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A point mass of `0.5` kg is moving along `x-` axis as `x=t^(2)+2t` , where, `x` is in meters and `t` is in seconds. Find the work done (in J) by all the forces acting on the body during the time interval `[0,2s]`.A. `6J`B. `7J`C. `8J`D. `9J` |
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Answer» Correct Answer - C Method `I:` `W_(all)=DeltaKE` `=(1)/(2)m(v_(2)^(2)-v_(0)^(2))` `v=(dx)/(dt)=2t+2=2(t+1)` `v_(2)=6m//s,v_(0)=2m//s` `W_(all)=(1)/(2)(1/2)(6^(2)-2^(2))=8J` Method `II :` `x=t^(2)+2t` `a=(d^(2)x)/(dt^(2))=2m//s^(2)=` constant so `F=ma=1N=` constant so, `W=Fd=(1)(8-0)=8J` |
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