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Physics Problem

  1. Dec 4, 2004 #1
    Alright, so here's the problem:

    A 75kg passenger in a van is wearing a seat belt when the van, moving at 15m/s, collides with the concrete wall. The front end of the van collapses 0.50m as ot comes to rest.

    a)What was the passenger's kinetic energy before the crash?

    b)What average force did the seat belt exert on the passenger during the crash?

    I would show my work but I don't even know where to start.
    And could anyone please explain to me briefly what is kinetic energy?

    Thanks
     
  2. jcsd
  3. Dec 4, 2004 #2
  4. Dec 4, 2004 #3
    Look for the Work Energy Theorem or Conservation of Mechanical Energy...
     
  5. Dec 4, 2004 #4
    Kinetic Energy is the energy possessed by a body due to its velocity
    KE = [tex]\frac{1}{2}mv^2[/tex]

    However, for this question, you will need somethind called momentum and impulse .

    Try to find them out!! o:)
     
  6. Dec 4, 2004 #5
    I know momentum is M=mv but I never heard of impulse.
    so KE would be:

    [tex]1/2(75)(15)^2[/tex]
    [tex]:632812.5[/tex]
     
  7. Dec 4, 2004 #6
    Did I do it right?
     
  8. Dec 4, 2004 #7
    Could someone please respond? I have to understand this for my unit test and I also need to hand in a assignment regarding this question.
    Thanks :)
     
  9. Dec 5, 2004 #8
    Yeah - u got the KE right :)
     
  10. Dec 5, 2004 #9
    for b ..

    I think it is...

    [tex]
    W = \sum F \Delta x = KE_f - KE_i
    [/tex]

    the final kinetic energy is 0 - cuz it comes to rest...
    the displacement is 0.50m - so then you should be able to solve for F.
     
  11. Dec 5, 2004 #10
    Could you explain it to me more about question B?
    I don't even know what the big, E-looking thing is or the triangle.
     
  12. Dec 5, 2004 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If I were your teacher I would NOT say that
    [tex]:632812.5[/tex]
    [tex]1/2(75)(15)^2[/tex]

    What are the units?
     
  13. Dec 5, 2004 #12
    I noticed I did A wrong. It's suppose to like this:

    A) [tex](1/2)(75)(15^2)[/tex]

    [tex](1/2)(16875)[/tex]

    [tex]8437.5J/1000[/tex]

    [tex]:8.43kJ[/tex]

    And I also figured out B, thanks futb0l!
    This is what I did:

    [tex]\frac{Ke}{D} = \frac{A}{1m}[/tex]

    Given:
    Ke:8.4kJ
    D:0.5m

    [tex]\frac{8.4kJ}{0.5m} = \frac{A}{1m}[/tex]

    Now, we can solve for A

    [tex]A=\frac{(8.4kJ)(1m)}{0.5m}[/tex]

    [tex]A=16.8kN[/tex]
     
    Last edited: Dec 5, 2004
  14. Dec 5, 2004 #13


    Usually p is used for momentum. Impulse is defined as I = Ft = change in momentum.
     
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