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 energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

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

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 in motin along the x-axis is given by U = 4(1-cos2x)J Here, x is in meter. The period of small osciallationis (in second) is