Chapter 4, Problem 45P

To determine

The stopping distance of the car measured along the incline.

Answer to Problem 45P

The stopping distance of the car measured along the incline is $\underset{\_}{25\text{m}}$.

Explanation of Solution

Figure 1 represents the free body diagram of car on level ground.

Figure 1 represents the free body diagram of car on the incline .

Write the kinematic equation.

    $v^2=v_0^2+2ax$        (I)

Here, $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration, and $x$ is the initial distance of the car.

Write the expression for net force acting on the car.

${\displaystyle \sum F=ma}$        (II)

Here, $m$ is the mass of the car, and $a$ is the acceleration of the car.

Write the expression for kinetic friction force.

    $F_{\text{friction}}={\mu }_{\text{k}}N$

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

The above equation can be written as

    $F_{\text{friction}}={\mu }_{\text{k}}mg$        (III)

The net force acting on the level car is opposite to the friction force.

    $m{a}_{\text{level}\text{car}}=-{\mu }_{\text{k}}mg$

Rearrange the above equation.

    $\begin{array}{c}m{a}_{\text{level}\text{car}}=-{\mu }_{\text{k}}mg\\ {\mu }_{\text{k}}=\frac{-a_{\text{level}\text{car}}}{g}\end{array}$        (IV)

Since the car on the level ground is at rest, equation (I) is written as

    $\begin{array}{c}{v}^2=v_0^2+2a_{\text{level car}}{x}_{\text{level car}}\\ a_{\text{level car }}=\frac{-v_0^2}{2x_{\text{level car }}}\end{array}$

Use the above equation in equation (IV).

    $\begin{array}{c}{\mu }_{\text{k}}=\frac{-\left(\frac{-v_0^2}{2x_{\text{level car }}}\right)}{g}\\ =\frac{v_0^2}{2g{x}_{\text{level car }}}\end{array}$        (V)

Write the acceleration of the object down an incline from equation.

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

Here, $a_{\text{incline}}$ is the acceleration of the incline, $g$ is acceleration due to gravity, ${\mu }_{\text{k}}$ is the coefficient of kinetic friction, and $\theta$ is the angle of inclination.

Use the above equation in equation (V).

    $a_{\text{incline}}=g\left(\sin \theta -\left(\frac{v_0^2}{2g{x}_{\text{level car }}}\right)\cos \theta \right)$        (VI)

Write the kinematic equation of the car in inclined motion.

    $v^2=v_0^2+2a_{\text{incline}}{x}_{\text{incline }}$

Here, $v$ is the final velocity of the car in inclined motion, $v_{\text{o}}$ is the initial velocity of the car in inclined motion, $a_{\text{incline}}$ is the acceleration of the car in incline plane, and $x_{\text{incline }}$ is the stopping distance of the car.

Since the final velocity of the car in inclined motion is zero, the above equation is written as

    $\begin{array}{c}0=v_0^2+2a_{\text{incline}}{x}_{\text{incline }}\\ v_0^2=-2a_{\text{incline}}{x}_{\text{incline }}\\ x_{\text{incline }}=\frac{v_0^2}{-2a_{\text{incline}}}\end{array}$        (VII)

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $8.0°$ for $\theta$, $25\text{m}/\text{s}$ for $v_0$, $22\text{m}$ for $v_{\text{level car }}$ in equation (VI), to find $a_{\text{incline }}$.

    $\begin{array}{c}{a}_{\text{incline}}=\left(9.8\text{m}/{\text{s}}^2\right)\left[\sin 8.0°-\left(\frac{{\left(25\text{m}/\text{s}\right)}^2}{2\left(9.8{\text{m}/\text{s}}^2\right)22\text{m}}\right)\cos 8.0°\right]\\ =-12\text{m}/{\text{s}}^2\end{array}$

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $25\text{m}/\text{s}$ for $v_0$, $-12.7\text{m}/{\text{s}}^2$ for $a_{\text{incline car}}$ in equation (VII), to find $x_{\text{incline }}$.

    $\begin{array}{c}{x}_{\text{incline }}=\frac{-{\left(25\text{m}/\text{s}\right)}^2}{2\left(-12.7\text{m}/{\text{s}}^2\right)}\\ =25\text{m}\end{array}$

Therefore, the stopping distance of the car measured along the incline is $\underset{\_}{25\text{m}}$.