[" The potential energy of a particle of mass "2kg" moving along the "x" -axis is given by "],[U(x)=16(x^(2)-2x)" joule.Its velocity at "x=1m" is "2m/s" .Then : "]

The potential energy of particle of mass 1kg moving along the x-axis is given by U(x) = 16(x^(2) - 2x) J, where x is in meter. Its speed at x=1 m is 2 m//s . Then,

The potential energy of particle of mass 1kg moving along the x-axis is given by U(x) = 16(x^(2) - 2x) J, where x is in meter. Its speed at x=1 m is 2 m//s . Then,

The potential energy of a particle of mass 2 kg moving along the x-axis is given by U(x) = 4x^2 - 2x^3 ( where U is in joules and x is in meters). The kinetic energy of the particle is maximum at

The potential energy of a particle of mass 1 kg moving along x-axis given by U(x)=[(x^(2))/(2)-x]J . If total mechanical speed (in m/s):-

The potential energy of a particle of mass 1 kg moving along x-axis given by U(x)=[(x^(2))/(2)-x]J . If total mechanical speed (in m/s):-

athe potential energy of a particle of mass 2 kg moving along the x-axis is given by U(x) = 4x^2 - 2x^3 ( where U is in joules and x is in meters). The kinetic energy of the particle is maximum at

The potential energy of a particle of mass 1 kg free to move along x-axis is given by U(x) =(x^(2)/2-x) joule. If total mechanical energy of the particle is 2J then find the maximum speed of the particle. (Assuming only conservative force acts on particle)

The potential energy U(x) of a particle moving along x - axis is given by U(x)=ax-bx^(2) . Find the equilibrium position of particle.

The potential energy U(x) of a particle moving along x - axis is given by U(x)=ax-bx^(2) . Find the equilibrium position of particle.