A force F=−K(yi+aj) (where K is a positive constant) acts on a particle

Discussion Forum : Work Power And Energy

Question -

A force $F = - K (y i + a j)$ (where K is a positive constant) acts on a particle moving in the xy-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 F on the particles is

Options:

A .

- 2 K a^{2}

B .

2 K a^{2}

C .

- K a^{2}

D .

K a^{2}

Answer: Option C
:
C

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0 $∴ \to F = - k x^j$

i.e. force is in negative y-direction while displacement is in positive x-direction.

$∴ W_{1} = 0$

Because force is perpendicular to displacement

Then particle moves from (a,0) to (a,a) along a line parallel to y-axis (x = +a) during this $\to F = - k (y^i + a^j)$

The first component of force, $- k y^i$ will not contribute any work because this components is along negative x-direction ( $-^i)$ while displacement is in positive y-direction (a,0) to (a,a).

The second component of force i.e. $- k a^j$

$∴ W_{2} = (- k a^j) (a^j) = (- k a) (a) = - k a^{2}$

So net work done on the particle $W = W_{1} + W_{2}$

$= 0 + (- k a^{2}) = - k a^{2}$

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