Chapter 4, Problem 172P

(a)

To determine

The distance travelled by Carlos and Shannon when Shannon catches up to Carlos.

Answer to Problem 172P

The distance travelled by Carlos and Shannon when Shannon catches up to Carlos is $\underset{\_}{77\text{m}}$.

Explanation of Solution

Given that the angle of inclination of the slope is $12°$ below the horizontal, the coefficient of friction for Carlos’s sled is $0.10$, the coefficient of friction for the Shannon’s supersled is $0.010$, and the initial distance of separation between the sleds is $5.0\text{m}$.

The forces acting on the system is shown in Figure 1.

Write the equilibrium condition for forces acting on the system in $x$ and $y$ direction.

${\displaystyle \sum F_x}=0$ (I)

${\displaystyle \sum F_y}=0$ (II)

Apply the condition in equation (I) to the system.

$m{a}_x=mg\sin \theta -f_{\text{k}}$ (III)

Here, $m$ is the mass of the sled, $a_x$ is the acceleration in $x$-direction, $g$ is the acceleration due to gravity, $\theta$ is the angle of inclination, and $f_{\text{k}}$ is the force of friction.

Write the expression for the force of friction $f_{\text{k}}$.

$f_{\text{k}}={\mu }_{\text{k}}N$ (IV)

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

Use equation (IV) in (III).

$m{a}_x=mg\sin \theta -{\mu }_{\text{k}}N$ (V)

Apply the condition in equation (II) to the system.

$N=mg\cos \theta$ (VI)

Use equation (VI) in (V) and solve for $a_{\text{x}}$.

$\begin{array}{c}m{a}_x=mg\sin \theta -{\mu }_{\text{k}}mg\cos \theta \\ a_x=g\left(\sin \theta -{\mu }_{\text{k}}\cos \theta \right)\end{array}$ (VII)

Write the expression for the distance moved by Shannon when she meet Carlos.

$\Delta x_{\text{S}}=\Delta x_{\text{1}}+\Delta x_{\text{C}}$ (VIII)

Here, $\Delta x_{\text{S}}$ is the displacement of Shannon, $\Delta x_{\text{1}}$ is the separation between Shannon and Carlos when Shannon just start moving, and $\Delta x_{\text{C}}$ is the distance moved by Carlos after Shannon started moving.

Write the expression for $\Delta x_{\text{S}}$ in terms of acceleration.

$\Delta x_{\text{S}}=\frac{1}{2}{a}_{\text{S}}{t}^2$ (IX)

Here, $a_{\text{S}}$ is the acceleration of Shannon, and $t$ is the time.

Write the expression for the expression for $\Delta x_{\text{C}}$.

$\Delta x_{\text{C}}=v_{\text{C}}t+\frac{1}{2}{a}_{\text{C}}{t}^2$ (X)

Here, $v_{\text{C}}$ is the speed of Carlos when Shannon just started moving, and $a_{\text{C}}$ is the acceleration of Carlos.

Use equation (IX) and (X) in (VIII).

$\begin{array}{c}\frac{1}{2}{a}_{\text{S}}{t}^2=\Delta x_1+v_{\text{C}}t+\frac{1}{2}{a}_{\text{C}}{t}^2\\ a_{\text{S}}{t}^2=2\Delta x_1+2v_{\text{C}}t+a_{\text{C}}{t}^2\end{array}$ (XI)

Write the expression for the speed of the Carlos’s sled when Shannon just begin to move.

$\begin{array}{c}{v}_{\text{C}}^2=2a_{\text{C}}\Delta x_1\\ v_{\text{C}}=\sqrt{2a_{\text{C}}\Delta x_1}\end{array}$ (XII)

Use equation (XII) in (IX).

$\begin{array}{c}{a}_{\text{S}}{t}^2=2\Delta x_1+2\left(\sqrt{2a_{\text{C}}\Delta x_1}\right)t+a_{\text{C}}{t}^2\\ 0=\left(a_{\text{S}}-a_{\text{C}}\right)t^2-t\sqrt{8a_{\text{C}}\Delta x_1}-2\Delta x_1\end{array}$ (XIII)

Solve the quadratic equation (XIII) for $t$.

$\begin{array}{l}0=\left(a_{\text{S}}-a_{\text{C}}\right)t^2-t\sqrt{8a_{\text{C}}\Delta x_1}-2\Delta x_1\\ t=\frac{\sqrt{8a_{\text{C}}\Delta x_1}\pm \sqrt{8a_{\text{C}}\Delta x_1+4\left(a_{\text{S}}-a_{\text{C}}\right)\left(2\Delta x_1\right)}}{2\left(a_{\text{S}}-a_{\text{C}}\right)}\end{array}$ (XIV)

Using equation (VII) write the expression for the acceleration of Carlos and Shannon.

$a_{\text{C}}=g\left(\sin \theta -{\mu }_{\text{C}}\cos \theta \right)$

$a_{\text{S}}=g\left(\sin \theta -{\mu }_{\text{S}}\cos \theta \right)$

Here, ${\mu }_{\text{C}}$ is the coefficient of friction of Carlos’s sled, and ${\mu }_{\text{S}}$ is the coefficient of friction of Shannon’s sled.

Deduce $\left(a_{\text{S}}-a_{\text{C}}\right)$ using the expressions of $a_{\text{S}}$ and $a_{\text{C}}$.

$\begin{array}{c}\left(a_{\text{S}}-a_{\text{C}}\right)=g\left(\sin \theta -{\mu }_{\text{S}}\cos \theta \right)-g\left(\sin \theta -{\mu }_{\text{C}}\cos \theta \right)\\ =g\cos \theta \left({\mu }_{\text{C}}-{\mu }_{\text{S}}\right)\end{array}$ (XV)

Use equation (XV) and the expression for $a_{\text{C}}$ in equation (XIV).

$t=\frac{\sqrt{8g\left(\sin \theta -{\mu }_{\text{C}}\cos \theta \right)\Delta x_1}\pm \sqrt{8g\left(\sin \theta -{\mu }_{\text{C}}\cos \theta \right)\Delta x_1+4g\cos \theta \left({\mu }_{\text{C}}-{\mu }_{\text{S}}\right)\left(2\Delta x_1\right)}}{2g\cos \theta \left({\mu }_{\text{C}}-{\mu }_{\text{S}}\right)}$ (XVI)

Write the expression for the distance travelled by Shannon when she meet Carlos in terms of acceleration of Shannon.

$\Delta x_{\text{S}}=\frac{1}{2}{a}_{\text{S}}{t}^2$ (XVII)

Use the expression for $a_{\text{S}}$ in equation (XVII).

$\Delta x_{\text{S}}=\frac{1}{2}g\left(\sin \theta -{\mu }_{\text{S}}\cos \theta \right)t^2$ (XVIII)

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $12.0°$ for $\theta$, $0.10$ for ${\mu }_{\text{C}}$, $0.010$ for ${\mu }_{\text{S}}$, and $5.0\text{m}$ for $\Delta x_1$ in equation (XVI) to find $t$.

$\begin{array}{c}t=\frac{\sqrt{8\left(9.80{\text{m/s}}^2\right)\left(\sin 12.0°-0.10\cos 12.0°\right)\left(5.0\text{m}\right)}\pm \sqrt{\begin{array}{l}8\left(9.80{\text{m/s}}^2\right)\left(\sin 12.0°-0.10\cos 12.0°\right)\left(5.0\text{m}\right)+\\ 4\left(9.80{\text{m/s}}^2\right)\left(\cos 12.0°\right)\left(0.10-0.010\right)\left(2\times 5.0\text{m}\right)\end{array}}}{2\left(9.80{\text{m/s}}^2\right)\left(\cos 12.0°\right)\left(0.10-0.010\right)}\\ \approx 8.915\text{s}\end{array}$

The negative value obtained as the solution of $t$ is extraneous and is neglected.

Substitute $9.80{\text{m/s}}^2$ for $g$, $12.0°$ for $\theta$, $0.010$ for ${\mu }_{\text{S}}$, and $8.915\text{s}$ for $t$ in equation (XVIII) to find $\Delta x_{\text{S}}$.

$\begin{array}{c}\Delta x_{\text{S}}=\frac{1}{2}\left(9.80{\text{m/s}}^2\right)\left(\sin 12.0°-0.010\cos 12.0°\right){\left(8.915\text{s}\right)}^2\\ =77\text{m}\end{array}$

Therefore, the distance travelled by Carlos and Shannon when Shannon catches up to Carlos is $\underset{\_}{77\text{m}}$.

(b)

To determine

The speed with which Shannon moving with respect to Carlos as she pass by.

Answer to Problem 172P

The speed with which Shannon moving with respect to Carlos as she pass by is $\underset{\_}{4.4\text{m/s}}$.

Explanation of Solution

Write the expression for the relative speed between Shannon and Carlos.

$\begin{array}{c}{v}_{\text{SC}}=a_{\text{S}}t-\left(v_{\text{C}}+a_{\text{C}}t\right)\\ =\left(a_{\text{S}}-a_{\text{C}}\right)t-v_{\text{C}}\end{array}$ (XIX)

Here, $v_{\text{SC}}$ is the speed of Shannon relative to Carlos.

Use equation (XII) and (XV), and the expression for $a_{\text{C}}$ in (XIX).

$v_{\text{SC}}=g\cos \theta \left({\mu }_{\text{C}}-{\mu }_{\text{S}}\right)t-\sqrt{2g\left(\sin \theta -{\mu }_{\text{C}}\cos \theta \right)\Delta x_1}$ (XX)

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $12.0°$ for $\theta$, $0.10$ for ${\mu }_{\text{C}}$, $0.010$ for ${\mu }_{\text{S}}$,$8.915\text{s}$ for $t$, and $5.0\text{m}$ for $\Delta x_1$ in equation (XX) to find $v_{\text{SC}}$.

$\begin{array}{c}{v}_{\text{SC}}=\left(9.80{\text{m/s}}^2\right)\left(\cos 12.0°\right)\left(0.10-0.010\right)\left(8.915\text{s}\right)-\sqrt{2\left(9.80{\text{m/s}}^2\right)\left(\sin 12.0°-0.10\cos 12.0°\right)\left(5.0\text{m}\right)}\\ =4.4\text{s}\end{array}$

Therefore, the speed with which Shannon moving with respect to Carlos as she pass by is $\underset{\_}{4.4\text{m/s}}$.