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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 ?

Text Solution

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Position `x=3t^(2)+5`
`:.` Velocity `v=(dx)/(dt)`
`=v=(d(3t^(2)+5))/(dt)`
`impliesv=6t+0`
At t=0 v=0
And at t=5 sec v=30m/s
According to work energy theorem `w=DeltaKE`
or `W=1/2 mv^(2)-0=1/2(2)(30)^(2)=900J`
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