The potential energy of 1kg particle free to move along the x-axis is U(x)=x^4/4-x^2/2 J (x in meters). if the total mechanical energy of the particles is 2J find the maximum speed of the particle.

we know that

mechanical energy = kinetic energy + potential energy

T = E + U = 1/2mv² + (x⁴/4 - x²/2) = 2 J

here, m = 1 kg

so,

(1/2)v² + (x⁴/4 - x²/2) = 2

or

v² + x⁴/2 - x² = 2

or

v² = x² - x⁴/2 + 2

so,

v = [ x² - x⁴/2 + 2]^1/2.............................(1)

now,

differentiating the above equation wrt time

dv/dt = (d/dt)[[ x² - x⁴/2 + 2]^1/2]

so,

for velocity to be maximum dv/dt = 0

so,

(d/dt)[[ x² - x⁴/2 + 2]^1/2] = 0

calculate the value of 'x' and substitute it in (1) to get maximum velocity.