Bowling Ball Rise: Find Speed at Top w/Conservation of Energy

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The discussion revolves around calculating the translational speed of a bowling ball at the top of a vertical rise of 0.662 m, given its speed of 3.70 m/s at the bottom. Participants emphasize using the conservation of energy principle, where the initial kinetic energy (KE) at the bottom should equal the sum of potential energy (U) gained and the final kinetic energy at the top. The equation set up is (KE + U)_1 = (KE + U)_2, but there is confusion regarding the correct application, particularly since the mass of the ball is not provided. Attempts to solve the problem yield incorrect results, indicating a misunderstanding of energy conservation in this context. The final calculated speed of 0.837 m/s is disputed, suggesting further clarification is needed on the energy transformation process.
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Homework Statement



A bowling ball encounters a h = 0.662 m vertical rise on the way back to the ball rack, as the figure below illustrates.

Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3.70 m/s at the bottom of the rise. Find the translational speed at the top.

Homework Equations



KE= 1/2mv^2

U=mgh

Conservation of energy and momentum?


The Attempt at a Solution


I tried basic conservation of energy but that didn't really work out. I'm not sure where to start on this one since it doesn't give the mass of the ball. Though that probably cancels out anyway. When I set KE = U the answer is incorrect. I'm not sure on what else I can do.
 
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The initial kinetic energy is not equal to the change in potential energy. The initial energy is equal to the change in potential energy plus the final kinetic energy.
(KE + U)_1 = (KE + U)_2
 
So:

1/2mvi2=mgh + 1/2mvf2?


When I plug in the values for this the answer is incorrect as well.
 
The answer is not 0.837 m/s?
 
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