Chapter 5, Problem 5.15P

Two forces, ${\stackrel{\to }{F}}_1=(-6.00\hat{i}-4.00\hat{j})N$ and ${\stackrel{\to }{F}}_2=(-3.00\hat{i}+7.00\hat{j})N$, act on a particle of mass 2.00 kg that is initially at rest at coordinates (−2.00 m, + 4.00 m). (a) What are the components of the particle’s velocity at t = 10.0 s? (b) In what direction is the particle moving at t = 10.0 s? (c) What displacement does the particle undergo during the first 10.0 s? (d) What are the coordinates of the particle at t = 10.0 s?

To determine

The components of the velocity of the particle at $t=10.0\sec$.

Answer to Problem 5.15P

The $x$ components of the particle’s velocity at $t=10.0\sec$ is $-45.0\text{m}/\text{s}$ and the $y$ component of particle’s velocity at $t=10\text{s}$ is $15.0\text{m}/\text{s}$.

Explanation of Solution

The mass of the particle is $2.00\text{kg}$ and the two forces acting on it are ${\overrightarrow{\text{F}}}_{\text{1}}=\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}$ and ${\overrightarrow{\text{F}}}_2=\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}$. The particle is initially at rest at $\left(-2.00\text{m, +4}\text{.00}\text{m}\right)$.

Write the formula to calculate net force acts on a particle

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

Here, ${\overrightarrow{\text{F}}}_{\text{net}}$ is the net force acting on a particle, ${\overrightarrow{\text{F}}}_1$ and ${\overrightarrow{F}}_{\text{2}}$ are the two given forces.

Write the formula to calculate acceleration of a particle

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

Here, $\overrightarrow{a}$ is the acceleration of the particle and $m$ is the mass of the particle.

Substitute ${\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2$ for ${\overrightarrow{\text{F}}}_{\text{net}}$ in above equation.

    $\overrightarrow{a}=\frac{{\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2}{m}$

Substitute $\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_{\text{1}}$, $\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}$ for ${\overrightarrow{\text{F}}}_2$ and $2.00\text{kg}$ for $m$ to find $\overrightarrow{a}$.

    $\begin{array}{c}\overrightarrow{a}=\frac{\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}+\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}}{2.00\text{kg}}\\ =\left(-4.50\widehat{i}+1.50\widehat{j}\right){\text{m/s}}^2\end{array}$

Write the formula to calculate the velocity of the particle

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

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

Conclusion:

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_{\text{i}}$, $\left(-4.50\widehat{i}+1.50\widehat{j}\right){\text{m/s}}^2$ for $\overrightarrow{a}$ and $10.0\text{s}$ for $t$ to find ${\overrightarrow{\text{v}}}_{\text{f}}$.

    $\begin{array}{c}{\overrightarrow{\text{v}}}_{\text{f}}=0+\left(-4.50\widehat{i}+1.50\widehat{j}\right){\text{m/s}}^2\times \left(10.0\text{s}\right)\\ =\left(-45.0\widehat{i}+15.0\widehat{j}\right)\text{m}/\text{s}\end{array}$

Therefore, the $x$ components of the particle’s velocity at $t=10.0\sec$ is $-45.0\text{m}/\text{s}$ and the $y$ component of particle’s velocity at $t=10\text{s}$ is $15.0\text{m}/\text{s}$.

To determine

The direction of the motion of the particle at $t=10.0\sec$.

Answer to Problem 5.15P

The direction of the particle’s motion at $t=10.0\sec$ is $162°$ counterclockwise from the $+x$ axis.

Explanation of Solution

Write the formula to calculate the direction of the moving particle

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

Conclusion:

Substitute $15.0\text{m}/\text{s}$ for $v_{\text{y}}$ and $-45.0\text{m}/\text{s}$ for $v_{\text{x}}$ to calculate $\theta$.

    $\begin{array}{c}\tan \theta =\left(\frac{15\text{m}/\text{s}}{-45\text{m}/\text{s}}\right)\\ \theta \approx 162°\end{array}$

Therefore, the direction of particle’s motion at $t=10.0\sec$ is $162°$ counterclockwise from the $+x$ axis.

To determine

The displacement of the particle during $10\sec$.

Answer to Problem 5.15P

The displacement of the particle during $10\sec$ is $\left(-225\widehat{i}+75\widehat{j}\right)\text{m}$.

Explanation of Solution

Write the formula to calculate the displacement of the particle

    $\overrightarrow{\text{s}}={\overrightarrow{\text{v}}}_{\text{i}}t+\frac{1}{2}\overrightarrow{a}{t}^2$

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_{\text{i}}$, $\left(-4.50\widehat{i}+1.50\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ and $10.0\text{s}$ for $t$ to find $\overrightarrow{\text{s}}$.

    $\begin{array}{c}\overrightarrow{\text{s}}=0\times t+\frac{1}{2}\left(-4.50\widehat{i}+1.50\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}{\left(10.0\text{s}\right)}^2\\ =\left(-255\widehat{i}+75.0\widehat{j}\right)\text{m}\end{array}$

Conclusion:

Therefore, the displacement of the particle during $10\sec$ is $\left(-255\widehat{i}+75.0\widehat{j}\right)\text{m}$.

To determine

The final coordinates of the particle at $t=10.0\sec$.

Answer to Problem 5.15P

The final coordinates of the particle at $t=10.0\sec$ is $\left(-227\widehat{i}+79.0\widehat{j}\right)\text{m}$.

Explanation of Solution

The initial position of the particle is $\left(-2.00\widehat{i}+4.00\widehat{j}\right)\text{m}$.

Write the expression for the final position coordinates.

  ${\overrightarrow{r}}_{\text{f}}={\overrightarrow{r}}_{\text{i}}+\overrightarrow{s}$

Here, ${\overrightarrow{r}}_{\text{i}}$ is the initial position and ${\overrightarrow{r}}_{\text{f}}$ is the final position coordinate

Conclusion:

Substitute $\left(-2.00\widehat{i}+4.00\widehat{j}\right)\text{m}$ for ${\overrightarrow{r}}_{\text{i}}$ and $\left(-255\widehat{i}+75.0\widehat{j}\right)\text{m}$ for $\overrightarrow{s}$ in the above equation to find ${\overrightarrow{r}}_{\text{f}}$

$\begin{array}{c}{\overrightarrow{r}}_{\text{f}}=\left(-255\widehat{i}+75.0\widehat{j}\right)\text{m}+\left(-2.00\widehat{i}+4.00\widehat{j}\right)\text{m}\\ =\left(-227\widehat{i}+79.0\widehat{j}\right)\text{m}\end{array}$

Therefore, the coordinates of the particle at $t=10.0\sec$ is $\left(-227\widehat{i}+79.0\widehat{j}\right)\text{m}$