Aparticle of mass m = 13 kg moves in space under the action of a conservative force. its potential energy is given by pe = 2xyz + 3z2 + 4yx + 16 where pe is in joules and x, y, and z are in meters. calculate the x-, y-, and z-components of the force on the particle when it is at the position x = 20 m, y = 1 m, and z = 4 m. x-component: y-component: z-component

= -12 N ,   = -80 N  and   = - 44 N

Explanation:

The force is related to the potential energy by the formula

      F = -Δ U = - ( i ^ + j ^ + k ^)

It indicates the potential energy

      U = 2xyz + 3z² + 4yx + 16

To solve this problem let's make the derivatives, to find each component of the force

        = 2yz + 0 + 4y  + 0 = 2yz + 4y

        = 2xz + 0 + 4x  + 0 = 2xz + 4x

      = 2xy + 3 2z + 0 +0 = 2xy + 6z

We look for the expression for the force in each axis

      = - 2yz - 4y

      = -2xz -4x

      = -2xy -6z

We calculate at the point P = (20 i ^ + 1j ^ + 4 k ^) m

        = - 2 1 4 - 4 1

        = -12 N

        = - 2 20 4 - 4 20

        = -80 N

        = - 20 1 - 6 4

        = - 44 N

We put together the expression for strength

        F = (-12 i ^ - 80j ^ -44k ^) N

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