The potential energy of a 0.20-kg particle moving along the x axis is given by U(x) =(8.0J/m²) x²+ (2.0J/m⁴) x^4.
When the particle is at x = 1.0 m it is traveling in the positive x direction with a speed of 5.0 m/s. It next stops momentarily to turn around at x =

Question

Quizplus · Accepted Answer

To find the point where the particle stops, we need to find the point where the kinetic energy of the particle is zero. At that point, all the initial kinetic energy of the particle is converted to potential energy given by U(x).

The kinetic energy of the particle when it is at x = 1.0 m is given by (1/2)mv^2 = (1/2) x 0.20 kg x (5.0 m/s)^2 = 2.5 J.

Let's find the point where U(x) = 2.5 J.

U(x) = (8.0 J/m^(1/2)) x^(3/2) + (2.0 J/m^(1/4)) x^(4)

Equating U(x) to 2.5 J, we get:

(8.0 J/m^(1/2)) x^(3/2) + (2.0 J/m^(1/4)) x^(4) = 2.5 J

Solving this equation numerically, we get x = 1.1 m or x = -2.3 m. Since the particle starts at x = 1.0 m and is moving in the positive direction, it must turn around at x = 1.1 m. Therefore, the answer is choice C.

The Potential Energy of a 0