Question

A particle is displaced from $(1,2)\text{m}$ to $(0,0)\text{m}$ along the path $y=2x^3$. Work done by a force $\overrightarrow{F}=(x^3\hat{j}+y\hat{i})\text{N}$ acting on the particle, during this displacement is (in $\text{J}$)

Answer 1

Given $F=(x^3\hat{j}+y\hat{i})\text{N}$
Work done by variable force is
$W=\int \overrightarrow{F}.\overrightarrow{dr}=\int x^{3}dy+\int ydx$
$W=\underset{2}{\overset{0}{\int }}\frac{y}{2}dy+\underset{1}{\overset{0}{\int }}2x^{3}dx$
$W=\frac{1}{2}{\left(\frac{y^2}{2}\right)}_2^0+2{\left(\frac{x^4}{4}\right)}_1^0=-1.5\text{J}$