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))`

Options

A. 2
B. 4
C. 6
D. 8

Answer

A

Explanation

(a) Given, `" "F=2hati+3hatj`
displacement `ds=dxhati+dyhatj+dzhatk`
work done, `W=intF.ds=int(2ds+3dy)`
Also`" "3y+kx=5 implies (3dy)/(dx)+k=0`
`implies 3dy=-kdx implies W=int(2dx-kdx)=0`
`implies 2x=kx implies k=2`
Alternately
`m_(1)m_(2)=-1`
`(3)/(2)(-(k)/(3))=-1 impliesk=2`