The figure here shows a plot of potential energy U versus position x of a 0.854
ID: 1359749 • Letter: T
Question
The figure here shows a plot of potential energy U versus position x of a 0.854 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 J. The particle is released at x = 4.5 m with an initial speed of 7.4 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 7.4 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 = 5.0 m?
Explanation / Answer
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Figure 8-48 shows a plot of potential energy U versus position x of a .90 kg particle that can travel only along an x axis.(Nonconservative forces are not involved.) Three values are U1=15J, U2=35J, and U3=45J. 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 = 1.0m, 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= 4.0m? Suppose,instead, the particle is headed in the positive x direction when it released at x = 4.5 m at speed 7.0 m/s, (d) If the particle can reach x = 7.0m, 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 =5.0m
answer:
It's just conservation of energy U + K = U + K with K = 1/2mv^2 and the U's from the figure.
a) 1/2(.9)7^2 + 15 = Initial total energy = 37.1 J which is enough to take it to the 35 J level.
37.1 J = 35 + 1/2 m v^2 solve for v.
b) as it goes to the left past 4 it is slowing down so it gets pushed to the right. The particle looses 20J of energy over the 2 m between 4 and 2 m. So work done on particle = 20J = F x d = F x 2 so F = 10 N
c) The particle has only 22.1 J of kinetic energy which is not enough to get it up to the 45J level. It will stop and come back. Again 37.1 J = 0 + Uf so Uf = 37.1. Since the slope on the right is (45-15J)/1m = 30 J/m, and it has to go from 15 J to 37.1J , it will do this in (37.1 - 15)/30 = 0.74 m from the 5m point. So it will turn back at 5.74 m. It is slowing down at 5m so it is getting pushed to the left. Since it looses 30 J per meter, work done = 30 J = F x 1m or F = 30 N