InterviewSolution
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InterviewSolution
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A particle of mass `0.5kg` travels in a straight line with a velocity `v=(5x^(5//2))m//s`. How much work is done by the net force during the displacement from `x=0` to `x=2m` ? |
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Answer» Here,`m=0.5kg` `v=5x^(5//2),W=?` When `x=0,v_(1)=0` When `x=2m, v_(2)=5(2)^(5//2)` Using work energy theorem, `W=` change in K.E. `=(1)/(2)mv_(2)^(2)-(1)/(2)mv_(1)^(2)` `W=(1)/(2)m(v_(2)^(2)-v_(1)^(2))=(1)/(2)xx0.5[{5(2)^(5//2)}^(2)-0]` `=(1)/(2)xx(1)/(2)xx25xx2^(5)=200J` |
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