(C) The work done is given by the line integral $W = \int \overrightarrow F \cdot d\overrightarrow r = \int (-K(y\hat i + x\hat j)) \cdot (dx\hat i +

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(C) The work done is given by the line integral $W = \int \overrightarrow F \cdot d\overrightarrow r = \int (-K(y\hat i + x\hat j)) \cdot (dx\hat i + dy\hat j) = -K \int (y\,dx + x\,dy) = -K \int d(xy)$.Path $1$: From $(0, 0)$ to $(a, 0)$. Here $y = 0$,so $dy = 0$. The work done $W_1 = -K \int_0^a (0) dx = 0$.Path $2$: From $(a, 0)$ to $(a, a)$. Here $x = a$,so $dx = 0$. The work done $W_2 = -K \int_0^a (a) dy = -K[ay]_0^a = -Ka^2$.Total work done $W = W_1 + W_2 = 0 + (-Ka^2) = -Ka^2$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Variable Force and Force-Displacement Graph

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

IIT 1998 All IIT

A
$ - 2Ka^2$
B
$2Ka^2$
C
$ - Ka^2$
D
$Ka^2$

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Work Done by Variable Force and Force-Displacement Graph

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