[" A particle of mass "2kg" is subjected to a two dimensional conservative "],[" force given by,"F_(x)=-2x+2y,F_(y)=2x-y^(2)(x,y" in "m" and "F" in "N" ).If the "],[" particle has kinetic energy of "(8/3)J" at point "(2,3)" ,find the speed of the "],[" particle when it reaches "(1,2)" in "(m/s)]

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The minimum speed of the particle is:

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The maximum speed of the particle is:

The potential energy of a particle of mass 5 kg moving in the x-y plane is given by U=(-7x+24y)J , where x and y are given in metre. If the particle starts from rest, from the origin, then the speed of the particle at t=2 s is

A particle of mass m free to move in the x - y plane is subjected to a force whose components are F_(x) = - kx and F_(y) = - ky , where k is a constant. The particle is released when t = 0 at the point (2, 3) . Prove that the subsequent motion is simple harmonic along the straight line 2y - 3x = 0 .

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:

[" 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 particle of mass of mass m = 1 kg is lying on x-axis experiences a force given by F = x(3x-2) Newton, where x is the x-coordinate of the particle in meters. The points where the particle is in equilibrium are-

If a function satisfies (x-y)f(x+y)-(x+y)f(x-y)=2(x^(2)y-y^(3))AA x,y in R and f(1)=2 ,then

A single conservative force F(x) acts on a 1.0-kg particle that moves along the x-axis. The potential energy U(x) is given by U(x)=20+(x-2)^2 where x is in meters. At x=5.0m , the particle has a kinetic energy of 20J . The maximum kinetic energy of the particle and the value of x at which maximum kinetic energy occurs are

A single conservative f_((x)) acts on a m = 1 kg particle moving along the x-axis. The potential energy U_((x)) is given by : U_((x)) = 20 + (x - 2)^(2) where x is in metres. At x = 5 m , a particle has kintetic energy of 20 J The maximum value of potential energy is: