Chapter 6, Problem 72P

(a)

To determine

To Show: The magnitude of the force is inversely proportional to $r^2$ and the direction of the force is opposite to the radius vector.

Explanation of Solution

Given:

The mass of the particle is $2.0\text{kg}$ . The force acting on the particle is $-\left(\frac{b}{r^3}\right)\left(x\text{}\widehat{i}+y\widehat{j}\right)$ .

Formula used:

The vector notation of vector $\overrightarrow{A}$ is $\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}$ .

Write the expression for magnitude of any vector.

  $\left|\overrightarrow{A}\right|=\sqrt{\left(A_x^2+A_y^2\right)}$

Here, $A_x$ is the $x-$ component of the vector and $A_y$ is the $y-$ component of the vector.

Calculation:

Write the expression for the magnitude of the given force.

  $\begin{array}{c}\left|\overrightarrow{F}\right|=\sqrt{{\left(-\frac{b}{r^3}x\right)}^2+{\left(-\frac{b}{r^3}y\right)}^2}\\ =\left(\frac{b}{r^3}\right)\sqrt{\left(x^2+y^2\right)}\end{array}$

Substitute $r$ for $\sqrt{\left(x^2+y^2\right)}$ in the above expression.

  $\left|\overrightarrow{F}\right|=\frac{b}{r^2}$

From the above expression it can be concluded that force is inversely proportional to square of radius.

  $\left|\overrightarrow{F}\right|\propto \frac{1}{r^2}$

Take the dot product between the force and $\left(x\widehat{i}+y\widehat{j}\right)$ .

  $\begin{array}{c}\overrightarrow{F}\cdot \left(x\widehat{i}+y\widehat{j}\right)=-\left(\left(\frac{b}{r^3}\right)\left(x\text{}\widehat{i}+y\widehat{j}\right)\right).\left(x\widehat{i}+y\widehat{j}\right)\\ =-\left(\frac{b}{r^3}\right)\left(x^2+y^2\right)\end{array}$

Substitute $r^2$ for $\left(x^2+y^2\right)$ in the above expression.

  $\overrightarrow{F}\cdot \left(x\widehat{i}+y\widehat{j}\right)=-\frac{b}{r}$   ........ (1)

The direction is in the opposite to $\overrightarrow{r}$ .

Conclusion:

Thus, the magnitude of the force is inversely proportional to $r^2$ and the direction of the force is opposite to the radius vector.

(b)

To determine

The work done by the force.

Explanation of Solution

Given:

The value of $b$ is $3.0\text{N}\cdot {\text{m}}^{\text{2}}$ .

The particle moves in $xy-$ plane from a point $\left(2.00\text{m, 0}\text{.00}\text{m}\right)$ to another point $\left(5.00\text{m ,}0.00\text{m}\right)$ .

Formula used:

Write the expression of work done by a particle.

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

Calculation:

The particle is moving along $x-$ axis.

Substitute $-\frac{b}{x^2}$ for the magnitude of $\overrightarrow{F}$ and $dx$ for the magnitude of $d\overrightarrow{s}$ in expression (1).

  $\begin{array}{l}W=-b{\displaystyle \underset{x_1}{\overset{x_2}{\int }}\frac{dx}{x^2}}\\ =b{\left[\frac{1}{x}\right]}_{x_1}^{x_2}\\ =b\left[\frac{1}{x_2}-\frac{1}{x_1}\right]\end{array}$

Substitute $5.00\text{m}$ for $x_2$ , $2.00\text{m}$ for $x_1$ and $3.0\text{N}\cdot {\text{m}}^{\text{2}}$ for b in the above expression.

  $\begin{array}{c}W=\left(3.0\text{N}\cdot {\text{m}}^{\text{2}}\right)\left(\frac{1}{5.00}-\frac{1}{2.00}\right){\text{m}}^{\text{-1}}\\ =-0.9\text{N}\cdot \text{m}\end{array}$

Conclusion:

Thus, the work done by the force is $-0.9\text{}\text{N}\cdot \text{m}$ .

(c)

To determine

The work done by the force on the particle.

Explanation of Solution

Given:

The particle moving around a circle of radius $7.0\text{m}$ .

Calculation:

Write the expression for the instantaneous position of the particle on the circumference of the circle.

  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}$

Write the expression for the work done by the force.

  $\begin{array}{c}W=\left(-\left(\frac{b}{r^3}\right)\left(x\text{}\widehat{i}+y\widehat{j}\right)\right).\left(x\widehat{i}+y\widehat{j}\right)\\ =-\frac{b}{r^3}\left(x^2+y^2\right)\end{array}$

Substitute $r^2$ for $\left(x^2+y^2\right)$ in the above expression.

  $W=-\frac{b}{r}$

Substitute $3.0\text{N}\cdot {\text{m}}^2$ for $b$ and $7.00\text{m}$ for $r$ in the above expression.

  $\begin{array}{c}W=-\frac{3\text{N}\cdot {\text{m}}^2}{7\text{m}}\\ =-0.428\text{N}\cdot \text{m}\end{array}$

Conclusion:

Thus, the work done by the force on the particle moving around a circle is $-0.428\text{N}\cdot \text{m}$ .