Question

An object moving in the xy-plane is subjected to the force $\overrightarrow{F}=\left(2xy\hat{i}+x^2\hat{j}\right)\mathrm{N}$, where x and y are in m.

a. The particle moves from the origin to the point with coordinates $\left(a,b\right)$by moving first along the x-axis to $\left(a,0\right)$, then parallel to the y-axis. How much work does the force do?

b. The particle moves from the origin to the point with coordinates $\left(a,b\right)$by moving first along the y-axis to $\left(0,b\right)$, then parallel to thex-axis. How much work does the force do?

c. Is this a conservative force?

Short answer

(a) The work done $=a^{2}b$

(b) The work done $=a^{2}b$

(c) The force is conservative. Since the work done is path independent.

Step-by-step solution

Given information (part a)

The force acting on the object is given as,

$\overrightarrow{F}=\left(2xy\hat{i}+x^2\hat{j}\right)\mathrm{N}$

Explanation (part a)

The work done by the force is

$$\begin{gathered} w={\int }_{x_i}^{x_f}{F}_{s}dx \\ {\text{wherex}}_i\text{istheinitialpositionand}{x}_f\text{isthefinalposition} \\ \overrightarrow{F}=\left(2xy\hat{i}+x^2\hat{j}\right)\mathrm{N} \\ w={\int }_{x_{\mathrm{i}}}^{x_{\mathrm{f}}}\left(2\mathrm{xy}\hat{\mathrm{i}}+{\mathrm{x}}^2\hat{\mathrm{j}}\right)dx \end{gathered}$$

The work done by the force while moving along the x-axis$\left(a,0\right)$:

$$\begin{gathered} w={\int }_{x_i}^{x_f}\left(2xy\hat{i}+x^2\hat{j}\right)dx\hat{i} \\ {x}_i=0to{x}_f=a \\ y=0\text{alongthex-axis} \\ w={\int }_0^a\left(x^2\hat{j}\right)dx\hat{i} \\ =0 \end{gathered}$$

Now the particle is moved parallel to the y-axis.

role="math" localid="1649407041412" $$\begin{gathered} w_2={\int }_{x_i}^{x_f}\left(2xy\hat{i}+x^2\hat{j}\right)dy\hat{j} \\ {x}_i=0to{x}_f=b \\ y=0\text{alongthex-axis} \\ w={\int }_0^b\left(2xy\hat{i}+x^2\hat{j}\right)dy\hat{j} \\ w={\int }_0^b\left(x^2\right)dy \\ x=a \\ w={\int }_0^b\left(a^2\right)dy \\ ={\left[a^{2}y\right]}_0^b \\ =\left(a^{2}b-0\right) \\ =a^{2}b \end{gathered}$$

Total work done $w=w_1+w_2$

$$\begin{gathered} w=0+a^{2}b \\ w=a^{2}b \end{gathered}$$

Given information (part b)

The force acting on the object is given as,

$\overrightarrow{F}=\left(2xy\hat{i}+x^2\hat{j}\right)\mathrm{N}$

Explanation (part b)

The work done by the force is while moving the particle parallel to the y-axis.

$$\begin{gathered} w={\int }_{x_i}^{x_f}\left(2xy\hat{i}+x^2\hat{j}\right)ds \\ {x}_i=initialpositionand{x}_f=finalposition \\ x=0alongthey-axis \\ {w}_1={\int }_0^b\left(0\right)dy\hat{j} \\ =0 \end{gathered}$$

Work done while moving parallel to the x-axis and reach to (a, b).

$$\begin{gathered} w={\int }_{x_i}^{x_f}\left(2xy\hat{i}+x^2\hat{j}\right)dx \\ {x}_i=0and{x}_f=a \\ y=b \\ w={\int }_0^a\left(2xb\hat{i}+x^2\hat{j}\right)dx\hat{i} \\ ={\int }_0^a\left(2xb\right)dx \\ ={\left[x^{2}b\right]}_0^a \\ =\left(a^{2}b-0\right) \\ =a^{2}b \end{gathered}$$

Total work done $w=w_1+w_2$

$$\begin{gathered} w=0+a^{2}b \\ w=a^{2}b \end{gathered}$$

Given information (part c)

The force acting on the object is given as,

$\overrightarrow{F}=\left(2xy\hat{i}+x^2\hat{j}\right)\mathrm{N}$

Explanation (part c)

The force is conservative since the work done by the force is independent of the path followed by the particle. Work done is the same in both the paths followed while the particle reaches $(a,b)$.