In an experiment, one of the forces exerted on a proton is $\overrightarrow{\mathbf{F}}\mathbf{=}{\mathbf{\alpha x}}^{\mathbf{2}}\mathbf{}\widehat{\mathbf{i}\mathbf{}}$, where $\mathbf{\alpha }\mathbf{=}\mathbf{12}\mathbf{}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$. (a) How much work does $\overrightarrow{\mathbf{F}}$ do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force $\overrightarrow{\mathbf{F}}$ conservative? Explain. If $\overrightarrow{\mathbf{F}}$is conservative, what is the potential-energy function for it? Let U = 0 when x = 0.

The force on the proton is $\overrightarrow{F}=\alpha x^2\hat{i}$

The constant for force is $\alpha =12\hspace{0.33em}N/m^2$

The initial position of the proton is $(x_1,y_1)=(0.10\hspace{0.33em}m,0)$

The intermediate position of the proton is $(x_2,y_2)=(0.10\hspace{0.33em}m,0.40\hspace{0.33em}m)$

The final position of the proton is $(x_3,y_3)=(0.30\hspace{0.33em}m,0)$

The force by which work is done is independent of the path is called conservative and the force for which the work done is dependent on the path is called non-conservative force.

The work done on proton by force is given as:

$$\begin{gathered} W_a={\int }_{x_2}^{x_1}\overrightarrow{F}·\left(\hat{i}dx\right) \\ {W}_a={\int }_{x_2}^{x_1}\left(\alpha x^2\hat{i}\right)\left(\hat{i}dx\right) \\ {W}_a={\int }_{x_2}^{x_1}{x}^{2}dx \\ {W}_a=\frac{\alpha }{3}\left(x_2^3-x_1^3\right) \end{gathered}$$

Here and are the initial and final positions of the proton.

Substitute all the values in the above equation and we get,

role="math" localid="1665042881143" $W_a=\frac{\left(12\hspace{0.33em}{\displaystyle \frac{N}{m^2}}\right)}{3}[{(0.10\hspace{0.33em}m)}^3-{(0.10\hspace{0.33em}m)}^3]W_a=0\hspace{0.33em}J$

Therefore, the work done on protons by force is 0 J .

The work done on proton by force is given as:

$$\begin{gathered} W_b={\int }_{x_1}^{x_3}\overrightarrow{F}·\left(\hat{i}dx\right) \\ {W}_b={\int }_{x_1}^{x_3}\left(\alpha x^{2}i\right)\left(\hat{i}dx\right) \\ {W}_b=\alpha {\int }_{x_1}^{x_3}{x}^2\left(\hat{i}dx\right) \\ {W}_b=\frac{\alpha }{3}\left(x_3^3-{\mathit{x}}_{\mathbf{1}}^{\mathbf{3}}\right) \end{gathered}$$

Here$x_1$ and $x_3$ are the initial and final positions of the proton.

Substitute all the values in the above equation and we get,

$W_b=\frac{\left(12\hspace{0.33em}{\displaystyle \frac{N}{m^2}}\right)}{3}[{(0.30\hspace{0.33em}m)}^3-{(0.10\hspace{0.33em}m)}^3]W_b=0.104\hspace{0.33em}J$

Therefore, the work done on protons by force is 0.104 J .

The work done on proton by force is given as:

$W_c=-W_b$

Substitute all the values in the above equation and we get,

$$\begin{gathered} W_c=-(0.104\hspace{0.33em}J) \\ {W}_c=-0.104\hspace{0.33em}J \end{gathered}$$

Therefore, the work done on protons by force is - 0.104 J .

The net work done on the proton is zero and depends on the initial and final positions of the proton so the force on the proton is conservative.

The potential energy due to force is calculated as:

$$\begin{gathered} U={\int }_0^x\overrightarrow{F}·\left(\hat{i}dx\right) \\ U={\int }_0^x\left(\alpha x^{2}i\right)\left(\hat{i}dx\right) \\ U=\alpha {\int }_0^x{x}^{2}dx \\ U=\frac{\alpha x^3}{3} \end{gathered}$$

Therefore, the force on the proton is conservative and the potential energy of the proton is $\frac{\alpha x^3}{3}$.