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)x⁴.When the particle is at x = 1.0 m the magnitude of its acceleration is: A)0 m/s² B)-8 m/s² C)8 m/s² D)-40 m/s² E)40 m/s²

Question

Quizplus · Accepted Answer

The magnitude of acceleration can be found by taking the derivative of the potential energy with respect to position twice and multiplying by the mass of the particle:

a = -m(d^2U/dx^2)

Taking the second derivative of U(x), we get:

d^2U/dx^2 = -16 J/m^3x + 6 J/m^4x^3

Plugging in x = 1.0 m and m = 0.20 kg, we get:

a = -(0.20 kg)(-16 J/m^3(1.0 m) + 6 J/m^4(1.0 m)^3)
  = -40 m/s^2

Therefore, the magnitude of the particle's acceleration is 40 m/s^2 in the negative direction, so the answer is D).

The Potential Energy of a 0