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

`F_(x)` = -12 N ,
`F_(y)` = -80 N and
`F_(z)` = - 44 N

Step-by-step explanation:

The force is related to the potential energy by the formula

F = -Δ U = - (
`(dU)/(dx)` i ^ +
`(dU)/(dy)` j ^ +
`(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

`(dU)/(dx)` = 2yz + 0 + 4y + 0 = 2yz + 4y

`(dU)/(dx)` = 2xz + 0 + 4x + 0 = 2xz + 4x

`(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