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
(1/2)v2 + (x4/4 - x2/2) = 2
v2 + x4/2 - x2 = 2
v2 = x2 - x4/2 + 2
v = [ x2 - x4/2 + 2]1/2.............................(1)
differentiating the above equation wrt time
dv/dt = (d/dt)[[ x2 - x4/2 + 2]1/2]
for velocity to be maximum dv/dt = 0
(d/dt)[[ x2 - x4/2 + 2]1/2] = 0
calculate the value of 'x' and substitute it in (1) to get maximum velocity.