1.

Under the action of a constant force, a 2 kg body moves such that its position along X-axis is given by x=(t^(2))/(3) where x is in metres and in seconds and x is function time. The work done in 2 sec is:

Answer»

1.6 J
`(16)/(9) J`
`(9)/(16) J`
160 J

Solution :`x=(t^(2))/(3).:. (DX)/(dt)=(2t)/(3)`
implies `(d^(2)x)/(dt)=2/3`
Since F=ma implies `F=2xx2/3=4/3N`
Displacement in TWO seconds is `x=((2)^(2))/(3)=4/3`.
`:.` W=FS
`=4/3xx4/3=(16)/(9)J`.


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