Displacement of a particle of mass 2 kg varies with time as `s=(2t^(2)-2t + 10)m`. Find total work done on the particle in a time interval from `t=0` to `t=2s`.
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Correct Answer – B::C
`v=(ds)/(dt) =(4t-2)`
`W_(“all”) =DeltaK=K_(f)-K_(i)=K_(2s)-K_(0s)
`=1/2m(v_(f)^(2)-v_(i)^(2))
`=1/2 xx 2[(4 xx 2-2)^(2)-(4 xx 0-2)^(2)]`
`=32J` .