Question: Figure shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Non

Figure shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Non conservative forces are not involved.) Three values are UA = 15.0J, UB = 35.0 J, and JC = 45.0 J. The particle is released at x = 4.5 m with an initial speed of 7.0 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? What are the

(b) Magnitude and

(c) Direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead,

the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s.

(d) If the particle can reach x = 7 .0 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 = 50m?

Ucr Up UA 6. x (m) ()0

Ucr Up UA 6. x (m) ()0

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From Fig 850 we see that at x 45 m the potential energy is U 15 J If the speed is v 70 ms then the k... View full answer

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