/{\text{s}}^{\text{2}}\right)\right)}}{2\left(\left(4.5\text{kg}\right){\left(9.8\text{m}/{\text{s}}^2\right)}^2\right)}\\ =\frac{-17.64\text{kg}\times {\left(\text{m}/{\text{s}}^2\right)}^2\pm 118.91\text{kg}\times {\left(\text{m}/{\text{s}}^2\right)}^2}{864.36\text{kg}\times {\left(\text{m}/{\text{s}}^2\right)}^2}\\ =0.117\\ \simeq 0.12\end{array}$

Therefore, the coefficient of kinetic friction between the sled and the plane is $\underset{\_}{0.12}$.