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The potential energyof 1kg particle, free tomove allongis givenby U(x) = ((x^(3))/3-(x^(2))/2) I .if its mechanicalenergyis 2J . Itsmaximum speedis …… ms^(-1) |
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Answer» `sqrt((13)/3)` For maximum kinetic energy( or maximum SPEED ) of a body, its potential energy should beminimum`( : . " On BASISOF lawof conservationofmechanical energy "= U+K= " CONSTANT" )` For minimumvalue of`U = (dU)/(dx) = 0 and (d^(2)U)/(dx^(2)) gt 0 ` Now , `(dU)/(dx) = ((3x^(2))/3 - (2x)/2)` Taking `(dU)/(dx) = 0 ` ` :. 0 = ((3x^(2))/3 - (2x)/3)` `:. 0 = (x^(2) - x)` ` :. x = 0 m ` or we get`x = +1 m ` Now ,`(d^(2)U)/(dx^(2)) ` is `2x-1 ` x=0 m , `(d^(2)U)/(dx^(2)) = 2xx0 - 1 = - 1 lt 0 ` and For`x = + 1 m , (d^(2)U)/(dx^(2)) = 2XX1- 1 = 1 gt 0 ` Hereonlyat ` x = +1 ` m potential energy will beminimum for a body. thereforeminimum potential energy . `U_(min) = ((1)^(3))/3 - ((1)^(2))/2 =1/3 - 1/2 = -1/6` Now , maximum K.E= (Total energymeansmechanical energy ) - (minimum potential energy ) `:. 1/2 mv_(max)^(2) = 2 - (-1/6) = 13/6 J ` ` :.v_(max)^(2) = 13/3 ""( :.m = 1 kg)` ` :. v_(max) = sqrt((13)/3)m//s` |
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