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Force acting on a particale is (2hat(i)+...

Force acting on a particale is `(2hat(i)+3hat(j))N`. Work done by this force is zero, when the particle is moved on the line `3y+kx=5`. Here value of k is `(`Work done `W=vec(F).vec(d))`

A

`2`

B

`4`

C

`6`

D

`8`

Text Solution

Verified by Experts

The correct Answer is:
A
Displacement `dvec(r)=dxhat(i)+dyhat(j)`
but `3y+kx=5` so `3dy+kdx=0`
`rArr dvec(r)=dxhat(i)-k/3 dx hat(j)=(hat(i)-k/3 hat(j))dx`
Work done is zero if `vec(F).dvec(r)=0`
`(2hat(i)+3hat(j)).(hat(i)-k/3 hat(j))dx=0 rArr (2-k)dx=0 rArr k=2`
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