Chapter 4, Problem 25P

(a)

To determine

The mass of the sailboat.

Answer to Problem 25P

The mass of the sailboat is $1.3\text{ kg}$ .

Explanation of Solution

Write the mathematical equation of Newton’s second law.

$\Sigma F=ma$

Here, $\Sigma F$ is the magnitude of the sum of the forces on sailboat, $m$ is the mass of the sailboat and $a$ is the acceleration of the sailboat

Rewrite the above equation for $m$ .

$m=\frac{\Sigma F}{a}$ (I)

Conclusion:

Given that the magnitude of the force on the sailboat is $0.375\text{ N}$ and the acceleration of the boat is $0.30{\text{ m/s}}^2$ at $28°$ south of west.

Substitute $0.375\text{ N}$ for $\Sigma F$ and $0.30{\text{ m/s}}^2$ for $a$ in equation (I) to find $m$ .

$\begin{array}{c}m=\frac{0.375\text{ N}}{0.30{\text{ m/s}}^2}\\ =1.3\text{ kg}\end{array}$

Therefore, the mass of the sailboat is $1.3\text{ kg}$ .

(b)

To determine

The new velocity of the boat after the $2.0\text{ s}$ gust of wind.

Answer to Problem 25P

The new velocity of the boat after the $2.0\text{ s}$ gust of wind is $0.90\text{ m/s at }18°\text{ south of west}$ .

Explanation of Solution

The diagram of the motion of the sailboat is shown in figure 1.

Write the equation for the additional velocity given to the boat by the wind.

$v_a=a\Delta t$

Here, $v_a$ is the additional velocity given to the boat by the wind, $a$ is the acceleration of the boat and $\Delta t$ is the time for which the wind gust

Substitute $0.30{\text{ m/s}}^2$ for $a$ and $2.0\text{ s}$ for $\Delta t$ in the above equation to find $v_a$ .

$\begin{array}{c}{v}_a=\left(0.30{\text{ m/s}}^2\right)\left(2.0\text{ s}\right)\\ =0.60\text{ m/s at }28°\text{ south of west}\end{array}$

Find the $x$ and $y$ components of $v_a$ .

$\begin{array}{c}{v}_{ax}=0.60\text{ m/s cos208}°\\ v_{ay}=0.60\text{ m/s sin208}°\end{array}$

The initial velocity of boat is only in $-x$ direction and its value is $0.33\text{ m/s}$ .

Find the $x$ and $y$ components of the new velocity of the boat.

$\begin{array}{c}{v}_x=-0.33\text{ m/s}+0.60\text{ m/s cos208}°\\ v_y=0.60\text{ m/s sin208}°\end{array}$

Here, $v_x$ is the $x$ component of the new velocity and $v_y$ is the $y$ component of the new velocity

Write the equation for the magnitude of the new velocity.

$\left|\overrightarrow{v}\right|=\sqrt{v_x^2+v_y^2}$ (II)

Here, $\left|\overrightarrow{v}\right|$ is the magnitude of the new velocity

Write the equation for the angle the new velocity makes.

$\theta ={\tan }^{-1}\frac{v_y}{v_x}$ (III)

Conclusion:

Substitute $\left(-0.33\text{ m/s}+0.60\text{ m/s cos208}°\right)$ for $v_x$ and $0.60\text{ m/s sin208}°$ for $v_y$ in equation (II) to find $\left|\overrightarrow{v}\right|$ .

$\begin{array}{c}\left|\overrightarrow{v}\right|\sqrt{={\left(-0.33\text{ m/s}+0.60\text{ m/s cos208}°\right)}^2+{\left(0.60\text{ m/s sin208}°\right)}^2}\\ =0.90\text{ m/s}\end{array}$

Substitute $\left(-0.33\text{ m/s}+0.60\text{ m/s cos208}°\right)$ for $v_x$ and $0.60\text{ m/s sin208}°$ for $v_y$ in equation (III) to find $\theta$ .

$\begin{array}{c}\theta ={\tan }^{-1}\frac{0.60\text{ m/s sin208}°}{\left(-0.33\text{ m/s}+0.60\text{ m/s cos208}°\right)}\\ =18°\end{array}$

$\theta$ should be in south of west direction since both the components are negative.

Therefore, the new velocity of the boat after the $2.0\text{ s}$ gust of wind is $0.90\text{ m/s at }18°\text{ south of west}$ .