Chapter 4, Problem 6P

(a)

To determine

To determine: The direction of the acceleration.

Answer to Problem 6P

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 $181.36°$.

(b)

To determine

To determine: The mass of the object.

Answer to Problem 6P

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 $11.2\text{kg}$.

(c)

To determine

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

Answer to Problem 6P

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 $37.5\text{m}/\text{s}$.

(d)

To determine

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

Answer to Problem 6P

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.