Question

(^-> stands for vector)
Given F^-> = (xy²) i^{^} + (x²y)j^{^} Newton .
Find the work done by F^-> when a particle is taken along the semicircular path
OAB from O to B where the co-ordinates of B are (4,0) and O is the origin.
(1) 65/3 joules
(2) 75/2 joules
(3) 73/4 joules
(4) 0 joules

Answer 1

Dear Student ,
Here in this case the solution of the question is as follows :
$$\begin{gathered} Force,\overrightarrow{F}=x{y}^2\hat{i}+x^{2}y\hat{j} \\ Sotheworkdonebytheforceontheparticletakenalongasemicircuilarpathis, \\ W={\int }_{\left(0,0\right)}^{\left(4,0\right)}\overrightarrow{F}·d\overrightarrow{r} \\ ={\int }_{\left(0,0\right)}^{\left(4,0\right)}\left(x{y}^2\hat{i}+x^{2}y\hat{j}\right)·\left(dx\hat{i}+dy\hat{j}\right) \\ ={\int }_{\left(0,0\right)}^{\left(4,0\right)}x{y}^{2}dx+{\int }_{\left(0,0\right)}^{\left(4,0\right)}{x}^{2}ydy \\ =0 \end{gathered}$$
Regards