The figure below shows a plot of potential energy U versus position x of a 0.76 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) In the graphs, thepotential energies are UA = 20.0 J, UB = 35.0 J, and Uc = 45.0 J.UcUBUA4х (m)The particle is released at x = 4.5 m with an initial speed of 6.0 m/s, headed in the negative x direction.(a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point?m(b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m?magnitudedirection+XSuppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 6.0 m/s.(c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point?m(d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m?magnitudedirection-X(r)n

Law of Conservation of Energy According to the law of conservation of energy, energy can neither be…

Answered: x (m) The particle is released at x =…

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