Chapter 5, Problem 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

To determine: The components of the particle’s velocity at $t=10.0\sec$.

Answer to Problem 15P

Answer: The $x$ components of the particle’s velocity at $t=10.0\sec$ is $-45\text{m}/\text{s}$ and the $y$ component of particle’s velocity at $t=10\text{s}$ is $15\text{m}/\text{s}$.

Explanation of Solution

Explanation:

Given information:

Two forces ${\overrightarrow{\text{F}}}_{\text{1}}=\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}$ and ${\overrightarrow{\text{F}}}_2=\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}$, act on a particle of mass $2.00\text{kg}$ that is initially at rest at coordinates $\left(-2.00\text{m, +4}\text{.00}\text{m}\right)$.

Formula to calculate net force acts on a particle is,

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

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

Formula to calculate acceleration of a particle is,

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

• $\overrightarrow{a}$ is the acceleration of the particle.
• $m$ 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 $\overrightarrow{a}$.

To determine

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

Answer to Problem 15P

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

Explanation of Solution

Explanation:

Given information:

Two forces ${\overrightarrow{\text{F}}}_{\text{1}}=\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}$ and ${\overrightarrow{\text{F}}}_2=\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}$, act on a particle of mass $2.00\text{kg}$ that is initially at rest at coordinates $\left(-2.00\text{m, +4}\text{.00}\text{m}\right)$.

Formula to calculate the direction of the moving particle is,

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

Substitute $15\text{m}/\text{s}$ for $v_{\text{y}}$ and $-45\text{m}/\text{s}$ for $v_{\text{x}}$ to calculate $\theta$.

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

Conclusion:

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

To determine

To determine: The displacement of the particle during $10\sec$.

Answer to Problem 15P

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

Explanation of Solution

Explanation:

Given information:

Two forces ${\overrightarrow{\text{F}}}_{\text{1}}=\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}$ and ${\overrightarrow{\text{F}}}_2=\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}$, act on a particle of mass $2.00\text{kg}$ that is initially at rest at coordinates $\left(-2.00\text{m, +4}\text{.00}\text{m}\right)$.

Formula to calculate the displacement of the particle is,

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

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_{\text{i}}$, $\left(-4.5\widehat{i}+1.5\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{a}$ and $10.0\text{s}$ for $t$ to find $\overrightarrow{\text{s}}$.

$\begin{array}{c}\overrightarrow{\text{s}}=0\times t+\frac{1}{2}\left(-4.5\widehat{i}+1.5\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}{\left(10.0\text{s}\right)}^2\\ =\left(-255\widehat{i}+75\widehat{j}\right)\text{m}\end{array}$

Conclusion:

Therefore, the displacement of the particle during $10\sec$ is $\left(-255\widehat{i}+75\widehat{j}\right)\text{m}$.

To determine

To determine: The coordinates of the particle at $t=10.0\sec$.

Answer to Problem 15P

Answer: The coordinates of the particle at $t=10.0\sec$ is $\left(-227\widehat{i}+79\widehat{j}\right)\text{m}$.

Explanation of Solution

Explanation:

Given information:

Two forces ${\overrightarrow{\text{F}}}_{\text{1}}=\left(-6.00\widehat{i}-4.00\widehat{j}\right)\text{N}$ and ${\overrightarrow{\text{F}}}_2=\left(-3.00\widehat{i}+7.00\widehat{j}\right)\text{N}$, act on a particle of mass $2.00\text{kg}$ that is initially at rest at coordinates $\left(-2.00\text{m, +4}\text{.00}\text{m}\right)$.

The initial position of the particle is $\left(-2\widehat{i}+4\widehat{j}\right)\text{m}$.

The final coordinates of the particle at $t=10.0\text{s}$ is,

Final coordinates $\begin{array}{l}=\left(-255\widehat{i}+75\widehat{j}\right)\text{m}+\left(-2\widehat{i}+4\widehat{j}\right)\text{m}\\ =\left(-227\widehat{i}+79\widehat{j}\right)\text{m}\end{array}$

Conclusion:

Therefore, the coordinates of the particle at $t=10.0\sec$ is $\left(-227\widehat{i}+79\widehat{j}\right)\text{m}$