The potential energy of a 0.20-kg particle moving along the x axis is given by U(x) =(8.0 J/m²) x²+ (2.0 J/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 will stop momentarily to turn around, we need to find the point where the particle's velocity becomes zero. 
Using the conservation of energy, we can write:

K1 + U1 = K2 + U2

where K is the kinetic energy, U is the potential energy, 1 denotes the initial state at x = 1.0 m, and 2 denotes the final state when the particle momentarily stops to turn around.

At x = 1.0 m, the particle has a kinetic energy given by:

K1 = (1/2)mv^2 = (1/2)(0.2 kg)(5.0 m/s)^2 = 5.0 J

The potential energy at x = 1.0 m is given by:

U1 = (8.0 J/m)(1.0 m)^2 + (2.0 J/m)(1.0 m)^4 = 10 J

At the point where the particle momentarily stops to turn around, it has zero kinetic energy, so:

K2 = 0

Therefore, we can write:

U2 = K1 + U1

(8.0 J/m)x^2 + (2.0 J/m)x^4 = 5.0 J + 10 J

Simplifying:

2x^4 + 8x^2 - 15 = 0

This is a quadratic equation in x^2, so we can solve for x^2 using the quadratic formula:

x^2 = (-8 ± sqrt(8^2 - 4(2)(-15))) / (2(2))
x^2 = (-8 ± sqrt(136)) / 4
x^2 = (-8 ± 2sqrt(34)) / 4

Taking the positive root and simplifying:

x = sqrt(17) / 2

Therefore, the particle momentarily stops to turn around at x = 1.04 m (rounded to two significant figures), which corresponds to choice C.

The Potential Energy of a 0