The figure here shows a plot of potential energy U versus position x of a 0.894 kgparticle that can travel only along an x axis. (Nonconservative forces are notinvolved.) Three values are UÃ = 15.0 J, Uß = 35.0 J and Uc = 45.0 J. The particle isreleased at x = 4.50 m with an initial speed of 7.49 m/s, headed in the negative xdirection. (a) If the particle can reach x = 1.00 m, what is its speed there, and if itcannot, what is its turning point? What are the (b) magnitude and (c) direction of theforce on the particle as it begins 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 at x = 4.50 m atspeed 7.49 m/s. (d) If the particle can reach x = 7.00 m, what is its speed there, and ifit cannot, what is its turning point? What are the (e) magnitude and (f) direction ofthe force on the particle as it begins to move to the right of x = 5.00 m?(ΔΩUclUB2(a) Number(b) Number(c)(d) Number(e) NumberI1x (m)MO6UnitUnitUnitUnit

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