Energy: Block sliding down frictionless ramp

AI Thread Summary
A block sliding down a frictionless ramp has speeds of 1.90 m/s at point A and 2.60 m/s at point B. The discussion involves calculating the speed at point B when the block's speed at point A is increased to 3.85 m/s. Participants reference kinetic energy (KE) and potential energy (PE) equations to analyze the problem. There is confusion regarding the height at point B, leading to requests for clarification on the solution process. The conversation emphasizes the importance of correctly applying energy conservation principles in this scenario.
basenne
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1. In Figure 8-49, a block is sent sliding down a frictionless ramp. Its speeds at points A and B are 1.90 m/s and 2.60 m/s, respectively. Next, it is again sent sliding down the ramp, but this time its speed at point A is 3.85 m/s. What then is its speed at point B?

W0150-N.jpg

Figure 8-49




2. KE = 1/2mv^2

PE = mg (h)



3. 1/2(m)(2.2^2) + m(9.8)(h) = 1/2(m)(2.6^2) + m(9.8)(.5h)

Yes, I know now that I can't assume that the height for point b is .5 of the height for point A. I'm totally lost on this question and I'd really appreciate a push in the right direction.
 
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Umm, I really appreciate the answer, however, would you care to elaborate how you got it?

And yeah, you're correct. (Although there was never any doubt that you wouldn't be :D)
 
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