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Work and Conservation

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A water skier is going to glide up a 2.0m frictionless ramp then sail over a 5.0m wide tank filled with sharks, she'll drop the tow rope at the base of the ramp. What minimum speed must she have as she reaches the ramp in order for her to clear the sharks.


    2. Relevant equations
    1/2mv^2=mgh
    Vf^2=Vi^2 +2ax
    t=sqrt of 2(height)/g


    3. The attempt at a solution

    figured out time to fall 2 meters = .63 seconds. So she'll need to cross the 5.0m wide tank in this amount of time. 5/.63 = 7.82 m/s this is the speed needed as she leaves the top of the ramp.

    1/2mv^2=mgh solved for v
    v=sqrt of 2gh
    this is the speed needed to make it from the the bottom of the ramp to the top.
    =6.26 m/s

    add these two speed together and it should be velocity at bottom of ramp 6.26+7.82 =14.1 m/s

    or so I thought. Book gives answer as 10 m/s.

    so which calculation is wrong? 10-7.82 =2.18 m/s this isn't enough to make it up the ramp.

    10-6.26 = 3.74 this isn't enough speed to make it over the sharks.

    All I can think is that maybe I am calculating the speed to get up the ramp incorrectly.

    Any help?

    Thanks.
     
  2. jcsd
  3. Oct 20, 2008 #2

    Doc Al

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    Staff: Mentor

    What's the angle of the ramp? I assume 2.0m is the height of the top of the ramp?
     
  4. Oct 20, 2008 #3
    No angle given. In the drawring it is not linear, but curved like a slightly flattened 's', or a local min concave up, inflection point, then concave down local max, if that makes sense.

    Yes 2.0m is the top of the ramp.
     
  5. Oct 21, 2008 #4

    Doc Al

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    Staff: Mentor

    All that matters is the angle she moves when she leaves the ramp. We assume it's horizontal. (And that gives the correct answer.)

    Good!

    That's the speed she'd need to just make it up the ramp, with zero speed at the top.

    Ah.... speeds don't add, but energy does.

    Use conservation of energy from start to finish. She starts off with some unknown KE and ends up with PE + KE.
     
  6. Oct 21, 2008 #5
    I always have trouble setting up the before and after, with conservation.....

    OK, PE+KE=KE so solving for Vi gives

    Vi=sqrt of Vf^2 -2gh

    but it should be

    Vi=sqrt of Vf^2+2gh

    But yes the second equation gives the right answer. Thank you very much.
     
    Last edited: Oct 21, 2008
  7. Oct 21, 2008 #6
    Ok, nevermind, it is because I had Vi and Vf on the wrong sides of the equation, it does work out to be the second equation!!!!

    Thanks again.
     
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