Question

A force in the x direction acts on a particle moving also in the x direction, producing a potential energy U(x)=Ax4 where A=0.630J/m4. Find the magnitude and direction of this force, when the particle is at x=-0.800m?

Answer 1

The magnitude of the force is 1.29*10^-3N in the positive x direction

Step-by-step explanation:

In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:

`F(x)=-(dU(x))/(dx)` (1)

where:

`U(x)=Ax^4\\\\A=0.0630(J)/(m^4)`

You replace the expression for U into the equation (1) and solve for F:

`F(x)=-(d)/(dx)[Ax^4]=-4Ax^3` (2)

The force on the particle, for x = -0.080m is:

`F=-4(0.630(J)/(m^4))(-0.0800m)^3=1.29*10^(-3)N`

The magnitude of the force is 1.29*10^-3N in the positive x direction