Question

An object moving in the xy-plane is subjected to the force $\overrightarrow{F}=(2xy\hat{ı}+3y\hat{ȷ})N$, where x and y are in m.

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

c. Is this a conservative force?

Short answer

(a) The work done $=\frac{3b^2}{2}$

(b) The work done $=\frac{3b^2}{2}+a^{2}b$

(c) The force is non-conservative. Since the work done depends on the path.

Step-by-step solution

Given information (part a)

The force acting on the object is given as,

$\overrightarrow{F}=(2xy\hat{i}+3y\hat{j})\mathrm{N}$

Explanation (part a)

The work done by the force is

$$\begin{gathered} w_1={\int }_{x_i}^{x_f}Fdx \\ where{x}_{i}and{x}_{f}isinitialandfinalpositionrespectively. \\ Theparticalismovedalongthex-axisfrom\left(0,0\right)to\left(a,0\right) \\ {x}_i=0,x_f=a \\ \overrightarrow{F}=(2xy\hat{i}+3y\hat{j})\mathrm{N} \\ \mathrm{y}=0\mathrm{for}\mathrm{particle}\mathrm{moving}\mathrm{along}\mathrm{x}-\mathrm{axis} \\ {w}_1={\int }_{x_{\mathrm{i}}}^{x_{\mathrm{f}}}(2\mathrm{xy}\hat{\mathrm{i}}+3\mathrm{y}\hat{\mathrm{j}})dx\hat{\mathrm{i}} \\ {w}_1={\int }_0^a\left(0\right)dx\hat{i} \\ =0 \end{gathered}$$

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

$$\begin{gathered} w_2={\int }_{y_i}^{y_f}(2xy\hat{i}+3y\hat{j})dy \\ Sincetheparticleismovingfrom\left(a,0\right)to\left(a,b\right) \\ {y}_i=0,y_f=b,x=a \\ {w}_2={\int }_0^b(2ay\hat{i}+3y\hat{j})dy\hat{j} \\ ={\int }_0^{b}3ydy \\ ={\left[\frac{3y^2}{2}\right]}_0^b \\ =\left(\frac{3b^2}{2}-0\right) \\ =\frac{3b^2}{2} \\ Totalworkdonew=w_1+w_2 \\ w=0+\frac{3b^2}{2} \\ w=\frac{3b^2}{2} \end{gathered}$$

Given information (part b)

The force acting on the object is given as,

$\overrightarrow{F}=(2xy\hat{i}+3y\hat{j})\mathrm{N}$

Explanation (part b)

The work done by the force

$$\begin{gathered} w={\int }_{y_i}^{y_f}(2xy\hat{i}+3y\hat{j})dy \\ {y}_{i}and{y}_{f}istheinitialandfinalpositionrespectively \\ Theparticleismovingalongthey-axisfrom\left(0,0\right)to\left(0,b\right) \\ {y}_i=0and{y}_i=b \\ w={\int }_0^b(2ay\hat{i}+3y\hat{j})dy\hat{j} \\ w={\int }_0^b\left[3y\right]dy \\ w={\left[\frac{3y^2}{2}\right]}_0^b \\ w=\left(\frac{3b^2}{2}-0\right) \\ w=\frac{3b^2}{2} \end{gathered}$$

Now the particle is moved parallel to the x-axis from $\left(0,b\right)$to $\left(a,b\right)$.

$$\begin{gathered} w_2={\int }_{x_i}^{x_f}(2xy\hat{i}+3y\hat{j})dx\hat{i} \\ {x}_i=0,x_f=a,y=b \\ {w}_2={\int }_0^a(2xb\hat{i}+3y\hat{j})dx\hat{i} \\ {w}_2={\int }_0^a\left(2xb\right)dx \\ {w}_2={\left[x^{2}b\right]}_0^a \\ {w}_2=a^{2}b \\ Totalworkdonew=w_1+w_2 \\ w=\frac{3b^2}{2}+a^{2}b \end{gathered}$$

Given information (part c)

The force acting on the object is given as,

$\overrightarrow{F}=(2xy\hat{i}+3y\hat{j})\mathrm{N}$

Explanation (part c)

The force is non-conservative since the work done by the force is dependent on the path followed by the particle. Work done is different in both the paths followed while the particle reach (a, b)