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?

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

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:

    (1)

where:

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

    (2)

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

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

compass can be used day and night, cloudy or sunny

not too sure thinking all accelerations are pointed down