1. The potential energy of a system of two particles separated by a distance r i

Question

1. The potential energy of a system of two particles separated by a distance r is given by the equation below, where A is a positive constant. U(r) = (A)/(r^11) Find the radial force vector F that each particle exerts on the other. (Use any variable or symbol stated above as necessary.) |vector F | = I already tried (11A/r^12) vector r and also tried (11A/r^12) vector r. 2. A single conservative force acts on a 4.90-kg particle within a system due to its interaction with the rest of the system. The equation Fx = 2x + 4 describes the force, where Fx is in newtons and x is in meters. As the particle moves along the x axis from x = 1.02 m to x = 6.60 m, calculate the following. (a) the kinetic energy the particle has at x = 6.60 m if its speed is 3.00 m/s at x = 1.02 m

Explanation / Answer

Write the equation as

U(r) = (A) x [r^(-4)]

Ask yourself "what power of r would I need to differentiate to get the expression [r^(-4)]?". If I differentiate [r^(-3)] I get
(-3)[r^(-4)], right?

So if I differentiate -(1/3)[r^(-3)], I will get
-(1/3){(-3)[r^(-4)]} = r^(-4), right?

Then -(1/3)[r^(-3)] is the antiderivative of r^(-4), and later you will say that the integral of
r^(-4) is -(1/3)[r^(-3)].

Using this concept, the integral of U(r) is the integral of
A/(r^4) = (A) x [r^(-4)] = (A) x {-(1/3)[r^(-3)]}, the force that each particle exerts on the other.

In the equation form you started with

F(r) = Integral U(r) = Integral (A)/(r^4) = -(A)/[3r^(3)]

By the way: you don't differentiate with respect to x since x is not the variable. Differentiate with respect to the variable r which is distance.

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