Question

By applying a force $F=(3xy-5z)j+4zk$ a particle is moved along the path $y=x^2$ from point $(0,0,0)$ to the point $(2,4,0)$. The work done by the F on the particle is (all values are in SI units)

• A. $\frac{280}{5}J$
• B. $\frac{140}{5}J$
• C. $\frac{232}{5}J$
• D. $\frac{192}{5}J$

Answer 1

The correct option is D $\frac{192}{5}J$

$\overline{F}=(3xy-5z)\hat{j}+4z\hat{k}$

$x^2=y$

$x=y^{1/2}$

Substitute in above equation

$\overline{F}=(3y^{3/2}-5z)\hat{j}-4z\hat{k}$

dw = F.dx

In $\left(\hat{j}\right)$ direction

$W{=}_0^4\int 3y^{3/2}dy$

$\to \frac{3y^{5/2}}{5/2}{|}_0^4$

$W=\frac{192}{5}$ joule

(in 'z' work done is always zero)