Chapter 4, Problem 85P

(a)

To determine

The distance travelled by the car before it comes to a stop.

Answer to Problem 85P

The distance travelled by the car before it comes to a stop is $\underset{\_}{58\text{m}}$.

Explanation of Solution

Write the expression for the Newton’s second law along horizontal axis.

    $\begin{array}{c}{\displaystyle \sum F_x}=-F_{\text{friction}}\\ =m{a}_x\end{array}$        (I)

Here, ${\displaystyle \sum F_x}$ is the force along horizontal direction, $F_{\text{friction}}$ is the frictional force, $m$ is the mass, $a_x$ is the acceleration along $x$ direction.

Write the expression for the Newton’s second law along vertical direction.

    $\begin{array}{c}{\displaystyle \sum F_y}=N-mg\\ =0\\ N=mg\end{array}$        (II)

Here, $N$ is the normal force.

Use equation (II) to write the expression for $F_{\text{friction}}$.

    $\begin{array}{c}{F}_{\text{friction}}={\mu }_{\text{k}}N\\ ={\mu }_{\text{k}}mg\end{array}$        (III)

Use equation (III) in (I) to solve for $a_x$.

    $\begin{array}{c}{a}_x=\frac{-F_{\text{friction}}}{m}\\ =-{\mu }_{\text{k}}g\end{array}$        (IV)

Write the expression for the kinematics equation to solve for the distance travelled by the car before it comes to a stop.

    $\begin{array}{c}{v}^2-v_0^2=2ax\\ x=\frac{v^2-v_0^2}{2a}\end{array}$        (V)

Here, $x$ is the distance travelled by the car, $v_0$ is the initial velocity, $v$ is the final velocity.

As the care comes to a stop, the final velocity $v=0\text{m}/\text{s}$.

Use equation (IV) in (V) to solve for $x$.

    $x=\frac{v_0^2}{2{\mu }_{\text{k}}g}$        (VI)

Conclusion:

Substitute $25\text{m}/\text{s}$ for $v_0$, $0.55$ for ${\mu }_{\text{k}}$, $9.8{\text{m}/\text{s}}^2$ for $g$ in equation (VI) to find $x$.

    $\begin{array}{c}x=\frac{{\left(25\text{m}/\text{s}\right)}^2}{2\left(0.55\right)\left(9.8{\text{m}/\text{s}}^2\right)}\\ =58\text{m}\end{array}$

Therefore, the distance travelled by the car before it comes to a stop is $\underset{\_}{58\text{m}}$.

(b)

To determine

As the same car is moving downhill along the road makes an angle, the distance travelled by the car before comes to a stop.

Answer to Problem 85P

As the same car is moving downhill along the road makes an angle, the distance travelled by the car before comes to a stop is $\underset{\_}{97\text{m}}$.

Explanation of Solution

Write the expression for the Newton’s second law along horizontal axis.

    $\begin{array}{c}{\displaystyle \sum F_x}=mg\sin \theta -F_{\text{friction}}\\ =m{a}_x\end{array}$        (VII)

Here, ${\displaystyle \sum F_x}$ is the force along horizontal direction, $F_{\text{friction}}$ is the frictional force, $m$ is the mass, $a_x$ is the acceleration along $x$ direction, $g$ is the acceleration due to gravity, $\theta$ is the inclination angle.

Write the expression for the Newton’s second law along vertical direction.

    $\begin{array}{c}{\displaystyle \sum F_y}=N-mg\cos \theta \\ =0\\ N=mg\cos \theta \end{array}$        (VIII)

Here, $N$ is the normal force.

Use equation (VIII) to write the expression for $F_{\text{friction}}$.

    $\begin{array}{c}{F}_{\text{friction}}={\mu }_{\text{k}}N\\ ={\mu }_{\text{k}}mg\cos \theta \end{array}$        (IX)

Use equation (IX) in (VII) to solve for $a_x$.

    $\begin{array}{c}{a}_x=\frac{mg\sin \theta -F_{\text{friction}}}{m}\\ =g\sin \theta -{\mu }_{\text{k}}g\cos \theta \end{array}$        (X)

As the care comes to a stop, the final velocity $v=0\text{m}/\text{s}$.

Write the expression for the kinematics equation to solve for the distance travelled by the car before it comes to a stop.

    $\begin{array}{c}{v}^2-v_0^2=2ax\\ x=\frac{v^2-v_0^2}{2g\left(\sin \theta -{\mu }_{\text{k}}\cos \theta \right)}\\ =\frac{v_0^2}{2g\left({\mu }_{\text{k}}\cos \theta -\sin \theta \right)}\end{array}$        (XI)

Here, $x$ is the distance travelled by the car, $v_0$ is the initial velocity, $v$ is the final velocity.

Conclusion:

Substitute $25\text{m}/\text{s}$ for $v_0$, $0.55$ for ${\mu }_{\text{k}}$, $9.8{\text{m}/\text{s}}^2$ for $g$, $12°$ for $\theta$ in equation (XI) to find $x$.

    $\begin{array}{c}x=\frac{{\left(25\text{m}/\text{s}\right)}^2}{2\left(9.8{\text{m}/\text{s}}^2\right)\left[\left(0.55\right)\cos 12°-\sin 12°\right]}\\ =97\text{m}\end{array}$

Almost double the stopping distance.

Therefore, the same car is moving downhill along the road makes an angle, the distance travelled by the car before comes to a stop is $\underset{\_}{97\text{m}}$.

(c)

To determine

As the same car is moving uphill along the road makes an angle, the distance travelled by the car before comes to a stop.

Answer to Problem 85P

As the same car is moving uphill along the road makes an angle, the distance travelled by the car before comes to a stop is $\underset{\_}{43\text{m}}$.

Explanation of Solution

Write the expression for the Newton’s second law along horizontal axis.

    $\begin{array}{c}{\displaystyle \sum F_x}=-mg\sin \theta -F_{\text{friction}}\\ =m{a}_x\end{array}$        (VII)

Write the expression for the Newton’s second law along vertical direction.

    $\begin{array}{c}{\displaystyle \sum F_y}=N-mg\cos \theta \\ =0\\ N=mg\cos \theta \end{array}$        (VIII)

Use equation (VIII) to write the expression for $F_{\text{friction}}$.

    $\begin{array}{c}{F}_{\text{friction}}={\mu }_{\text{k}}N\\ ={\mu }_{\text{k}}mg\cos \theta \end{array}$        (IX)

Use equation (IX) in (VII) to solve for $a_x$.

    $\begin{array}{c}{a}_x=\frac{-mg\sin \theta -F_{\text{friction}}}{m}\\ =-\left(g\sin \theta +{\mu }_{\text{k}}g\cos \theta \right)\end{array}$        (X)

As the care comes to a stop, the final velocity $v=0\text{m}/\text{s}$.

Write the expression for the kinematics equation to solve for the distance travelled by the car before it comes to a stop.

    $\begin{array}{c}{v}^2-v_0^2=2ax\\ x=\frac{v^2-v_0^2}{-2\left(g\sin \theta +{\mu }_{\text{k}}g\cos \theta \right)}\\ =\frac{v_0^2}{2\left(g\sin \theta +{\mu }_{\text{k}}g\cos \theta \right)}\end{array}$        (XI)

Conclusion:

Substitute $25\text{m}/\text{s}$ for $v_0$, $0.55$ for ${\mu }_{\text{k}}$, $9.8{\text{m}/\text{s}}^2$ for $g$, $12°$ for $\theta$ in equation (XI) to find $x$.

    $\begin{array}{c}x=\frac{{\left(25\text{m}/\text{s}\right)}^2}{2\left(9.8{\text{m}/\text{s}}^2\right)\left[\left(0.55\right)\cos 12°+\sin 12°\right]}\\ =43\text{m}\end{array}$

Therefore, the same car is moving uphill along the road makes an angle, the distance travelled by the car before comes to a stop is $\underset{\_}{43\text{m}}$.