Chapter 6, Problem 67P

(a)

To determine

To express two vectors in unit vector notation.

Answer to Problem 67P

The two vectors in unit vector notation is ${\overrightarrow{F}}_1=\left(20.5\widehat{i}+14.3\widehat{j}\right)N$, and ${\overrightarrow{\text{F}}}_2=\left(-36.4\widehat{i}+21.0\widehat{j}\right)N$.

Explanation of Solution

Write the expression for force from Figure P6.67.

    ${\overrightarrow{F}}_1=\left(25.0\text{N}\right)\left(\cos 35.0°\widehat{i}+\sin 35.0°\widehat{j}\right)$

Simplify the above equation.

    ${\overrightarrow{F}}_1=\left(20.5\widehat{i}+14.3\widehat{j}\right)N$        (I)

Write the expression for force from Figure P6.67.

    ${\overrightarrow{F}}_2=\left(42.0\text{N}\right)\left(\cos 150°\widehat{i}+\sin 150°\widehat{j}\right)$

Simplify the above equation.

    ${\overrightarrow{\text{F}}}_2=\left(-36.4\widehat{i}+21.0\widehat{j}\right)N$        (II)

Conclusion:

Therefore, the two vectors in unit vector notation is ${\overrightarrow{F}}_1=\left(20.5\widehat{i}+14.3\widehat{j}\right)N$, and ${\overrightarrow{\text{F}}}_2=\left(-36.4\widehat{i}+21.0\widehat{j}\right)N$.

(b)

To determine

The total force exerted on the object.

Answer to Problem 67P

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

Explanation of Solution

Write the total force exerted on the object.

    ${\displaystyle \sum \overrightarrow{\text{F}}=}{\overrightarrow{\text{F}}}_1+{\overrightarrow{\text{F}}}_2$        (III)

Use equation (I) and (II) in equation (III), to find the net force.

    $\begin{array}{c}{\displaystyle \sum \overrightarrow{\text{F}}=}\left(20.5\widehat{i}+14.3\widehat{j}\right)N+\left(-36.4\widehat{i}+21.0\widehat{j}\right)N\\ =\left(-15.9\widehat{\text{i}}\text{+}35.3\widehat{\text{j}}\right)\text{N}\end{array}$        (IV)

Conclusion:

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

(c)

To determine

The acceleration of the object.

Answer to Problem 67P

The acceleration of the object is $\underset{\_}{\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for acceleration of the object.

    $\overrightarrow{\text{a}}=\frac{{\displaystyle \sum \overrightarrow{F}}}{m}$        (V)

Here, $\overrightarrow{\text{a}}$ is the acceleration, $m$ is the mass, and ${\displaystyle \sum \overrightarrow{F}}$ is the net force on the object.

Use equation (IV) in (V).

    $\overrightarrow{\text{a}}=\frac{\left(-15.9\widehat{\text{i}}\text{+}35.3\widehat{\text{j}}\right)\text{N}}{m}$        (VI)

Given that the mass of the object is $5\text{kg}$. Apply this condition in equation (VI).

    $\overrightarrow{\text{a}}=\frac{\left(-15.9\widehat{\text{i}}\text{+}35.3\widehat{\text{j}}\right)\text{N}}{5\text{kg}}$

Rearrange the above equation.

    $\overrightarrow{\text{a}}=\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$        (VII)

Conclusion:

Therefore, the acceleration of the object is $\underset{\_}{\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2}$.

(d)

To determine

The velocity of the object.

Answer to Problem 67P

The velocity of the object is $\underset{\_}{\left(-5.54\widehat{\text{i}}\text{+}23.7\widehat{\text{j}}\right)\text{m}/\text{s}}$.

Explanation of Solution

Write the velocity of the object.

    ${\overrightarrow{v}}_f={\overrightarrow{v}}_i+\overrightarrow{a}t$        (VIII)

Here, ${\overrightarrow{v}}_f$ is the final velocity, ${\overrightarrow{v}}_i$ is the initial velocity, $\overrightarrow{a}$ is the acceleration, and $t$ is the time.

Conclusion:

Substitute $\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_i}$, $\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$ for $\overrightarrow{\text{a}}$, and $3.00\text{s}$ for $t$ in equation (VIII), to find ${\overrightarrow{v}}_f$.

    $\begin{array}{c}{\overrightarrow{v}}_f=\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}+\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2\times 3.00\text{s}\\ \text{=}\left(-5.54\widehat{\text{i}}\text{+}23.7\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Therefore, the velocity of the object is $\underset{\_}{\left(-5.54\widehat{\text{i}}\text{+}23.7\widehat{\text{j}}\right)\text{m}/\text{s}}$.

(e)

To determine

The position of the object at time $3.00\text{s}$.

Answer to Problem 67P

The position of the object at $3.00\text{s}$ is $\underset{\_}{\left(-2.30\widehat{\text{i}}\text{+}39.3\widehat{\text{j}}\right)\text{m}}$.

Explanation of Solution

Given that the initial position of the object is zero.

Write the expression for position using equation of motion.

    ${\overrightarrow{\text{r}}}_f={\overrightarrow{\text{r}}}_i+{\overrightarrow{\text{v}}}_{i}t+\frac{1}{2}\overrightarrow{\text{a}}{t}^2$        (IX)

Here, ${\overrightarrow{\text{r}}}_f$ is the final position, ${\overrightarrow{\text{r}}}_i$ is the initial position, ${\overrightarrow{\text{v}}}_i$ is the initial velocity, $t$ is the time, and, $\overrightarrow{\text{a}}$ is the acceleration.

Conclusion:

Substitute $\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_i}$, $\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2$ for $\overrightarrow{\text{a}}$, $3.00\text{s}$ for $t$, $0\text{m}$ for ${\overrightarrow{\text{r}}}_i$, in equation (IX), to find ${\overrightarrow{\text{r}}}_f$.

    $\begin{array}{c}{\overrightarrow{\text{r}}}_f=\left(0\text{m}\right)+\left(4.00\widehat{\text{i}}+2.50\widehat{\text{j}}\right)\text{m}/\text{s}\left(3.00\text{s}\right)+\frac{1}{2}\left(-3.18\widehat{\text{i}}\text{+}7.07\widehat{\text{j}}\right)\text{m}/{\text{s}}^2\times {\left(3.00\text{s}\right)}^2\\ =\left(-2.30\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}\end{array}$

Therefore, the position of the object at $3.00\text{s}$ is $\underset{\_}{\left(-2.30\widehat{\text{i}}\text{+}39.3\widehat{\text{j}}\right)\text{m}}$.

(f)

To determine

The final kinetic energy of the object.

Answer to Problem 67P

The final kinetic energy of the object is $\underset{\_}{1.48\text{kJ}}$.

Explanation of Solution

Write the expression for kinetic energy.

    $K_f=\frac{1}{2}m{v}_f^2$        (X)

Here, $K_f$ is the final kinetic energy, $m$ is the mass, and $v_f$ is the final velocity.

From subpart (d) the final velocity vector is given by

  ${\overrightarrow{v}}_f=\left(-5.54\widehat{\text{i}}\text{+}23.7\widehat{\text{j}}\right)\text{m}/\text{s}$

Write the magnitude of final velocity vector.

    $\left|v_f\right|=\sqrt{{\left(5.54\right)}^2+{\left(23.7\right)}^2}\text{m}/\text{s}$        (XI)

Conclusion:

Substitute $5.00\text{kg}$ for $m$, and $\sqrt{{\left(5.54\right)}^2+{\left(23.7\right)}^2}\text{m}/\text{s}$ for $v_f$ in equation (X), to find $K_f$.

    $\begin{array}{c}{K}_f=\frac{1}{2}\left(5.00\text{kg}\right){\left(\sqrt{{\left(5.54\right)}^2+{\left(23.7\right)}^2}\text{m}/\text{s}\right)}^2\\ =1480.95\text{J}\\ \text{=}1480.95\text{J}\times \frac{1\text{kJ}}{{10}^3\text{J}}\\ =1.48\text{kJ}\end{array}$

Therefore, the final kinetic energy of the object is $\underset{\_}{1.48\text{kJ}}$.

(g)

To determine

The final kinetic energy of the object using the equation $K_f=\frac{1}{2}m{v}_f^2+{\displaystyle \sum \overrightarrow{\text{F}}.\Delta \overrightarrow{\text{r}}}$.

Answer to Problem 67P

The final kinetic energy of the object using the equation $K_f=\frac{1}{2}m{v}_f^2+{\displaystyle \sum \overrightarrow{\text{F}}.\Delta \overrightarrow{\text{r}}}$ is $\underset{\_}{1.48\text{kJ}}$.

Explanation of Solution

Write the expression for kinetic energy.

  $K_f=\frac{1}{2}m{v}_f^2+{\displaystyle \sum \overrightarrow{\text{F}}.\Delta \overrightarrow{\text{r}}}$        (XII)

From subpart (b) the net force ${\displaystyle \sum \overrightarrow{\text{F}}=}\left(-15.9\widehat{\text{i}}\text{+}35.3\widehat{\text{j}}\right)\text{N}$, and from subpart (e) the change in position is $\Delta \overrightarrow{\text{r}}=\left(-2.30\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}$.

Apply the above condition in equation (XII).

  $K_f=\frac{1}{2}m{v}_f^2+\left[\left(-15.9\widehat{\text{i}}\text{+}35.3\widehat{\text{j}}\right)\text{N}\cdot \left(-2.30\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}\right]$        (XIII)

Conclusion:

Substitute $5.00\text{kg}$ for $m$, and $\sqrt{{\left(5.54\right)}^2+{\left(23.7\right)}^2}\text{m}/\text{s}$ for $v_f$ in equation (XIII), to find $K_f$.

    $\begin{array}{c}{K}_f=\frac{1}{2}\left(5.00\text{kg}\right){\left(\sqrt{{\left(5.54\right)}^2+{\left(23.7\right)}^2}\text{m}/\text{s}\right)}^2+\left[\left(-15.9\widehat{\text{i}}\text{+}35.3\widehat{\text{j}}\right)\text{N}\cdot \left(-2.30\widehat{\text{i}}+39.3\widehat{\text{j}}\right)\text{m}\right]\\ =55.6\text{J+1426}\text{J}\\ \text{=1480}\text{J}\times \frac{1\text{kJ}}{{10}^3\text{J}}\\ =1.48\text{kJ}\end{array}$

Therefore, the final kinetic energy of the object using the equation $K_f=\frac{1}{2}m{v}_f^2+{\displaystyle \sum \overrightarrow{\text{F}}.\Delta \overrightarrow{\text{r}}}$ is $\underset{\_}{1.48\text{kJ}}$.

(h)

To determine

To compare the answers in subpart (f) and subpart (g).

Answer to Problem 67P

The answers in subpart (f) and subpart (g) is same, and it proves that the work energy theorem is consistent with the newton’s second law of motion.

Explanation of Solution

From subpart (f) and (g), the kinetic energy is obtained as $1.48\text{kJ}$.

The work energy theorem states that the change in kinetic energy is equal to the external work done, and by rearranging the terms the final kinetic energy of the object is obtained in subpart (g).

Since the answers in subpart (g) and (f) are same, which proves that the work energy theorem is consistent with the newton’s second law of motion.

Conclusion:

Therefore, the answers in subpart (f) and subpart (g) is same, and it proves that the work energy theorem is consistent with the newton’s second law of motion.