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. `(1)/(sqrt(2))`B. `2`C. `(3)/(sqrt(2))`D. `sqrt(2)` |
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Answer» Correct Answer - C Total mechanical energy of particle, `E_(T) = 2J` When kinetic energy is maximum, the potential energy should be minimum. The potential of the particle is given by `V (x) = (x^(4))/(4) -(x^(2))/(2)` or `(dV)/(dx) =(4x^(3))/(4)-(2x)/(2) = x^(3) - x = x (x^(2) -1)` For `V` to be minimum, `(dV)/(dx) = 0` `:. x(x^(2) -1) = 0`, or `x = 0 pm 1` At `x = 0, V (x) = 0` At `x = "pm" 1, V (x) = - (1)/(4) J` `:. ("Kinetic energy")_(max) = E_(T) - V_(min)`. or `("Kinetic energy")_(max) = 2 2-(-(1)/(4)) = (9)/(4) J` or `(1)/(2) mv_(m)^(2) = (9)/(4)` or `v_(m)^(2) = (9xx2)/(mxx4)` or `v_(m)^(2) = (9xx2)/(1xx4) = (9)/(2)` `rArr v_(m) = (3)/(sqrt(2)) m//s`. |
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