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Forces/conservation of energy of slide/free fall

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Two children stand on a platform at the top of a curving
    slide next to a backyard swimming pool. At the same
    moment the smaller child hops off to jump straight down
    into the pool, the bigger child releases herself at the top
    of the frictionless slide. During their motions from the platform
    to the water, the average acceleration of the smaller child
    compared with that of the larger child is (a) greater (b) less
    (c) equal.

    2. Relevant equations
    F = ma


    3. The attempt at a solution

    I originally though it was equal because if you use F= ma, the only force acting on them is gravity right? so it it would be mg = ma, therefore a = g, but i don't understand how the answer could be a. could someone help me with this and represent it mathematically...thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 18, 2012 #2
    Consider the change in the potential energy and the length of the path traveled by either child.
     
  4. Oct 18, 2012 #3
    Ok so initially both of them have energy of mgh, and Mgh (for the bigger one)...but they both travel the same height, so the change in potential energy would be larger for the child with more mass.
     
  5. Oct 18, 2012 #4
    They travel the same height, but not the same path. But when I think about this approach again, it seems too complex. Let's try something else.

    Imagine a child somewhere on the slide. What are the forces acting on him, and what is his acceleration?
     
  6. Oct 18, 2012 #5
    Y-Forces would be N*cos(theta) - mg = ma

    X-Forces would be N*sin(theta) = ma

    I think?
     
  7. Oct 18, 2012 #6
    No, that's not right. Use the coordinates normal and tangential to the slope.
     
  8. Oct 18, 2012 #7
    Ok sooooo....

    Y-forces would be F = mg*cos(theta) = ma
    X-forces would be F = mg*sin(theta) = ma
     
  9. Oct 18, 2012 #8
    Where is the reaction?
     
  10. Oct 18, 2012 #9
    the reaction of the force would along the horizontal
    so F = sqrt((mgcos(theta)^2 + mgsin(theta)2)) = 2mg?
     
  11. Oct 18, 2012 #10
    I do not follow.

    Do a FBD. Write down the components normal and tangential to the slide. Do not forget the reaction. What is the acceleration?
     
  12. Oct 18, 2012 #11
    Ax = gsin(theta)
    Ay = gcos(theta)

    i'm sure this is wrong though
     
  13. Oct 18, 2012 #12
    Again, use normal/tangential axes. Acceleration can only be tangential.
     
  14. Oct 18, 2012 #13
    i don't get it honestly...acceleration has to have 2 components to it when it is on a slope if your asking for the resultant acceleration then i suppose that would be sqrt(Ax^2 + Ay^2) = g
     
  15. Oct 18, 2012 #14
    Do you understand what "normal" and "tangential" mean?
     
  16. Oct 18, 2012 #15
    normal is perpendicular to the surface tangential is along the surface:

    the acceleration is along the tangential so that would be the x-component of the force so it would be mg*sin(theta)
     
  17. Oct 18, 2012 #16
    Correct. Now compare that with acceleration of the child plunging straight down.
     
  18. Oct 18, 2012 #17
    so mg*sin(theta) < mg....therefore because of a larger force the acceleration would also have to be larger?
     
  19. Oct 18, 2012 #18
    Force and acceleration are very simply related, aren't they?

    Note the formula above has a problem: it uses the same "m", while you have two different masses.
     
  20. Oct 18, 2012 #19
    ok so if we don't know the size of each m how could you even tell which is greater?
     
  21. Oct 18, 2012 #20
    What you have is forces; you are asked to compare accelerations.
     
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