Answer Happy

Show Attempt History Current Attempt in Progress Your answer is partially correct The figure here shows a plot of potential energy U versus position x of a 0.894 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are U - 15.0J, U - 35.0) and Uc - 45.0 J. The particle is released at X - 4.50 m with an initial speed of 740 m/s, headed in the negative x direction (a) If the particle can reach x = 1.00 m, what is its speed 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 it begins to move to the left of x = 4.00m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.50 m at speed 7.40m/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 (direction of the force on the particle as it begins to move to the right of x-5.00 m? Uch UM UOT UM 6 4 x (m) (a) Number 3.165 Unit m/s (b) Number 10 Unit N (c) positive (+) (d) Number 5214 Unit m (e) Number 30 Unit N (f) negative)