Multiple Choice
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 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 =
A) 0 m
B) -1.1 m
C) 1.1 m
D) -2.3 m
E) 2.3 m
Correct Answer:

Verified
Correct Answer:
Verified
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