The potential energy of 1 kg particle , free to move allong is given by `U(x) = ((x^(3))/3-(x^(2))/2) I . ` if its mechanical energy is 2J . Its maximum speed is …… `ms^(-1)`

Options

A. `sqrt((13)/3)`
B. `sqrt((9)/7)`
C. `sqrt((7)/9)`
D. `sqrt((7)/6)`

Answer

A

Explanation

Potential energy `U(x) = (x^(3))/3 - (x^(2))/2 `
For maximum kinetic energy ( or maximum speed ) of a body , its potential energy should be minimum `( : . " On basis of law of conservation of mechanical energy "= U+K = " constant" )`
For minimum value 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 `
Here only at ` x = +1 ` m potential energy will be minimum for a body . therefore minimum potential energy .
`U_(min) = ((1)^(3))/3 - ((1)^(2))/2 = 1/3 - 1/2 = -1/6`
Now , maximum K.E = (Total energy means mechanical 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`