Chapter 6, Problem 64P

(a)

To determine

The work done by the force acting on the particle and the velocity of the particle.

Explanation of Solution

Given:

The force acting on the particle is $\left(2.0\text{N}/{\text{m}}^{\text{2}}\right)x^2\widehat{i}$ . The particle moves along a straight line from point $(2.0\text{m}\text{,}2.\text{0}\text{m)}$ topoint $(2.0\text{m}\text{,}7.\text{0}\text{m)}$ . The particle starts moving from rest.

Formula used:

Write the expression of work done by a particle.

  $W={\displaystyle \underset{x_1}{\overset{x_2}{\int }}\overrightarrow{F}.d\overrightarrow{s}}$   ........ (1)

Here, $W$ is the work done, $\overrightarrow{F}$ is the force acting on the particle, $x_1$ is the initial position of the particle and $x_2$ is the final position of the particle.

Write the expression for the work-energy theorem.

  $W=K_f-K_i$   ........ (2)

Here, $W$ is the work done, $K_f$ is the final kinetic energy of the particle and $K_i$ is the initial kinetic energy of the particle.

Calculation:

Write the expression for the displacement of the particle. $$

  $d\overrightarrow{s}=\left(x_2-x_1\right)\widehat{i}+\left(y_2-y_1\right)\widehat{j}$

Here, $ds$ is the magnitude of displacement, $\left(x_1,y_1\right)$ are the initial position coordinates of the particle and $\left(x_2,y_2\right)$ are the final position coordinates of the particle.

Substitute $0\text{m}$ for $\left(x_2-x_1\right)$ and $5\text{m}$ for $\left(y_2-y_1\right)$ in the above expression.

  $d\overrightarrow{s}=5\widehat{j}$

Substitute $\left(2.0\text{N}/{\text{m}}^{\text{2}}\right)x^2\widehat{i}$ for $\overrightarrow{F}$ , $5\widehat{j}$ for $d\overrightarrow{x}$ , 2 for $x_1$ and 7 for $x_2$ inexpression (1).

  $$$\begin{array}{c}W={\displaystyle \underset{2}{\overset{7}{\int }}\left(\left(2.0\text{N}/{\text{m}}^{\text{2}}\right)x^2\widehat{i}\right).\left(5\widehat{j}\right)}\\ =0\end{array}$

Write the expression for the kinetic energy of the particle.

  $K=\frac{1}{2}m{v}^2$

Here, $k$ is the kinetic energy, $m$ is the mass of the particle and $v$ is the velocity of the particle.

Write the expression of the final kinetic energy of the particle.

  $K_f=\frac{1}{2}m{v}_f^2$   ........ (3)

Here, $v_f$ is the final velocity of the particle and $K_f$ is the final kinetic energy.

Substitute $\frac{1}{2}m{v}_f^2$ for $K_f$ , 0 for $K_i$ and 0 for $W$ in expression (2).

  $0=\frac{1}{2}m{v}_f^2-0$

Rearrange the above expression.

  $v_f=0$

Conclusion:

Thus, the work done is $0$ and the final speed of the particle is 0.

(b)

To determine

The work done by the force acting on the particle and the velocity of the particle.

Explanation of Solution

Given:

The particle moves along a straight line from point $(2.0\text{m}\text{,}2.\text{0}\text{m)}$ to point $(5.0\text{m}\text{,}6.\text{0}\text{m)}$ .

Calculation:

Write the expression for the displacement of the particle. $$

  $d\overrightarrow{s}=\left(dx\right)\widehat{i}+\left(dy\right)\widehat{j}$

Here, $dx$ is the magnitude of displacement, $dx$ is the displacement of the particle along $x-$ axis and $dy$ is the displacement of the particle along $y-$ axis.

Substitute $\left(2.0\text{N}/{\text{m}}^{\text{2}}\right)x^2\widehat{i}$ for $\overrightarrow{F}$ , $\left(d{x}_1\widehat{i}+d{y}_1\widehat{j}\right)$ for $d\overrightarrow{x}$ , 2 for $x_1$ and 5 for $x_2$ inexpression (1).

  $$$\begin{array}{c}W={\displaystyle \underset{2}{\overset{5}{\int }}\left(\left(2.0\text{N}/{\text{m}}^{\text{2}}\right)x^2\widehat{i}\right).\left(dx\widehat{i}+dy\widehat{j}\right)}\\ =\left(2.0\text{N}/{\text{m}}^{\text{2}}\right){\displaystyle \underset{2}{\overset{5}{\int }}{x}^{2}dx}\\ =\left(2.0\text{N}/{\text{m}}^{\text{2}}\right){\left[\frac{x^3}{3}\right]}_2^5\\ =78\text{J}\end{array}$

Substitute $\frac{1}{2}m{v}_f^2$ for $K_f$ , 0 for $K_i$ and $78\text{J}$ for $W$ in expression (2).

  $78\text{J}=\frac{1}{2}m{v}_f^2-0$

Rearrange the above expression.

  $\begin{array}{c}{v}_f=\sqrt{\frac{2\left(78\text{}\text{J}\right)}{m}}\\ =\sqrt{\frac{156\text{J}}{m}}\end{array}$

Here, $m$ is the mass of the particle.

Conclusion:

Thus, the work done is $78\text{}\text{J}$ and the final speed of the particle is $\sqrt{\frac{156\text{J}}{m}}$ .