[" 5.The potential energy function for a two dimensional force is "],[U=+2x^(3)-4x^(2)y+6y^(2)" where "U" is in joules,"x" and "y" are in metres.The force (in "N)],[" acting on particle at "(3m,2m)" is "(Ahat i+Bhat j)" newtons.Find "A+B?]

A potential energy function for a two-dimensional force is the form U=3x^2y-7x . Find the force that acts at the point (x,y) .

The potential energy function of a particle in a region of space is given as U=(2xy+yz) J Here x,y and z are in metre. Find the force acting on the particle at a general point P(x,y,z).

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

Potential energy of a particle along x-axis varies as U = - 30 + (X-2)^2 , where U is in joule and x is in meter. Hence the particle is in

If the potential energy function of a particle is given by U=-(x^2+y^2+z^2) J, whre x,y and z are in meters. Find the force acting on the particle at point A(1m,3m,5m) .

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 .

The potential energy (in joules ) function of a particle in a region of space is given as: U=(2x^(2)+3y^(2)+2x) Here x,y and z are in metres. Find the maginitude of x compenent of force ( in newton) acting on the particle at point P ( 1m, 2m, 3m).

The potential energy of a partical of mass 2 kg moving in y-z plane is given by U=(-3y+4z)J where y and z are in metre. The magnitude of force ( in newton ) on the particle is :

Potential energy of a particle moving along x-axis is by U=(x^(3)/3-4x + 6) . here, U is in joule and x in metre. Find position of stable and unstable equilibrium.

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: