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 = 1/2mv^{2} + (x^{4}/4 - x^{2}/2) = 2 J

here, m = 1 kg

so,

(1/2)v^{2} + (x^{4}/4 - x^{2}/2) = 2

or

v^{2} + x^{4}/2 - x^{2} = 2

or

v^{2} = x^{2} - x^{4}/2 + 2

so,

v = [ x^{2} - x^{4}/2 + 2]^{1/2}.............................(1)

now,

differentiating the above equation wrt time

dv/dt = (d/dt)[[ x^{2} - x^{4}/2 + 2]^{1/2}]

so,

for velocity to be maximum dv/dt = 0

so,

(d/dt)[[ x^{2} - x^{4}/2 + 2]^{1/2}] = 0

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

**
**