Energy: Block sliding down frictionless ramp

  • Thread starter basenne
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  • #1
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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 cant 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.
 

Answers and Replies

  • #2
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Note: answer deleted by moderator.
 
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  • #3
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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)
 
  • #4
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Note: detailed solution deleted by moderator.
 
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