The figure shows the potential energy U(x) of a solid ball that can roll along a
ID: 2186583 • Letter: T
Question
The figure shows the potential energyU(x)of a solid ball that can roll along anxaxis. The ball is uniform, rolls smoothly, and has a mass of 0.69 kg. It is released atx= 7.0 mheaded in the negative direction of thexaxis with a mechanical energy of 62 J.(a)If the ball can reachx= 0 m, what is its speed there, and if it cannot, what is its turning point? Suppose, instead, it is headed in the positive direction of thexaxis when it is released atx= 7.0 mwith 62 J.(b)If the ball can reachx= 13 m, what is its speed there, and if it cannot, what is its turning point?
Chapter 11, Problem 8Explanation / Answer
Ui + Ki = Uf + Kf initially at x = 7.0 m Ui = 0 J and Ki = 62 J Can ball reach x = 0 in negative direction? Uf = 100 J solve for Kf Kf = Ui + Ki - Uf = 0J + 62 J - 100 J = -38 J The ball can't reach x = 0 because it needs 38 J of energy to make it What is turning point? Kf = 0 Ui + Ki = Uf + Kf solve for Uf, Uf = Ui + Ki - Kf = 0 J + 62 J + 0J = 62 J Uf = 62 J is the turning point So if instead it travels in positive direction the ball can make it to 14 m because only 60 J of mechanical energy is needed, then ball starts of with 62 J of mechanical energy so it will have 2 J left over.
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