The figure here shows a plot of potential energy U versus position x of a 0.898 kg particle that can travel only along an x axis.(Nonconservative forces are not involved.) Three values are UA = 15.0 J, UB = 35.0 J and Uc = 45.0 J. The particle is released atx = 4.50 m with an initial speed of 7.52 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.00 m, what is itsspeed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as itbegins to move to the left of x = 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released atx = 4.50 m at speed 7.52 m/s. (d) If the particle can reach x = 7.00 m, what is its speed there, and if it cannot, what is its turning point?What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.00 m?UcUBUA24x (m)(a) NumberiUnit(b) NumberiUnit(c)(d) NumberiUnit(e) NumberiUnit(f)

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