Chapter 5, Problem 5.22P

Review. Three forces acting on an object are given by ${\stackrel{\to }{F}}_1=(-2.00\hat{i}-2.00\hat{j})N$, and ${\stackrel{\to }{F}}_1=(-5.00\hat{i}-3.00\hat{j})N$, and ${\stackrel{\to }{F}}_1=(-45.0\hat{i})N$. The object experiences an acceleration of magnitude 3.75 m/s². (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?

To determine

The direction of the acceleration.

Answer to Problem 5.22P

The direction of the acceleration is $181°$.

Explanation of Solution

The forces acting on the object are ${\overrightarrow{\text{F}}}_1=\left(-2.00\widehat{\text{i}}+2.00\widehat{\text{j}}\right)\text{N}$, ${\overrightarrow{\text{F}}}_2=\left(5.00\widehat{\text{i}}-3.00\widehat{\text{j}}\right)\text{N}$ and ${\overrightarrow{\text{F}}}_3=\left(-45.00\widehat{\text{i}}\right)\text{N}$. The magnitude of acceleration of the object is $3.75\text{m}/{\text{s}}^{\text{2}}$.

Write the formula to calculate net force act on a object

    ${\overrightarrow{\text{F}}}_{\text{net}}={\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2+{\overrightarrow{\text{F}}}_3$

Here, ${\overrightarrow{\text{F}}}_{\text{net}}$ is the net force acting on an object, ${\overrightarrow{\text{F}}}_1$, ${\overrightarrow{\text{F}}}_2$ and are the given forces.

Substitute $\left(-2.00\widehat{\text{i}}+2.00\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_1$, $\left(5.00\widehat{\text{i}}-3.00\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_2$ and $\left(-45.00\widehat{\text{i}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_{\text{3}}$ to find ${\overrightarrow{\text{F}}}_{\text{net}}$.

    $\begin{array}{c}{\overrightarrow{\text{F}}}_{\text{net}}=\left(-2.00\widehat{\text{i}}+2.00\widehat{\text{j}}\right)\text{N}+\left(5.00\widehat{\text{i}}-3.00\widehat{\text{j}}\right)\text{N}+\left(-45.00\widehat{\text{i}}\right)\text{N}\\ =\left(-42.0\widehat{\text{i}}-1.00\widehat{\text{j}}\right)\text{N}\end{array}$

Write the formula to calculate direction of force

    $\tan \theta =\left(\frac{F_{\text{y}}}{F_{\text{x}}}\right)$

Here, $F_{\text{y}}$ is the y-component of the force and $F_{\text{x}}$ is the x-component of the force.

Conclusion:

Substitute $-42.0\text{N}$ for $F_{\text{x}}$ and $-1.00\text{N}$ for $F_{\text{y}}$ in the above equation to calculate $\theta$.

    $\begin{array}{c}\tan \theta =\left(\frac{-1.00\text{N}}{-42.0\text{N}}\right)\\ \theta =1.36°\end{array}$

The direction of force is equal to the direction of acceleration of object and the value of $\theta$ lies at third quadrant so the direction of acceleration with $+x$ axis

    $\begin{array}{c}\theta =180°+1.36°\\ =181.36°\\ \approx 181°\end{array}$

Therefore, the direction of the acceleration is $181°$.

To determine

The mass of the object.

Answer to Problem 5.22P

The mass of the object is $11.2\text{kg}$.

Explanation of Solution

Write the formula to calculate magnitude of net force act of an object

    $F=\sqrt{F_{\text{x}}{}^2+F_{\text{y}}{}^2}$

Write the formula to calculate mass of the object

    $m=\frac{F}{a}$

Here, $a$ is the magnitude of acceleration of an object, $m$ is the mass of the object and $F$ is the magnitude of net force acting on the object.

Substitute $\sqrt{F_{\text{x}}{}^2+F_{\text{y}}{}^2}$ for $F$ in the above equation.

    $m=\frac{\sqrt{F_{\text{x}}{}^2+F_{\text{y}}{}^2}}{a}$

Conclusion:

Substitute $-42.0\text{N}$ for $F_{\text{x}}$ and $-1.00\text{N}$ for $F_{\text{y}}$  and $3.75\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to find $m$.

    $\begin{array}{c}m=\frac{\sqrt{{\left(-42.0\text{N}\right)}^2+{\left(1.00\right)}^2}}{3.75\text{m}/{\text{s}}^{\text{2}}}\\ =11.2\text{kg}\end{array}$

Therefore, the mass of the object is $11.2\text{kg}$.

To determine

The speed of the object after $10\sec$.

Answer to Problem 5.22P

The speed of the object after $10\sec$ is $37.5\text{m}/\text{s}$.

Explanation of Solution

Write the formula to calculate speed of an object

    $v_{\text{f}}=v_{\text{i}}+at$

Here, $v_{\text{f}}$ is the final speed of an object, $v_{\text{i}}$ is the initial speed of an object and $t$ is time.

Conclusion:

Substitute $0$ for $v_{\text{i}}$, $10\sec$ for $t$ and $3.75\text{m}/{\text{s}}^{\text{2}}$ for $a$ to find $v_{\text{f}}$.

    $\begin{array}{c}{v}_{\text{f}}=0+3.75\text{m}/{\text{s}}^{\text{2}}\times 10\sec \\ =37.5\text{m}/\text{s}\end{array}$

Therefore, the speed of the object after $10\sec$ is $37.5\text{m}/\text{s}$.

To determine

The velocity components of the object after $10\sec$.

Answer to Problem 5.22P

The $x$ and $y$ components of the velocity are $-37.5\text{m}/\text{s}$ and $0.893\text{m}/\text{s}$ respectively.

Explanation of Solution

Write the formula to calculate velocity of an object

    ${\overrightarrow{\text{v}}}_{\text{f}}={\overrightarrow{\text{v}}}_{\text{i}}+\overrightarrow{a}t$                          (I)

Here, ${\overrightarrow{\text{v}}}_{\text{f}}$ is the final velocity and ${\overrightarrow{\text{v}}}_{\text{i}}$ is the initial velocity.

Write the formula to calculate mass of the object

    $\overrightarrow{a}=\frac{{\overrightarrow{\text{F}}}_{\text{net}}}{m}$

Substitute $\frac{{\overrightarrow{\text{F}}}_{\text{net}}}{m}$ for $\overrightarrow{a}$ in equation (I).

    ${\overrightarrow{\text{v}}}_{\text{f}}={\overrightarrow{\text{v}}}_{\text{i}}+\frac{{\overrightarrow{\text{F}}}_{\text{net}}}{m}t$

Substitute $\left(-42.0\widehat{\text{i}}-1.00\widehat{\text{j}}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_{\text{net}}$, $11.2\text{kg}$ for $m$, $0$ for ${\overrightarrow{\text{v}}}_{\text{i}}$ and $10\sec$ for $t$ to find ${\overrightarrow{\text{v}}}_{\text{f}}$.

    $\begin{array}{c}{\overrightarrow{\text{v}}}_{\text{f}}=0+\frac{\left(-42.0\widehat{\text{i}}-1.00\widehat{\text{j}}\right)\text{N}}{11.2\text{kg}}10\sec \\ =\left(-37.5\widehat{\text{i}}-0.893\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, $x$ and $y$ components of the velocity are $-37.5\text{m}/\text{s}$ and $0.893\text{m}/\text{s}$ respectively.