Question

A particle 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

Answer 1

Answer:

$F_{x}$ = -12 N ,  $F_{y}$ = -80 N  and  $F_{z}$ = - 44 N

Explanation:

The force is related to the potential energy by the formula

      F = -Δ U = - ( $\frac{dU}{dx}$ i ^ + $\frac{dU}{dy}$ j ^ + $\frac{dU}{dz}$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

       $\frac{dU}{dx}$ = 2yz + 0 + 4y  + 0 = 2yz + 4y

       $\frac{dU}{dx}$ = 2xz + 0 + 4x  + 0 = 2xz + 4x

      $\frac{dU}{dx}$ = 2xy + 3 2z + 0 +0 = 2xy + 6z

We look for the expression for the force in each axis

      $F_{x}$ = - 2yz - 4y

      $F_{y}$ = -2xz -4x

      $F_{z}$ = -2xy -6z

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

        $F_{x}$ = - 2 1 4 - 4 1

        $F_{x}$ = -12 N

        $F_{y}$ = - 2 20 4 - 4 20

        $F_{y}$ = -80 N

        $F_{z}$= - 20 1 - 6 4

        $F_{z}$ = - 44 N

We put together the expression for strength

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