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The potential energy of a 0.20-kg particle moving along the x axis is given byU(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:

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 =

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