In case of negative wark, the angle between the force and displacement is

Options

A. 0
B. `45^(@)`
C. `90^(@)`
D. `180^(@)`

Answer

D

Explanation

(a) Work done `W = F.d cos0^(@)`
`:."Work done at "theta=0^(@),W=F.dcos0^(@)" " (":.cos0^(@)=1)`
`rArrW=F.d`
For angle `theta=0^(@),`
Work done is positive, so it is not true.
(b) We know that work done, `W=F.dcostheta`
`:."Work done at "theta=45^(@)`
`W=F.dcos45^(@)" "(":'cos45^(@)=(1)/(sqrt2))`
`W=(F.d)/(sqrt2)`
For angle `theta=45^(@),`
work done is positive, so it is not true.
(c) We know that work done, `W=F.dcostheta`
`:."Work done at "theta=90^(@),W=F.dcos90^(@)" "(":.cos90^(@)=0)`
W = 0
So, it is not true,
(d) Work done at `theta=180^(@),W=F.dcostheta" "(":'cos180^(@)=-1)`
W =- F.d
For negative work, the angle between the force and displacement should be `180^(@).` (i.e.,force and displacement are antiparallel to each other) So, it is true.