Chapter 9, Problem 51AP

Review. There are (one can say) three coequal theories of motion for a single particle: Newton’s second law, stating that the total force on the particle causes its acceleration; the work–kinetic energy theorem, stating that the total work on the particle causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on the panicle causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 3.00-kg object has velocity $7.00\hat{j}\text{m}/\text{s}$. Then, a constant net force $12.0\hat{i}\text{N}$ acts on the object for 5.00 s. (a) Calculate the object’s final velocity, using the impulse–momentum theorem. (b) Calculate its acceleration from $\stackrel{\to }{a}=({\stackrel{\to }{v}}_f-{\stackrel{\to }{v}}_i)/\Delta t$. (c) Calculate its acceleration from $\stackrel{\to }{a}={\displaystyle \sum \stackrel{\to }{F}/m}$. (d) Find the object’s vector displacement from $\Delta \stackrel{\to }{r}={\stackrel{\to }{v}}_{i}t+\frac{1}{2}\stackrel{\to }{a}{t}^2$ (e) Find the work done on the object from $W=\stackrel{\to }{F}\cdot \Delta \stackrel{\to }{r}$. (f) Find the final kinetic energy from $\frac{1}{2}m{v}_f{}^2=\frac{1}{2}m{\stackrel{\to }{v}}_f\cdot {\stackrel{\to }{v}}_f$. (g) Find the final kinetic energy from $\frac{1}{2}m{v}_i{}^2+W$. (h) State the result of comparing the answers to parts (b) and (c), and the answers to parts (f) and (g).

To determine

The final velocity of the object.

Answer to Problem 51AP

The final velocity of the object is $\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}$.

Explanation of Solution

The mass of the object is $3\text{kg}$, the velocity of the object is $7\widehat{\text{j}}\text{m}/\text{s}$ and the net force acting on the object is $12\widehat{\text{i}}\text{N}$. The time duration is $5\text{s}$.

Write the expression of impulse momentum equation.

    $m\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)=\overrightarrow{\text{F}}\cdot \Delta t$        (1)

Here, $m$ is the mass of the object, $\overrightarrow{{\text{v}}_{\text{f}}}$ is the final velocity of the object, $\overrightarrow{{\text{v}}_{\text{i}}}$ is the initial velocity of the object, $\overrightarrow{\text{F}}$ is the net force acting on the object and $\Delta t$ is the time duration.

Conclusion:

Substitute $3\text{kg}$ for $m$, $7\widehat{\text{j}}\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_{\text{i}}}$, $12\widehat{\text{i}}\text{N}$ for $\overrightarrow{\text{F}}$ and $5\text{s}$ for $\Delta t$ in equation (1) to find $\overrightarrow{{\text{v}}_{\text{f}}}$.

    $\begin{array}{c}3\text{kg}\left(\overrightarrow{{\text{v}}_{\text{f}}}-7\widehat{\text{j}}\text{m}/\text{s}\right)=12\widehat{\text{i}}\text{N}\times 5\text{s}\\ \overrightarrow{{\text{v}}_{\text{f}}}=\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}\end{array}$

Thus, the final velocity of the object is $\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}$.

To determine

The acceleration of the object.

Answer to Problem 51AP

The acceleration of the object is $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression to calculate the acceleration of the object.

    $\overrightarrow{\text{a}}=\frac{\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)}{\Delta t}$        (2)

Here,

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

Substitute $7\widehat{\text{j}}\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_{\text{i}}}$, $\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_{\text{f}}}$ and $5\text{s}$ for $\Delta t$ in equation (2) to find $\overrightarrow{\text{a}}$.

    $\begin{array}{c}\overrightarrow{\text{a}}=\frac{\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}-7\widehat{\text{j}}\text{m}/\text{s}}{5\text{s}}\\ =4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Thus, the acceleration of the object is $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$.

Conclusion:

Therefore, the acceleration of the object is $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$.

To determine

The acceleration of the object.

Answer to Problem 51AP

The acceleration of the object is $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the expression to calculate the acceleration of the object.

    $\overrightarrow{\text{a}}=\frac{{\displaystyle \sum \overrightarrow{\text{F}}}}{m}$        (3)

Substitute $12\widehat{\text{i}}\text{N}$ for ${\displaystyle \sum \overrightarrow{\text{F}}}$ and $3\text{kg}$ for $m$ in equation (3) to find $\overrightarrow{\text{a}}$.

    $\begin{array}{c}\overrightarrow{\text{a}}=\frac{12\widehat{\text{i}}\text{N}}{3\text{kg}}\\ =4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Thus, the acceleration of the object is $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$.

Conclusion:

Therefore, the acceleration of the object is $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$.

To determine

The vector displacement of the object.

Answer to Problem 51AP

The vector displacement of the object is $\left(50\widehat{\text{i}}+35\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

Write the expression to calculate the vector displacement of the object.

    $\overrightarrow{\text{r}}=\overrightarrow{{\text{v}}_{\text{i}}}t+\frac{1}{2}\overrightarrow{\text{a}}{t}^2$        (4)

Here,

$\overrightarrow{\text{r}}$ is the vector displacement of the object.

Substitute $7\widehat{\text{j}}\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_{\text{i}}}$, $4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}$ for $\overrightarrow{\text{a}}$ and $5\text{s}$ for $t$ in equation (4) to find $\overrightarrow{\text{r}}$.

    $\begin{array}{c}\overrightarrow{\text{r}}=7\widehat{\text{j}}\text{m}/\text{s}\times 5\text{s}+\frac{1}{2}\times 4\widehat{\text{i}}\text{m}/{\text{s}}^{\text{2}}\times {\left(5\text{s}\right)}^2\\ =\left(50\widehat{\text{i}}+35\widehat{\text{j}}\right)\text{m}\end{array}$

Thus, the vector displacement of the object is $\left(50\widehat{\text{i}}+35\widehat{\text{j}}\right)\text{m}$.

Conclusion:

Therefore, the vector displacement of the object is $\left(50\widehat{\text{i}}+35\widehat{\text{j}}\right)\text{m}$.

To determine

The work done on the object.

Answer to Problem 51AP

The work done on the object is $\text{600}\text{J}$.

Explanation of Solution

Write the expression to calculate the work done on the object.

    $W=\overrightarrow{\text{F}}\cdot \Delta \overrightarrow{\text{r}}$        (5)

Here,

$W$ is the work done on the object.

Substitute $12\widehat{\text{i}}\text{N}$ for $\overrightarrow{\text{F}}$ and $\left(50\widehat{\text{i}}+35\widehat{\text{j}}\right)\text{m}$ for $\Delta \overrightarrow{\text{r}}$ in equation (5) to find $W$.

    $\begin{array}{c}W=12\widehat{\text{i}}\text{N}\cdot \left(50\widehat{\text{i}}+35\widehat{\text{j}}\right)\text{m}\\ \text{=600}\text{J}\end{array}$

Thus, the work done on the object is $\text{600}\text{J}$.

Conclusion:

Therefore, the work done on the object is $\text{600}\text{J}$.

To determine

The final kinetic energy of the object.

Answer to Problem 51AP

The final kinetic energy of the object is $\text{674}\text{J}$.

Explanation of Solution

Write the expression to calculate the final kinetic energy of the object.

    $\begin{array}{c}E=\frac{1}{2}m{v}_{\text{f}}^{\text{2}}\\ =\frac{1}{2}m\overrightarrow{{\text{v}}_{\text{f}}}\cdot \overrightarrow{{\text{v}}_{\text{f}}}\end{array}$        (6)

Substitute $3\text{kg}$ for $m$ and $\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_{\text{f}}}$ in equation (6) to find $E$.

    $\begin{array}{c}E=\frac{1}{2}3\text{kg}\left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}\cdot \left(20\widehat{\text{i}}+7\widehat{\text{j}}\right)\text{m}/\text{s}\\ =\frac{1}{2}3\text{kg}\left(400+49\right)\text{m}/\text{s}\\ =673.5\text{J}\\ \approx \text{674}\text{J}\end{array}$

Thus, the final kinetic energy of the object is $\text{674}\text{J}$.

Conclusion:

Therefore, the final kinetic energy of the object is $\text{674}\text{J}$.

To determine

The final kinetic energy of the object.

Answer to Problem 51AP

The final kinetic energy of the object is $\text{674}\text{J}$.

Explanation of Solution

Write the expression to calculate the final kinetic energy of the object.

    $E=\frac{1}{2}m{v}_{\text{i}}^{\text{2}}+W$        (7)

Substitute $3\text{kg}$ for $m$, $600\text{J}$ for $W$ and $7\widehat{\text{j}}\text{m}/\text{s}$ for $\overrightarrow{{\text{v}}_{\text{i}}}$ in equation (7) to find $E$.

    $\begin{array}{c}E=\frac{1}{2}3\text{kg}\left(7\widehat{\text{j}}\text{m}/\text{s}\right)\left(7\widehat{\text{j}}\text{m}/\text{s}\right)+600\text{J}\\ =673.5\text{J}\\ \approx \text{674}\text{J}\end{array}$

Thus, the final kinetic energy of the object is $\text{674}\text{J}$.

Conclusion:

Therefore, the final kinetic energy of the object is $\text{674}\text{J}$.

To determine

The result of comparison of the answers in part (b), (c) and (f), (g).

Answer to Problem 51AP

The value of acceleration in part (b), (c) and kinetic energy in part (f), (g) are same.

Explanation of Solution

Write the expression to calculate the acceleration of the object.

    $\overrightarrow{\text{a}}=\frac{\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)}{\Delta t}$        (2)

Write the expression to calculate the acceleration of the object.

    $\overrightarrow{\text{a}}=\frac{{\displaystyle \sum \overrightarrow{\text{F}}}}{m}$        (3)

According to the second law of motion,

    ${\displaystyle \sum \overrightarrow{\text{F}}}=m\frac{\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)}{\Delta t}$

Substitute $m\frac{\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)}{\Delta t}$ for ${\displaystyle \sum \overrightarrow{\text{F}}}$ in equation (2).

    $\begin{array}{c}\overrightarrow{\text{a}}=\frac{m\frac{\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)}{\Delta t}}{m}\\ =\frac{\left(\overrightarrow{{\text{v}}_{\text{f}}}-\overrightarrow{{\text{v}}_{\text{i}}}\right)}{\Delta t}\end{array}$        (8)

The equation (2) and (8) are same therefore, the value of acceleration in part (b) and (c) are same.

Write the expression to calculate the work done on the object,

    $\begin{array}{c}W=\text{change}\text{in}\text{kinetic}\text{energy}\\ =\frac{1}{2}m{v}_{\text{f}}^{\text{2}}-\frac{1}{2}m{v}_{\text{i}}^{\text{2}}\end{array}$        (9)

Substitute $\frac{1}{2}m{v}_{\text{f}}^{\text{2}}-\frac{1}{2}m{v}_{\text{i}}^{\text{2}}$ for $W$ in equation (9).

    $\begin{array}{c}E=\frac{1}{2}m{v}_{\text{i}}^{\text{2}}+\frac{1}{2}m{v}_{\text{f}}^{\text{2}}-\frac{1}{2}m{v}_{\text{i}}^{\text{2}}\\ =\frac{1}{2}m{v}_{\text{f}}^{\text{2}}\end{array}$        (10)

The equation (10) and (6) are same.

Thus, the value of kinetic energy in part (f) and (g) are same.

Conclusion:

Therefore, the value of acceleration in part (b), (c) and kinetic energy in part (f), (g) are same.