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Tne potential energy if a 1kg particle free to move along the x -axis is given by V (x)=(x^4/4-x^2/2)The total mechanical energy of the particle is 2J .then the maximum speed is?
Given,K.E. = 2 – (x^4/4-x^2/2)for maximum speed k.e. should be maximum that is:dK/dx = 0 – (x^3 – x) = 0x(x^2 – 1) = 0x = 1at x = 1K.E = 2 – (1/4 – 1/2) => 2 + ¼ = > 2.25 Joule.v = root(2*2.25) = > root(4.50) = > 2.12 m/s
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