Chapter 4, Problem 43P

(a)

To determine

The magnitude and direction of the sled’s acceleration.

Answer to Problem 43P

The magnitude of the sled’s acceleration is $\underset{\_}{1.3\text{m}/{\text{s}}^2}$ and direction of the sled’s acceleration is $\underset{\_}{0.47°\text{ below the }x\text{ axis}}$.

Explanation of Solution

Write the expression for the Newton’s second law of motion acting on the surface.

    ${\displaystyle \sum \overrightarrow{F}}=m\overrightarrow{a}$        (I)

Here, $\overrightarrow{F}$ is the force, $m$ is the mass, $\overrightarrow{a}$ is the acceleration.

Since the surface is frictionless, the only forces in the plane of the motion are the children’s force.

Using the figure $1$ and write the expression for the force along the horizontal direction.

    $\begin{array}{c}{\displaystyle \sum F_x}=m{a}_x\\ =15\text{Ncos30}°\text{+10}\text{Ncos50}°\\ =19.4\text{N}\end{array}$        (II)

Here, ${\displaystyle \sum F_x}$ is the force along the horizontal direction, $a_x$ is the acceleration along the horizontal direction.

Using the figure $1$ and write the expression for the force along the vertical direction.

    $\begin{array}{c}{\displaystyle \sum F_y}=m{a}_y\\ =15\text{Nsin30}°-\text{10}\text{Nsin50}°\\ =-0.160\text{N}\end{array}$        (III)

Here, ${\displaystyle \sum F_y}$ is the force along the vertical direction, $a_y$ is the acceleration along the vertical direction.

Use equation (II) to solve for $a_x$.

    $a_x=\frac{19.4\text{N}}{m}$        (IV)

Use equation (III) to solve for $a_y$.

    $a_y=\frac{-0.160\text{N}}{m}$        (V)

Write the expression for the magnitude of the acceleration.

    $\left|\overrightarrow{a}\right|=\sqrt{a_x^2+a_y^2}$        (VI)

Here, $\left|\overrightarrow{a}\right|$ is the magnitude of acceleration.

Write the expression for the direction of the acceleration.

    $\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$        (VII)

Conclusion:

Substitute $15\text{kg}$ for $m$ in equation (IV) to find $a_x$.

    $\begin{array}{c}{a}_x=\frac{19.4\text{N}}{15\text{kg}}\\ =1.29\text{m}/{\text{s}}^2\end{array}$

Substitute $15\text{kg}$ for $m$ in equation (V) to find $a_y$.

    $\begin{array}{c}{a}_y=\frac{-0.160\text{N}}{15\text{kg}}\\ =-1.07\times {10}^{-2}\text{m}/{\text{s}}^2\end{array}$

Substitute $1.29\text{m}/{\text{s}}^2$ for $a_x$, $-1.07\times {10}^{-2}\text{m}/{\text{s}}^2$ for $a_y$ in equation (VI) to find $\left|\overrightarrow{a}\right|$.

    $\begin{array}{c}\left|\overrightarrow{a}\right|=\sqrt{{\left(1.29\text{m}/{\text{s}}^2\right)}^2+{\left(-1.07\times {10}^{-2}\text{m}/{\text{s}}^2\right)}^2}\\ =1.3\text{m}/{\text{s}}^2\end{array}$

Substitute $1.29\text{m}/{\text{s}}^2$ for $a_x$, $-1.07\times {10}^{-2}\text{m}/{\text{s}}^2$ for $a_y$ in equation (VII) to find $\theta$.

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-1.07\times {10}^{-2}\text{m}/{\text{s}}^2}{1.29\text{m}/{\text{s}}^2}\right)\\ =-0.48°\end{array}$

Below the $x$ axis.

Therefore, the magnitude of the sled’s acceleration is $\underset{\_}{1.3\text{m}/{\text{s}}^2}$ and direction of the sled’s acceleration is $\underset{\_}{0.47°\text{ below the }x\text{ axis}}$.

(b)

To determine

The time taken by the sled to reach a speed of $10\text{m}/\text{s}$.

Answer to Problem 43P

The time taken by the sled to reach a speed of $10\text{m}/\text{s}$ is $\underset{\_}{7.7\text{s}}$.

Explanation of Solution

Write the expression for the equation of motion.

    $v=v_0+at$        (VIII)

Here, $v$ is the final velocity, $v_0$ is the initial velocity, $t$ is the time.

The sled is initial at rest with velocity $v_0=0$.

Use equation (VIII) to solve for $t$.

    $\begin{array}{c}v=at\\ t=\frac{v}{a}\end{array}$        (IX)

Conclusion:

Substitute $10\text{m}/\text{s}$ for $v$, $1.3\text{m}/{\text{s}}^2$ for $a$ in equation (IX) to find $t$.

    $\begin{array}{c}t=\frac{10\text{m}/\text{s}}{1.3\text{m}/{\text{s}}^2}\\ =7.7\text{s}\end{array}$

Therefore, the time taken by the sled to reach a speed of $10\text{m}/\text{s}$ is $\underset{\_}{7.7\text{s}}$.