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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`. |
Answer» 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` . |
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