InterviewSolution
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InterviewSolution
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The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `sqrt(2)`B. `1//sqrt(2)`C. `2`D. `3//sqrt(2)` |
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Answer» Correct Answer - C Max. `K.E.=(1)/(2)m upsilon_(max)^(2)0`total mechanical energy `:. (1)/(2)xx1xxupsilon_(max)^(2)=2` `upsilon_(max)=sqrt(2xx2)=2m//s` |
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