A force is given by Fx=Ax^-3, where A=8.0 N*m^3.(a) For positive values of x,does the potential energy associated with this force increase or decrease with increasing x?(You can determine the answer to this question by imagining what happens to a particle that is placed at rest at soke point x and is then released.) (b) Find the potential energy function U associated with this forse such that U approaches zero as x approaches infinity.

a) force= A/x^3 as x increases, it is positive, even though decreasing, it adds energy to a system.

b)

Potential energy= INT Fdx
= int Adx/x^3= -A/2x^2 +C
and apply the limits.

(a) To determine whether the potential energy associated with the force increases or decreases with increasing x, we need to analyze the behavior of the force as x increases.

Given that Fx = Ax^-3, we can see that the force is inversely proportional to the cube of x. As x increases, the value of Ax^-3 decreases, which means the force magnitude decreases.

Imagine a particle placed at rest at some point x and then released. As the particle moves away from the point of release, the force it experiences decreases. This implies that the force is doing negative work on the particle, slowing it down and causing it to lose potential energy. Therefore, the potential energy associated with this force decreases with increasing x.

(b) To find the potential energy function U associated with this force, we can integrate the force function with respect to x.

Given that Fx = Ax^-3, we integrate this expression with respect to x to find U:
∫ Fx dx = ∫ (Ax^-3) dx

Integrating with the appropriate power rule, we get:
U = -A/(2x^2) + C

Where C is a constant of integration. We can determine the value of C by approaching U to zero as x approaches infinity. This will ensure that the potential energy approaches zero at infinity.

As x approaches infinity, the term -A/(2x^2) becomes negligible compared to the constant C. To satisfy U approaching zero as x approaches infinity, we set C = 0.

Therefore, the potential energy function U associated with this force is:
U = -A/(2x^2), where A = 8.0 N*m^3.