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Force problem

  1. Aug 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A 45.1-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.47 m/s. Her hands are in contact with the wall for 0.983 s. Ignore friction and wind resistance. Find the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her). Note that this force has direction, which you should indicate with the sign of your answer.

    2. Relevant equations
    Fsw=Fws
    where s is the skater and w is the wall



    3. The attempt at a solution
    no really sure how to go about solving this problem, Im assuming that you are meant to find the force the skater pushes with since we are given all these variables for the skate. Then the magnitude of the force the wall exerts on the skater will be the same but opposite direction.
    Where do i start with this problem?


     
  2. jcsd
  3. Aug 1, 2016 #2

    billy_joule

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    The skater starts at rest then reaches -1.47 m/s in 0.983 s, what is the skaters acceleration? So, via Newtons second law, what must the force be?
     
  4. Aug 1, 2016 #3
    could i use the kinematic equation vf=vo+at where vo is 0 and use the magnitude of the skaters final velocity or would i have to include the - sign ? without it we get a= 1.47/0.983 a=1.495m/s2
    with that acceleration value in find the force is going to be his mass times the acceleration, so we have F=45.1⋅1.495 = 67.42N correct??
     
  5. Aug 1, 2016 #4

    billy_joule

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    Dropping the negative sign just happened to give the correct answer, you won't always be so lucky. Drawing a free body diagram is a fool proof way of ensuring directions/signs are correct.
     
  6. Aug 1, 2016 #5

    haruspex

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    That is not strictly valid here. We are not told that the force is constant, which is why we are asked for average force. Use momentum.
     
  7. Aug 2, 2016 #6

    billy_joule

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    Isn't the math identical? using the average acceleration will give the average force?
     
  8. Aug 2, 2016 #7
    I got the correct answer with my working ay
     
  9. Aug 2, 2016 #8

    haruspex

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    If you change it to average acceleration, yes.
     
  10. Aug 2, 2016 #9

    James R

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    The idea of impulse is useful for solving this problem.
     
  11. Aug 2, 2016 #10
    The easiest way to solve this is wuth momentum(p) and impulse(J). It says the velocity is -1.47 m/s so the negative direction is away from the wall.

    It is important that J=Δp and p= m·v

    p0=0 kg·m/s

    J=Δp=m·Δv-0=45.1·(-1.47)=-66.297 kg·m/s

    J=Δp=F·Δt=-66.297 kg·m/s

    -66.297=F·Δt

    F=-66.297/0.983 ≈ -67.444 N

    That is the force done by the wall on the skater. The force done by the skater will be 67.444. The sign is positive because it's in the opposite direction of movement, which is negative.
     
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