Question

A force $F=-K(y\hat{i}+x\hat{j})$, (where $K$ is a +ve constant) acts on a particle moving in xy plane starting from origin, the particle is taken along the positive x-axis to the point ($a,0$) and then parallel to y axis to the point ($a,a$). The total work done by force $F$ on the particle is

• A. $-2K{a}^2$
• B. $2K{a}^2$
• C. $-K{a}^2$
• D. $K{a}^2$

Answer 1

Answer: (C)

$-K{a}^2$

Work was done $\int Fdt$

When particle moves from $0$ to $a$ along x-axis

Then $W_1={\int }_0^a-\left(Ky\hat{i}\right)dx=0$ $\because y=0$

Now, when the particle moves parallel to the y-axis

$W_2={\int }_0^a-\left(Ka\hat{j}\right)dy$ $\because x=a$
$\therefore W_2=-Ka{\int }_0^{a}dy=-K{a}^2$

$\therefore W=W_1+W_2=-K{a}^2$