Question

Which of the following is/are conservative force(s)?

• A. $\overrightarrow{F}=2r^3\hat{r}$
• B. $\overrightarrow{F}=-\frac{5}{r}\hat{r}$
• C. $\overrightarrow{F}=\frac{3(x\hat{i}+y\hat{j})}{(x^2+y^2{)}^{\frac{3}{2}}}$
• D. $\overrightarrow{F}=\frac{3(y\hat{i}+x\hat{j})}{(x^2+y^2{)}^{\frac{3}{2}}}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\overrightarrow{F}=2r^3\hat{r}$
B $\overrightarrow{F}=-\frac{5}{r}\hat{r}$
C $\overrightarrow{F}=\frac{3(x\hat{i}+y\hat{j})}{(x^2+y^2{)}^{\frac{3}{2}}}$

Since: $W=\int \overrightarrow{F}.\overrightarrow{dr}$
Clearly for forces (A) and (B) the integration do not require any information of the path taken it depends only on initial and final positions.
For (C): $W_c=\int \frac{3(x\hat{i}+y\hat{j})}{(x^2+y^2{)}^{\frac{3}{2}}}(dx\hat{i}+dy\hat{j})=3\int \frac{xdx+ydy}{(x^2+y^2{)}^{\frac{3}{2}}}$
Taking : $x^2+y^2=t$
$\text{2xdx + 2y d y = dt}$
$\Rightarrow xdx+ydy=\frac{dt}{2}\Rightarrow W_c=3\int \frac{\frac{dt}{2}}{t^{\frac{3}{2}}}=\frac{3}{2}\int \frac{dt}{t^{\frac{3}{2}}}=\frac{-3}{\sqrt{x^2+y^2}}+C$
Work done is independent of the path.
Hence (A), (B)and (C) are conservative forces.
For (D):
$W_d=\int \frac{3(y\hat{i}+x\hat{j})}{(x^2+y^2{)}^{\frac{3}{2}}}(dx\hat{i}+dy\hat{j})=3\int \frac{ydx+xdy}{(x^2+y^2{)}^{\frac{3}{2}}}$
(D) requires some more information on path followed by particle to calculate work done. Hence non-conservative in nature.