The potential energy of a particle is determined by the expression `U=alpha(x^(2)+y^(2))`, where `alpha` is a positive constant. The particle begins to move from a point with the co-ordinates (3, 3) only under the action of potential fields force. When it reaches the point (1, 1) its kinetic energy is `4 Kalpha`. Find the value of K.

The potential energy of a particle is determined by the expression U=alpha(x^2+y^2) , where alpha is a positive constant. The particle begins to move from a point with coordinates (3, 3) , only under the action of potential field force. Then its kinetic energy T at the instant when the particle is at a point with the coordinates (1,1) is

The potential energy of a particle with displacement X is U(X) . The motion is simple harmonic, when (K is a positive constant)

The potential energy of a particle under a conservative force is given by U ( x) = ( x^(2) -3x) J. The equilibrium position of the particle is at

The potential energy of a particle under a conservative force is given by U(x)=(x^(2)-3x)J . The equilibrium position of the particle is at x m. The value of 10 x will be

A particle is acted upon by a force given by F = - alphax^(3) -betax^(4) where alpha and beta are positive constants. At the point x = 0, the particle is –

The potential energy of a point particle is given by the expression V(x) = -alpha x + beta sin (x//gamma) . A dimensionless combination of the constant alpha, beta and gamma is–

The potential energy of a particle of mass 1 kg in a conservative field is given as U=(3x^(2)y^(2)+6x) J, where x and y are measured in meter. Initially particle is at (1,1) & at rest then:

The potential energy function of a particle in the x-y plane is given by U =k(x+y) , where (k) is a constant. The work done by the conservative force in moving a particlae from (1,1) to (2,3) is .