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Dynamics Question #2

  1. May 15, 2008 #1
    1. The problem statement, all variables and given/known data
    The diagram shows the forces acting on a child on a playground slide. Air resistance is negligible.
    [​IMG]

    (a) The mass of the child is 40kg.
    (i) Calculate the size of the force W.
    (ii) Describe the force that, along with W, makes up the 'equal and opposite pair' of forces described by Newton's third law.

    (b) The size of the friction force is 90N.
    (i) Calculate the component of W parallel to the slide.
    (ii) Calculate the acceleration of the child.

    (c) The slide is 5.5m long. If the child maintains this acceleration, calculate her speed at the bottom.

    2. Relevant equations
    [tex]F = m.a[/tex]
    Frictional Force [tex]= F - mg\sin\theta[/tex]
    Normal Force = [tex]N_{normal force} = mg\cos\theta[/tex]

    3. The attempt at a solution
    (a)(i) [tex]m = 40kg[/tex]
    [tex]F = m.a[/tex]
    [tex]F = 40 \times 9.81[/tex]
    [tex]F = 392.4N[/tex]

    (ii)???

    (b)(i) Frictional Fore [tex]= 90N[/tex]
    F [tex]= mg\sin\theta[/tex]
    F [tex]= 40 \times 9.81 \times \sin{24}[/tex]
    F [tex]= 159.60[/tex]
    F [tex]= 159.60 - 90[/tex]
    F [tex]= 69.6N[/tex]

    Are my answers correct? I need to be sure to continue...
     
  2. jcsd
  3. May 15, 2008 #2

    Hootenanny

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    Without checking your arithmetic, your methods thus far are good.
     
  4. May 15, 2008 #3
    Thnx Hootenanny,

    (b)(ii) [tex]F = m.a[/tex]

    [tex]a = \frac{F}{m} = \frac{69.6}{40} = 1.7475 ms^{-2}[/tex]

    (c) [tex]v^2 = u^2 + 2as[/tex]
    [tex]v^2 = 2 \times 1.7475 \times 5.5[/tex]
    [tex]v = \sqrt{2 \times 1.7475 \times 5.5}[/tex]
    [tex]v = 4.38 ms^{-1}[/tex]


    Are my answers correct?
     
  5. May 15, 2008 #4

    Hootenanny

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    Again your methods are correct, but I ain't checking your arithmetic for you :wink:
     
  6. May 15, 2008 #5
    Thnx Hootenanny, just wanted to make sure that in part (b)(ii) using sin was correct rather than cos?
     
  7. May 15, 2008 #6

    Hootenanny

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    Yes, sine of the angle with give you the component parallel to the slope, whereas cosine of the angle will give you the component perpendicular to the slope.
     
  8. May 15, 2008 #7
    The vector N in the diagram has no purpose in this question?!
     
  9. May 15, 2008 #8

    Hootenanny

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    Nope it is not relevant for this question, it is merely shown for completeness I guess.
     
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