Potential Energy and Total Mechanical Energy

The potential energy of a particle moving along the x-axis is given by the equation U = 8x^2 - 2x^4, where U is in joules and x is in meters. In this problem, we are given that the total mechanical energy is 9 J. We need to determine the limits of motion based on this information.

Understanding Potential Energy and Total Mechanical Energy

In classical mechanics, the total mechanical energy of a system is the sum of the potential energy (U) and the kinetic energy (K) of the system. Mathematically, it can be expressed as E = U + K, where E is the total mechanical energy.

The potential energy is dependent on the position of the particle in the system. It represents the energy stored in the system due to its position or configuration. The kinetic energy, on the other hand, is dependent on the velocity of the particle and represents the energy associated with its motion.

Finding the Limits of Motion

To find the limits of motion for the particle, we need to analyze the potential energy equation and the given total mechanical energy.

First, let's set up the equation for the total mechanical energy:
E = U + K

Since the potential energy equation is given as U = 8x^2 - 2x^4, we can substitute it into the total mechanical energy equation:
9 = (8x^2 - 2x^4) + K

Simplifying the equation, we have:
9 = 8x^2 - 2x^4 + K

To find the limits of motion, we need to find the values of x that satisfy the equation. In other words, we need to find the values of x where the potential energy plus the kinetic energy equals the total mechanical energy.

Identifying the Key Points

To solve the equation, we need to determine the key points of interest. These key points include the maximum and minimum potential energy points as well as the turning points where the kinetic energy is zero.

1. Maximum and Minimum Potential Energy Points:
To find the maximum and minimum potential energy points, we can take the derivative of the potential energy equation with respect to x and set it equal to zero. The resulting values of x will be the x-coordinates of the maximum and minimum points. Taking the derivative, we have:

dU/dx = 16x - 8x^3

Setting this equal to zero, we get:
16x - 8x^3 = 0

Factoring out x, we have:
x(16 - 8x^2) = 0

Setting each factor equal to zero, we find:
x = 0, or x^2 = 2

Solving for x^2, we get:
x = 0, or x = ±√2

2. Turning Points:
To determine the turning points, we need to find the values of x where the kinetic energy is zero. At these points, the particle changes direction, and the kinetic energy is momentarily zero.

The kinetic energy is given by the equation:
K = E - U

Substituting the given values, we have:
K = 9 - (8x^2 - 2x^4)

Setting K equal to zero, we get:
0 = 9