Question

We move a particle along an x axis, first outward from to x=1.0cmand $\mathit{x}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$then back to $\mathit{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$, while an external force acts on it. That force is directed along the x axis, and its x components can have different values for the outward trip and for the return trip. Here are the values (in newton) for four situations where x is in metres:

Find the work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

Short answer

Net work done in the given situations

a) The work done is 18 J

b) The work done is 0J

c) The work done is 30J

d) The work done is 0J

e) The forces in part b and d are conservative.

Step-by-step solution

The given data

The particle along an x axis, first outward from to x=1.0cm and $x=4.0cm$then back to $x=1.0cm$, while an external force acts on it.

Understanding the concept of work done

In an isolated system where any conservative force causes energy changes, the net work done by the system is zero for conservative forces.

Formula:

The work done by an applied force from one point to the other,$W={\int }_{ri}^{rf}{F}_{x}dx$ (i)

a) Calculation of the work done for case (a)

We can find the work done by using the given data in equation (i) as follows:

$$\begin{gathered} W={\int }_{ri}^{rf}3dx+{\int }_{ri}^{rf}-3dx \\ =\left[3x\right]\begin{array}{c}4\\ 1\end{array}+\left[-3x\right]\begin{array}{c}1\\ 4\end{array} \\ =3\left(4\right)-3\left(1\right)+\left(-3\times 1\right)-\left(-3\times 4\right) \\ =12-3-3+12 \end{gathered}$$

Solve further as:

$$\begin{gathered} W=9+9 \\ =18J \end{gathered}$$

Hence, the value of the work done is 18 J.

b) Calculation of the work done for case (b)

We can find the work done by using the given data in equation (i) as follows:

$$\begin{gathered} W={\int }_1^{4}5dx+{\int }_4^{1}5dx \\ =\left[5x\right]\begin{array}{c}4\\ 1\end{array}+\left[5x\right]\begin{array}{c}4\\ 1\end{array} \\ =5\left(4\right)-5\left(1\right)+5\left(1\right)-5\left(4\right) \\ =20-5+5-20 \end{gathered}$$

Solve further as:

W=0 J

Hence, the value of the work done is 0 J.

c) Calculation of the work done for case (c)

We can find the work done by using the given data in equation (i) as follows:

$$\begin{gathered} W={\int }_1^{4}2xdx+{\int }_4^1-2xdx \\ =\left[2\frac{x^2}{2}\right]\begin{array}{c}4\\ 1\end{array}+\left[-2\frac{x^2}{2}\right]\begin{array}{c}4\\ 1\end{array} \\ =4^2-1^2+\left[-\left(1^2\right)-{\left(-4\right)}^2\right] \\ =16-1+\left[-1+16\right] \end{gathered}$$

Solve further as:

W=30 J

Hence, the value of the work done is 30 J.

d) Calculation of the work done for case (d)

We can find the work done by using the given data in equation (i) as follows:

$$\begin{gathered} W={\int }_1^{4}3xdx+{\int }_4^{1}3xdx \\ =\left[3\frac{x^2}{2}\right]\begin{array}{c}4\\ 2\end{array}+\left[3\frac{x^2}{2}\right]\begin{array}{c}1\\ 4\end{array} \\ =\frac{3}{2}\times 1^2-\frac{3}{2}\times 4^2+\frac{3}{2}\times 1^2-\frac{3}{2}\times 4^2 \\ =0J \end{gathered}$$

Hence, the value of the work done is 0 J.

e) Calculation of the conservative forces from above cases

As the value of the work done in parts (b) and (d) is$W_{net}=0J$

Hence, the forces in part b and d are conservative.