Chapter 5, Problem 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

To determine: The direction of the acceleration.

Answer to Problem 22P

Answer: The direction of the acceleration is $181.36°$.

Explanation of Solution

Explanation:

Given information:

Three forces acting on a 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 object experiences an acceleration of magnitude $3.75\text{m}/{\text{s}}^{\text{2}}$.

Formula to calculate net force act on a object is,

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

• ${\overrightarrow{\text{F}}}_{\text{net}}$ is the net force acting on a object.

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 ${\overrightarrow{\text{F}}}_{\text{net}}$.

$\begin{array}{c}{\overrightarrow{\text{F}}}_{\text{net}}=\left(-2.00\widehat{\text{i}}+2.00\widehat{\text{j}}\right)\text{N}+\left(5.00\widehat{\text{i}}-3.00\widehat{\text{j}}\right)\text{N}+\left(-45.00\widehat{\text{i}}\right)\text{N}\\ =\left(-42\widehat{\text{i}}-1\widehat{\text{j}}\right)\text{N}\end{array}$

Formula to calculate direction of force is,

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

Substitute $-42\text{N}$ for $F_{\text{x}}$ and $-1\text{N}$ for $F_{\text{y}}$ to calculate $\theta$.

$\begin{array}{c}\tan \theta =\left(\frac{-1\text{N}}{-42\text{N}}\right)\\ \theta =1.36°\end{array}$

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 is,

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

Conclusion:

Therefore, the direction of the acceleration is $181.36°$.

To determine

To determine: The mass of the object.

Answer to Problem 22P

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

Explanation of Solution

Explanation:

Given information:

Three forces acting on a 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 object experiences an acceleration of magnitude $3.75\text{m}/{\text{s}}^{\text{2}}$.

Formula to calculate magnitude of net force act of an object is,

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

Formula to calculate mass of the object is,

$m=\frac{F}{a}$

• $a$ is the magnitude of acceleration of an object.
• $m$ is the mass of the object.
• $F$ 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.

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

Substitute $-42\text{N}$ for $F_{\text{x}}$ and $-1\text{N}$ for $F_{\text{y}}$  and $3.75\text{m}/{\text{s}}^{\text{2}}$ for $a$ to find $m$.

$\begin{array}{c}m=\frac{\sqrt{{\left(-42\text{N}\right)}^2+{\left(1\right)}^2}}{3.75\text{m}/{\text{s}}^{\text{2}}}\\ =11.2\text{kg}\end{array}$

Conclusion:

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

To determine

To determine: The speed of an object after $10\sec$.

Answer to Problem 22P

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

Explanation of Solution

Explanation:

Given information:

Three forces acting on a 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 object experiences an acceleration of magnitude $3.75\text{m}/{\text{s}}^{\text{2}}$.

Formula to calculate speed of an object is,

$v_{\text{f}}=v_{\text{i}}+at$

• $v_{\text{f}}$ is the final velocity of an object.
• $v_{\text{i}}$ is the initial speed of an object.
• $t$ is time.

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

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

Conclusion:

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

To determine

To determine: The velocity components of an object after $10\sec$.

Answer to Problem 22P

Answer: 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

Explanation:

Given information:

Three forces acting on a 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 object experiences an acceleration of magnitude $3.75\text{m}/{\text{s}}^{\text{2}}$.

Formula to calculate velocity of an object is,

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

Formula to calculate mass of the object is,

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

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 $\left(-42\widehat{\text{i}}-1\widehat{\text{j}}\right)\text{N}$ 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 ${\overrightarrow{\text{v}}}_{\text{f}}$.

$\begin{array}{c}{\overrightarrow{\text{v}}}_{\text{f}}=0+\frac{\left(-42\widehat{\text{i}}-1\widehat{\text{j}}\right)\text{N}}{11.2\text{kg}}10\sec \\ =\left(-37.5\widehat{\text{i}}-0.893\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Conclusion:

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