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How many times as high to achieve twice speed

  1. Nov 11, 2009 #1
    A block initially at rest is allowed to slide down a frictionless ramp and attains a speed of v at the bottom. In order to achieve a speed of 2v instead at the bottom, how many times as high must the new ramp be?

    What laws does this pertain to? I don't really know where to start.
     
  2. jcsd
  3. Nov 11, 2009 #2

    rl.bhat

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    Homework Helper

    It pertains to the conservation of energy.
    mgh = 1/2*m*v^2.
     
  4. Nov 11, 2009 #3
    But I don't understand...velocity is only on one side of that equation. So how would you work it out for this problem?
     
  5. Nov 11, 2009 #4

    Pengwuino

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    Gold Member

    Exactly! What you have is the height of an object as a function of it's speed at the end. So if you know you must be a height, h, to achieve a velocity v, what must h be in order to achieve 2v?
     
  6. Nov 11, 2009 #5
    2h, but it says the answer is 4 times as high...that's what's confusing me...I don't know how to reach this final answer of 4.
     
  7. Nov 11, 2009 #6
    Nevermind, haha, I got it. Thanks a lot! Sorry for being a bit slow on the uptake : P
     
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