A force F = - K(y i + x j) (where K is a positive constant) acts … - Vidyadip

MCQ

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 forces $\overrightarrow F $ on the particle is

A
$ - 2\,K{a^2}$
B
$2\,K{a^2}$
✓
$ - K{a^2}$
D
$K{a^2}$

✓

Answer

Correct option: C.

$ - K{a^2}$

c
(c) For motion of the particle from $(0, 0)$ to $(a, 0)$

$\overrightarrow F = - K(0\,\hat i + a\,\hat j)$ $ \Rightarrow \,\,\overrightarrow F = - Ka\hat j$

Displacement $\overrightarrow {r\,} = (a\,\hat i + 0\,\hat j) - (0\,\hat i + 0\,\hat j) = a\hat i$

So work done from $(0, 0)$ to $(a, 0)$ is given by

$W = \overrightarrow F \,.\,\overrightarrow {r\,} $$ = - Ka\hat j\,.\,a\hat i = 0$

For motion $(a, 0)$ to $(a, a)$

$\overrightarrow F = - K(a\hat i + a\hat j)$ and displacement

$\overrightarrow {r\,} = (a\hat i + a\hat j) - (a\hat i + 0\hat j) = a\hat j$

So work done from $(a, 0)$ to $(a, a)$ $W = \overrightarrow F \,.\,\overrightarrow {r\,} $

$ = - K(a\hat i + a\hat j)\,.\,a\hat j = - K{a^2}$

So total work done$ = - K{a^2}$

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