x (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? magnitude N direction +x Suppose, 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? magnitude N direction

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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, the
potential energies are UA = 20.0 J, UB = 35.0 J, and Uc = 45.0 J.
Uc
UB
UA
4
х (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?
magnitude
direction
+X
Suppose, 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?
magnitude
direction
-X
(r)n
Transcribed Image Text: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, the potential energies are UA = 20.0 J, UB = 35.0 J, and Uc = 45.0 J. Uc UB UA 4 х (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? magnitude direction +X Suppose, 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? magnitude direction -X (r)n
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