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Newton's Laws -- Resting hockey puck hit with a force

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A resting hockey puck is hit with a force of 15.3 N. The frictional force slowing the puck down is 1.0 N.
    a) Find the net horizontal force acting on the puck while the stick is in contact with the puck.
    b) If the puck leaves the stick with a velocity of 45 m/s, how far will the puck travel in 3.0 s?

    2. Relevant equations
    Fnet (h) = Fapplied + Ff
    a = F/m
    Δd = v1Δt + 1/2 aΔt2


    3. The attempt at a solution
    a) Fnet (h) = Fapplied + Ff

    Fnet (h)= (15.3 N [forward]) + (1.0 N [backward ])

    Fnet (h)= (15.3 N [forward]) + (-1.0 N [forward])

    = 14.3 N [forward]

    Fnet (h) = 14 N [forward]

    c)m = 164 g = 0.164 kg

    a = F/m

    a = -1.0 N [forward] / 0.164 kg

    a= -6.1 m/s2 [forward]

    Δd = v1Δt + 1/2 aΔt2

    Δd = (45 m/s) (3.0 s) + ½ (-6.1 m/s2) (3.0 s)2

    Δd = 108 m (should I add a direction here, if yes then would it be 108 m [forward]?)

    The puck will travel 108 m in 3.0 s.

    Is this correct?
     
  2. jcsd
  3. May 7, 2016 #2

    TSny

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    Your work looks correct to me. I don't think you need to include a direction for the distance traveled.
     
  4. May 7, 2016 #3
    Ok, thanks for verifying my answer :)
     
  5. May 7, 2016 #4

    haruspex

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    In your answer to a), why did you round to 14N?
     
  6. May 8, 2016 #5
    I just thought I should round it to two sig-figs... is it more accurate to leave it as 14.3 N?
     
  7. May 8, 2016 #6

    haruspex

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    The given data were both quoted to one decimal place, so you are justified (here) in doing so in the answer. In more complicated algebraic relationships it is not quite that simple.
     
  8. May 9, 2016 #7
    So does that mean I should leave it at 14 N? or change it to 14.3 N?
     
  9. May 9, 2016 #8

    haruspex

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    I would submit the answer as 14.3N.
     
  10. May 9, 2016 #9
    Okay, thanks for the help :)
     
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