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)`.

[" 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?]

The potential energy of a certain particle is given by U = 30x^2 - 20 y^2 . Find the force acting on the particle.

Figure given below shows the variation of potential energy function U(x), corresponding to a particle lying in a one dimensional force field. The force acting on the particle at x = 2 m is:

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 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 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 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 potnetial energy for a force field vecF is given by U(x, y)=cos(x+y) . The force acting on a particle at the position given by coordinates (0, pi//4) is

The potential energy of configuration changes in x and y directions as U=kxy , where k is a positive constant. Find the force acting on the particle of the system as the function of x and y.