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Question

A force F=K(y^i+x^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). Find the total work done by the forces F on the particle.


A

Ka2

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B

-2 Ka2

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C

2 ka2

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D

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Solution

The correct option is D


F=k(y^i+x^j)

dx=(dx^i+dy^j+dz^k)

w=F.dS

= k(y^i+x^j)(dx^i+dy^j+dz^k)

w=[k y dxk x dy]

Now particle is starting from (0, 0) origin and going to (a, 0)

So x varies from 0 - a

Y varies from 0 - 0
So w=ky dx+kx dy

= kya0 dxkx00 dy

(here y is not changing (0 - 0))

K0.[x]20k x [y]00

00=0

Now particle goes from (a, 0) to (a, a)

So x varies from a - a

Y varies from 0 - a

w=ky dx+kx dy

kyaakxa0 dy

= ky[x]aakx[y]a0

=0k a a

W=ka2


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