Chapter 7, Problem 45AP

(a)

To determine

The expression for the two forces in unit vector notation.

Answer to Problem 45AP

The expression for the first force in unit vector notation is $\left(\text{20}\text{.5}\widehat{\text{i}}+14.3\widehat{\text{j}}\right)\text{N}$ and the expression for the second force in unit vector notation is $\left(-\text{36}\text{.4}\widehat{\text{i}}+21.0\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

The mass of an object is $5.00\text{kg}$, the magnitude of the first force is $25.0\text{N}$ at an angle of $35.0°$ and the magnitude of the second force is $42.0\text{N}$ at an angle of $150°$. The velocity of the object at origin is $\overrightarrow{v_{\text{i}}}=\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}$.

Write the formula to calculate the expression for the first force in unit vector notation is

    $\overrightarrow{{\text{F}}_{\text{1}}}=\left(F_1\cos {\theta }_1\right)\widehat{\text{i}}+\left(F_1\sin {\theta }_1\right)\widehat{\text{j}}$                                                          (I)

Here, $\overrightarrow{{\text{F}}_{\text{1}}}$ is the unit vector notation of the first force, $F_1$ is the magnitude of the first force and ${\theta }_1$ is the angle made by the first force from the horizontal.

Write the formula to calculate the expression for the second force in unit vector notation

    $\overrightarrow{{\text{F}}_{\text{2}}}=\left(F_2\cos {\theta }_2\right)\widehat{\text{i}}+\left(F_2\sin {\theta }_2\right)\widehat{\text{j}}$                                                         (II)

Here, $\overrightarrow{{\text{F}}_{\text{2}}}$ is the unit vector notation of the second force, $F_2$ is the magnitude of the second force and ${\theta }_2$ is the angle made by the second force from the horizontal.

Conclusion:

Substitute $25.0\text{N}$ for $F_1$ and $35.0°$ for ${\theta }_1$ in equation (I) to find $\overrightarrow{{\text{F}}_{\text{1}}}$.

    $\begin{array}{c}\overrightarrow{{\text{F}}_{\text{1}}}=\left(\left(25.0\text{N}\right)\cos 35.0°\right)\widehat{\text{i}}+\left(\left(25.0\text{N}\right)\text{sin}35.0°\right)\widehat{\text{j}}\\ =\left(\text{20}\text{.5}\widehat{\text{i}}+14.3\widehat{\text{j}}\right)\text{N}\end{array}$

Substitute $42.0\text{N}$ for $F_2$ and $150°$ for ${\theta }_2$ in equation (II) to find $\overrightarrow{{\text{F}}_{\text{2}}}$

    $\begin{array}{c}\overrightarrow{{\text{F}}_{\text{2}}}=\left(\left(42.0\text{N}\right)\cos 150°\right)\widehat{\text{i}}+\left(\left(42.0\text{N}\right)\text{sin150}°\right)\widehat{\text{j}}\\ =\left(-\text{36}\text{.4}\widehat{\text{i}}+21.0\widehat{\text{j}}\right)\text{N}\end{array}$

Therefore, the expression for the first force in unit vector notation is $\left(\text{20}\text{.5}\widehat{\text{i}}+14.3\widehat{\text{j}}\right)\text{N}$ and the expression for the second force in unit vector notation is $\left(-\text{36}\text{.4}\widehat{\text{i}}+21.0\widehat{\text{j}}\right)\text{N}$.

(b)

To determine

The total force exerted on the object.

Answer to Problem 45AP

The total force exerted on the object is $\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

Write the formula to calculate the total force exerted on the object

    $\overrightarrow{\text{F}}=\overrightarrow{{\text{F}}_{\text{1}}}+\overrightarrow{{\text{F}}_{\text{2}}}$

Here, $\overrightarrow{\text{F}}$ is the total force exerted on the object.

Conclusion:

Substitute $\left(\text{20}\text{.5}\widehat{\text{i}}+14.3\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{{\text{F}}_{\text{1}}}$ and $\left(-\text{36}\text{.4}\widehat{\text{i}}+21.0\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{{\text{F}}_{\text{2}}}$.

    $\begin{array}{c}\overrightarrow{\text{F}}=\left(\text{20}\text{.5}\widehat{\text{i}}+14.3\widehat{\text{j}}\right)\text{N}+\left(-\text{36}\text{.4}\widehat{\text{i}}+21.0\widehat{\text{j}}\right)\text{N}\\ \text{=}\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}\end{array}$

Therefore, the total force exerted on the object is $\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}$.

(c)

To determine

The acceleration on the object.

Answer to Problem 45AP

The acceleration on the object is $\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}$.

Explanation of Solution

Write the formula to calculate the acceleration of the object

    $\overrightarrow{\text{F}}=m\overrightarrow{\text{a}}$

Here, $\overrightarrow{\text{a}}$ is the acceleration of the object.

Conclusion:

Substitute $\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $5.00\text{kg}$ for $\text{m}$ to find $\overrightarrow{\text{a}}$.

    $\begin{array}{c}\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}=\left(5.00\text{kg}\right)\overrightarrow{\text{a}}\\ \overrightarrow{\text{a}}=\frac{1}{5.00\text{kg}}\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}\\ \text{=}\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}\end{array}$

Therefore, the acceleration on the object is $\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}$.

(d)

To determine

The velocity on the object.

Answer to Problem 45AP

The velocity on the object is $\left(-\text{5}\text{.54}\widehat{\text{i}}+23.7\widehat{\text{j}}\right)\text{m}/\text{s}$.

Explanation of Solution

Write the formula to calculate the velocity of the object at $t=3.00\text{s}$

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

Here, $\overrightarrow{v_{\text{f}}}$ is the velocity vector of the object at $t=3.00\text{s}$, $\overrightarrow{v_{\text{i}}}$ is the velocity vector of the object at origin and $t$ is the time interval during the motion of the object.

Conclusion:

Substitute $\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}$ for $\overrightarrow{v_{\text{i}}}$, $\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{a}$ and $3.00\text{s}$ for $\text{t}$ to find $\overrightarrow{v_{\text{f}}}$.

    $\begin{array}{c}\overrightarrow{{\text{v}}_{\text{f}}}=\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}+\left[\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}\right]\left(3.00\text{s}\right)\\ =\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}+\left(-\text{9}\text{.54}\widehat{\text{i}}+21.18\widehat{\text{j}}\right)\text{m}/\text{s}\\ =\left(-\text{5}\text{.54}\widehat{\text{i}}+23.7\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Therefore, the velocity on the object is $\left(-\text{5}\text{.54}\widehat{\text{i}}+23.7\widehat{\text{j}}\right)\text{m}/\text{s}$.

(e)

To determine

The position on the object.

Answer to Problem 45AP

The position on the object is $\left(-2.3\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

Write the formula to calculate the position of the object

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

Here, $\overrightarrow{\text{r}}$ is the position vector of the object at $t=3.00\text{s}$.

Substitute $\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}$ for $\overrightarrow{v_{\text{i}}}$, $\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{a}$ and $3.00\text{s}$ for $\text{t}$ to find $\overrightarrow{\text{r}}$.

    $\begin{array}{c}\overrightarrow{\text{r}}=\left[\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}\right]\left(3.00\text{s}\right)+\frac{1}{2}\left[\left(-3.18\widehat{\text{i}}+7.06\widehat{\text{j}}\right)\text{N}\right]{\left(3.00\text{s}\right)}^2\\ =\left(12.0\widehat{\text{i}}+7.50\widehat{\text{j}}\right)\text{m}/\text{s}+\left(-14.31\widehat{\text{i}}+31.77\widehat{\text{j}}\right)\text{m}/\text{s}\\ =\left(-2.3\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the position on the object is $\left(-2.3\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}$.

(f)

To determine

The kinetic energy of the object from the formula $\frac{1}{2}m{v}_{\text{f}}^2$.

Answer to Problem 45AP

The kinetic energy of the object from the formula $\frac{1}{2}m{v}_{\text{f}}^2$ is

Explanation of Solution

Write the formula to calculate the magnitude of the final velocity of the object

    $v_{\text{f}}=\sqrt{v_{\text{fx}}^2+v_{\text{fy}}^2}$

Here, $v_{\text{f}}$ is the magnitude of the final velocity of the object.

Substitute $-5.54\text{m}/\text{s}$ for $v_{\text{fx}}$ and $23.7\text{m}/\text{s}$ for $v_{\text{fy}}$ to find $v_{\text{f}}$.

    $\begin{array}{c}{v}_{\text{f}}=\sqrt{{\left(-5.54\text{m}/\text{s}\right)}^2+{\left(23.7\text{m}/\text{s}\right)}^2}\\ =24.34\text{m}/\text{s}\end{array}$

Write the formula to calculate the kinetic energy of the object

    $K_{\text{f}}=\frac{1}{2}m{v}_{\text{f}}^2$

Conclusion:

Substitute $5.00\text{kg}$ for $m$ and $24.34\text{m}/\text{s}$ for ${\text{v}}_{\text{f}}$ to find $K_{\text{f}}$.

    $\begin{array}{c}{K}_{\text{f}}=\frac{1}{2}\times 5.00\text{kg}\times {\left(24.34\text{m}/\text{s}\right)}^2\\ =1480.95{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{1\text{J}}{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}\times \frac{{10}^{-3}\text{kJ}}{1\text{J}}\\ =1.48\text{kJ}\end{array}$

Therefore, the kinetic energy of the object from the formula $\frac{1}{2}m{v}_{\text{f}}^2$ is $1.48\text{kJ}$.

(g)

To determine

The kinetic energy of the object from the formula $\frac{1}{2}m{v}_{\text{i}}^2+{\displaystyle \sum \overrightarrow{\text{F}}\cdot \Delta \overrightarrow{\text{r}}}$.

Answer to Problem 45AP

The kinetic energy of the object from the formula $\frac{1}{2}m{v}_{\text{i}}^2+{\displaystyle \sum \overrightarrow{\text{F}}\cdot \Delta \overrightarrow{\text{r}}}$ is $1.48\text{kJ}$.

Explanation of Solution

Write the formula to calculate the magnitude of the initial velocity of the object

    $v_{\text{i}}=\sqrt{v_{\text{ix}}^2+v_{\text{iy}}^2}$

Here, $v_{\text{i}}$ is the magnitude of the final velocity of the object.

Substitute $4.00\text{m}/\text{s}$ for $v_{\text{ix}}$ and $2.50\text{m}/\text{s}$ for $v_{\text{iy}}$ to find $v_{\text{i}}$.

    $\begin{array}{c}{v}_{\text{i}}=\sqrt{{\left(4.00\text{m}/\text{s}\right)}^2+{\left(2.50\text{m}/\text{s}\right)}^2}\\ =4.72\text{m}/\text{s}\end{array}$

Write the formula to calculate the final kinetic energy of the object

    $K_{\text{f}}=\frac{1}{2}m{v}_{\text{i}}^2+{\displaystyle \sum \overrightarrow{\text{F}}\cdot \Delta \overrightarrow{\text{r}}}$

Conclusion:

Substitute $5.00\text{kg}$ for $m$, $4.72\text{m}/\text{s}$ for ${\text{v}}_{\text{f}}$ and $\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{\text{F}}$ and $\left(-2.3\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{r}}$ to find $K_{\text{f}}$ in above expression.

    $\begin{array}{c}{K}_{\text{f}}=\frac{1}{2}\times 5.00\text{kg}\times {\left(4.72\text{m}/\text{s}\right)}^2+\left[\left(\left(-15.9\widehat{\text{i}}+35.3\widehat{\text{j}}\right)\text{N}\right)\cdot \left(\left(-2.3\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}\right)\right]\\ =55.625\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}+\left[36.57+1387.29\right]\text{N}\cdot \text{m}\times \frac{1{\text{kgm}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{N}\cdot \text{m}}\\ =1479.5\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\times \frac{1\text{J}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\times \frac{{10}^{-3}\text{kJ}}{1\text{J}}\\ =1.48\text{kJ}\end{array}$

Therefore, the kinetic energy of the object from the formula $\frac{1}{2}m{v}_{\text{f}}^2$ is $1.48\text{kJ}$.

(h)

To determine

The conclusion by comparing the answer of part (f) and (g).

Explanation of Solution

Newton gave the law for the constant acceleration motion while the work energy theorem relates the work done by the object to its energy.

In part (f) the kinetic energy of the object is calculated with the help of Newton’s law while the kinetic energy in part (g) is calculated by the work energy theorem. Since in both the parts the kinetic energy of the object comes out to be same that conclude both the law and theorem are relevant to each other. The work energy theorem is consistent with the Newton’s law.

Conclusion:

Therefore, the work energy theorem is consistent with the Newton’s law of equation.