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A force acts on a 2 kg object so that it...

A force acts on a 2 kg object so that its position is given as a function of time as `x=3t^(2)+5.` What is the work done by this force in first 5 seconds ?

A

850 J

B

900 J

C

875 J

D

950 J

Text Solution

Verified by Experts

The correct Answer is:
A
`v = (dx)/(dt) = d/(dt)(3t^(2)+5)=6t`
`K.E_(i) = 1/2 xx2xx (0)^(2) = 0 and K.E_(f)`
`=1/2 xx2xx(30)^(2) = 900`
From work energy theorem
`W = K.E_(f) - K.E_(i) = 900 J `
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