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Seat belt force

  1. Nov 29, 2006 #1
    Hey, I have the following problem:

    A car going 15m/s is brought to rest in a distance of 2.0m as it strikes a mound of dirt. How large an average force is exerted by the seatbelts on a 90kg passenger as the car is stopped.

    I have found one formula... F = (mass * change in velocity)/time ... and I have found stopping time to be approx (2.0/15) = 1.33 (unless I am wrong) so F = (90 (not sure about this) * 15)/1.33 = 10125N

    The only problem is I have the right answer from a friend that is approx 5100N. Can someone tell me whats going wrong? Thanks!,
    -Mike
     
  2. jcsd
  3. Nov 29, 2006 #2

    PhanthomJay

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    see comments above.
     
  4. Feb 16, 2009 #3
    As he had said, 15 is not the Vavg. Instead of using v to calculate, you should be using acceleration. Use the formula Vf ^ 2 = Vi ^2 + 2a DeltaX . Vf will be zero(because it comes to rest), Vi is 15, and delta X(displacement) is 2. Through the calculation, we get -56.25m/s^2 for acceleration. Force is mass times acceleration. The passenger's mass is 90kg, so we multiply the two figures, and get 5062.5. Because there's 2 significant figures, we round the number to 5100N.
     
    Last edited: Feb 16, 2009
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