The potential energy of 1kg particle free to move along x-axis is given by U(x)=[x^4/4-x^2/2]J. The total mechanical energy of the particle is 2J. The maximum speed of the particle is

The potential energy of a 1kg particle free to move along the x-axis is given by V(x)=(x^(4)/4-x^(2)/2)J The total mechanical energy of the particle is 2J then the maximum speed (in m//s) is

The potential energy of a 2 kg particle free to move along the x- axis is given by V(x)[(x^(4))/(4)-(x^(2))/(2)]jou l e The total mechanical energy of the particle is 0.75 J . The maximum speed of the particle ( in m//s) is :

The potential energy of a 1 kg particle free to move along the x- axis is given by V(x) = ((x^(4))/(4) - x^(2)/(2)) J The total mechainical energy of the particle is 2 J . Then , the maximum speed (in m//s) is

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 PE of a 2 kg particle, free to move along x-axis is given by V(x)=((x^(3))/(3)-(x^(2))/(2))J. The total mechanical energy of the particle is 4 J. Maximum speed (in ms^(-1) ) is

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 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 of a particle of mass m is given by U=1/2kx^(2) for x lt0and U=0 for x ge0. If total mechanical energy of the particle is E, is speed at x=sqrt((2E)/(k)) is