A particle of mass `m=1kg` lying on x-axis experiences a force given by law
`F=x(3x-2)` Newton.
Where x is the x-coordinate of the particle in meters.
`vec(x=0x=4`
(a) Locate the point on x-axis where the particle is in equilibrium.
(b) Draw the graph of variation of force F (y-axis) with x-coordinate of the particle (x-axis). Hence or otherwise indicate at which positions the particle is in stable or unstable equilibrium.
(c ) What is the minimum speed to be imparted to the particle placed at `x=4` meters such that it reaches tha origin.

Explanation

(a) The particle is at equilibrium at `x=0` and `x=(2)/(3)m`.
(b)

The particle is in stable equilibrium at `x=0` metre and unstable equilibrium at `x=(2)/(3)` metre
(c ) The minimum speed inparted to the particle should be such that it just reaches `x=(2)/(3)` from there on it shall automatically reach`x=0`
`(1)/(2)mv^2=-int_4^((2)/(3))Fdx=-int_-4^((2)/(3))`x(3x-2)dx=(1300)/(27)`
or `v=sqrt((2600)/(27))(m)/(s)`