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Homework Help: Newton's Laws Problem- Question on upcoming TEST

  1. Oct 17, 2008 #1
    Newton's Laws Problem- Question on upcoming TEST!!

    1. The problem statement, all variables and given/known data

    A small block is placed at height h on a frictionless 30 degree ramp. Upon being released, the block slides down the ramp then falls 1.0 m to the floor. A small hole is located 1.0 m from the end of the ramp. From what height h should the block be released in order to land in the hole?

    Physics.jpg


    2. Relevant equations
    Newton's Laws
    Kinematics Equations


    3. The attempt at a solution
    I did the problem assuming there would be no initial Y-velocity.
    And figured out the time it would take for the object to fall to the ground after leaving the ramp, and the X-velocity it would need to get to the whole.
    The time was .45 seconds and the X-velocity was 2.21 m/s.
    With that it should be really easy to find the time to get to that velocity, and then the height needed to start from....

    But I realized there would be an initial velocity in the y direction when it reaches the end of the ramp. I've spent nearly 3 hours trying to solve this problem, and it might be on an upcoming test. I have tried the problem with different axis, and I just can not figure it out. Please help, my test is this week!
     
    Last edited: Oct 17, 2008
  2. jcsd
  3. Oct 17, 2008 #2
    Re: Newton's Laws Problem- Question on upcoming TEST!!

    After leaving the plane, the block will have an unknown speed (S), which can be broken into x,y components. The x,y kinematics are:

    x - 1
    x0 - 0
    V - ?
    V0 - Scos(-30)
    a - 0
    t - t

    y - 0
    y0 - 1
    V - ?
    V0 - Ssin(-30)
    a - -9.8
    t - t
    (note: time will be the same in both)

    In both cases, V is what we don't care about. We then use x=x0+v0t+.5at2
    in the x case:
    1=0+Scos(-30)+.5(0)t2
    Solving for t gives t=1/Scos(-30)

    in the y case, with t-substitution:
    0=1+Ssin(-30)*1/Scos(-30)+.5(-9.8)(1/Scos(-30))2

    In the middle velocity term, S cancels out. Multiplying all known numbers as well as squaring the third term gives:
    0=1-.5774-6.5333/S2

    Solving for S gives S = 3.9319 m/s

    Now with a target final ramp speed, we can set up a 3rd kinematics equation for the slide. The acceleration will be modified from gravity: Slide force = 9.8*sin(30) = 4.9 m/s2.
    x - ?
    x0 - 0
    V - 3.9319
    V0 - 0
    a - 4.9
    t - ?
    We don't care about time in this one, so the equation we use is V2 = V02+2a(x-x0).
    3.93192=0+2*4.9*(x-0)
    Solving for x gives x=1.5775 m up the ramp. They said they wanted height and not ramp distance, so we find the y component of the ramp length: 1.5775*sin(30) = .78875 m 'high' on the ramp.

    Brutal!
     
  4. Oct 17, 2008 #3
    Re: Newton's Laws Problem- Question on upcoming TEST!!

    Yes! That was totally it. Brutal problem, indeed!
     
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