Chapter 6, Problem 37P

(a)

To determine

The kinetic energy of the object at $t=0$.

Answer to Problem 37P

The kinetic energy of the object at $t=0$ is $97.8\text{J}$.

Explanation of Solution

Given Information:

The mass of the object is $5.75\text{kg}$, the x-component of velocity of the object at $t=0$ is $5.00\text{m}/\text{s}$ and the y-component of velocity of the object at $t=0$ is $-3.00\text{m}/\text{s}$.

Formula to calculate the magnitude of the resultant velocity is,

$v_{\text{i}}=\sqrt{v_{\text{xi}}^2+v_{\text{yi}}^2}$

• $v_{\text{i}}$ is the magnitude of resultant velocity at $t=0$.
• $v_{\text{xi}}$ is the x-component of velocity at $t=0$.
• $v_{\text{yi}}$ is the y-component of velocity at $t=0$.

Substitute $5.00\text{m}/\text{s}$ for $v_{\text{xi}}$ and $-3.00\text{m}/\text{s}$ for $v_{\text{yi}}$ to find $v_{\text{i}}$.

$\begin{array}{c}{v}_{\text{i}}=\sqrt{{\left(5.00\text{m}/\text{s}\right)}^2+{\left(-3.00\text{m}/\text{s}\right)}^2}\\ =5.83\text{m}/\text{s}\end{array}$

Formula to calculate the kinetic energy of the object is,

$K.E=\frac{1}{2}m{v}_{\text{i}}^2$

• $K.E$ is the kinetic energy of the object.
• $m$ is the mass of the object.

Substitute $5.75\text{kg}$ for $m$ and $5.83\text{m}/\text{s}$ for $v_{\text{i}}$ to find $K.E$.

$\begin{array}{c}K.E=\frac{1}{2}\left(5.75\text{kg}\right){\left(5.83\text{m}/\text{s}\right)}^2\\ =97.8\text{kgm}/{\text{s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kgm}/{\text{s}}^{\text{2}}}\\ =97.8\text{J}\end{array}$

Conclusion:

Therefore, the kinetic energy of the object at $t=0$ is $97.8\text{J}$.

(b)

To determine

The constant force acted on the object during the time interval.

Answer to Problem 37P

The constant force acted on the object during the time interval is $\left(-\text{4}\text{.31}\widehat{\text{i}}+31.6\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

Given Information:

The mass of the object is $5.75\text{kg}$, the position of the object in x-direction at $t=2\text{s}$ is $8.50\text{m}$ and the position of the object in y-direction at $t=2\text{s}$ is $5.0\text{m}$.

Formula to calculate the acceleration of the object in x-direction is,

$x=v_{\text{xi}}t+\frac{1}{2}{a}_{\text{x}}{t}^2$

• $x$ is the position of the object at $t=2\text{s}$.
• $t$ is the time.
• $a_{\text{x}}$ is the acceleration of the object in x-direction.

Substitute $5.0\text{m}/\text{s}$ for $v_{\text{xi}}$, $8.50\text{m}$ for $x$ and $2\text{s}$ for $t$ to find $a_{\text{x}}$.

$\begin{array}{c}8.50\text{m}=\left(5.0\text{m}/\text{s}\right)\left(2\text{s}\right)+\frac{1}{2}{a}_{\text{x}}{\left(2\text{s}\right)}^2\\ \frac{1}{2}{a}_{\text{x}}{\left(2\text{s}\right)}^2=-1.5\text{m}\\ a_{\text{x}}=\frac{-1.5\text{m}}{2{\text{s}}^{\text{2}}}\\ =-0.75\text{m}/{\text{s}}^{\text{2}}\end{array}$

Formula to calculate the acceleration of the object in x-direction is,

$y=v_{\text{yi}}t+\frac{1}{2}{a}_{\text{y}}{t}^2$

• $y$ is the position of the object in y-direction at $t=2\text{s}$.
• $a_{\text{y}}$ is the acceleration of the object in y-direction.

Substitute $-3.00\text{m}/\text{s}$ for $v_{\text{yi}}$, $5.00\text{m}$ for $x$ and $2\text{s}$ for $t$ to find $a_{\text{y}}$.

$\begin{array}{c}5.00\text{m}=\left(-3.00\text{m}/\text{s}\right)\left(2\text{s}\right)+\frac{1}{2}{a}_{\text{y}}{\left(2\text{s}\right)}^2\\ \frac{1}{2}{a}_{\text{y}}{\left(2\text{s}\right)}^2=11\text{m}\\ a_{\text{y}}=\frac{11\text{m}}{2{\text{s}}^{\text{2}}}\\ =5.5\text{m}/{\text{s}}^{\text{2}}\end{array}$

Formula to calculate the force on the object in x-direction is,

$F_{\text{x}}=m{a}_{\text{x}}$

Formula to calculate the force on the object in y-direction is,

$F_{\text{y}}=m{a}_{\text{y}}$

Formula to calculate the constant force vector on the object is,

$\overrightarrow{F}=F_{\text{x}}\widehat{\text{i}}+F_{\text{y}}\widehat{\text{j}}$

Substitute $m{a}_{\text{x}}$ for $F_{\text{x}}$ and $m{a}_{\text{y}}$ for $F_{\text{y}}$.

$\overrightarrow{F}=\left(m{a}_{\text{x}}\right)\widehat{\text{i}}+\left(m{a}_{\text{y}}\right)\widehat{\text{j}}$

Substitute $5.75\text{kg}$ for $m$, $-0.75\text{m}/{\text{s}}^2$ for $a_{\text{x}}$ and $5.5\text{m}/{\text{s}}^2$ for $a_{\text{y}}$.

$\begin{array}{c}\overrightarrow{F}=\left(5.75\text{kg}\times \left(-0.75\text{m}/{\text{s}}^2\right)\right)\widehat{\text{i}}+\left(5.75\text{kg}\times 5.5\text{m}/{\text{s}}^2\right)\widehat{\text{j}}\\ \text{=}\left(-\text{4}\text{.31}\widehat{\text{i}}+31.6\widehat{\text{j}}\right)\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =\left(-\text{4}\text{.31}\widehat{\text{i}}+31.6\widehat{\text{j}}\right)\text{N}\end{array}$

Conclusion:

Therefore, the constant force acted on the object during the time interval is $\left(-\text{4}\text{.31}\widehat{\text{i}}+31.6\widehat{\text{j}}\right)\text{N}$.

(c)

To determine

The speed of the particle at $t=2\text{s}$.

Answer to Problem 37P

The speed of the particle at $t=2\text{s}$ is $8.73\text{m}/\text{s}$.

Formula to calculate the speed of the object in x-direction is,

$v_{\text{xf}}=v_{\text{xi}}+a_{\text{x}}t$

• $v_{\text{xf}}$ is the x-component of velocity at $t=2\text{s}$.

Substitute $5.0\text{m}/\text{s}$ for $v_{\text{xi}}$, $-0.75\text{m}/s^2$ for $a_{\text{x}}$ and $2\text{s}$ for $t$ to find $v_{\text{xf}}$.

$\begin{array}{c}{v}_{\text{xf}}=5.0\text{m}/\text{s}+\left(-0.75\text{m}/s^2\right)\left(2\text{s}\right)\\ =3.5\text{m}/\text{s}\end{array}$

Explanation of Solution

Given Information:

The mass of the object is $5.75\text{kg}$, the position of the object in x-direction at $t=2\text{s}$ is $8.50\text{m}$ and the position of the object in y-direction at $t=2\text{s}$ is $5.0\text{m}$.

Formula to calculate the speed of the object in y-direction is,

$v_{\text{yf}}=v_{\text{yi}}+a_{\text{y}}t$

• $v_{\text{yf}}$ is the y-component of velocity at $t=2\text{s}$.

Substitute $-3.0\text{m}/\text{s}$ for $v_{\text{yi}}$, $5.5\text{m}/s^2$ for $a_{\text{x}}$ and $2\text{s}$ for $t$ to find $v_{\text{yf}}$.

$\begin{array}{c}{v}_{\text{xf}}=-3.0\text{m}/\text{s}+\left(5.5\text{m}/s^2\right)\left(2\text{s}\right)\\ =8\text{m}/\text{s}\end{array}$

Formula to calculate the magnitude of the resultant velocity is,

$v_{\text{f}}=\sqrt{v_{\text{xf}}^2+v_{\text{yf}}^2}$

• $v_{\text{f}}$ is the magnitude of resultant velocity at $t=2\text{s}$.

Substitute $3.5\text{m}/\text{s}$ for $v_{\text{xf}}$ and $8.5\text{m}/\text{s}$ for $v_{\text{yf}}$ to find $v_{\text{f}}$.

$\begin{array}{c}{v}_{\text{i}}=\sqrt{{\left(3.5\text{m}/\text{s}\right)}^2+{\left(8\text{m}/\text{s}\right)}^2}\\ =8.73\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the particle at $t=2\text{s}$ is $8.73\text{m}/\text{s}$.