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Help Stuck on problem for 2 days.

  1. Oct 29, 2006 #1
    Help!! Stuck on problem for 2 days.

    A 75.0 kg stunt man jumps from a balcony and falls 26.0 m before colliding with a pile of mattresses. If the mattresses are compressed 0.950 m before he is brought to rest, what is the average force exerted by the mattresses on the stuntman? N

    I am really confuse on how to do this problem. I tried finding V = sqrt(2gh)
    so i can find the KE but seem to still get it wrong these are some of the steps i did

    sqrt(2(9.8)(26)) = 22.57 then i find sqrto(2(9.8)(.950)) = 4.32

    from there i use KE = 1/2mv^2
    1/2(75)(22.57)^2 - 1/2(75)(4.32)^2
    19102.7 - 699.84 = 18402.86 N which is completely wrong

    I think i used the wrong mass for the second KE but not sure at all.
    or would i have to set up the equation like this
    1/2m_1v_1^2 = 1/2m_2v_2^2 and find for m_2 then solve it KE1 - KE2

    any help would be great.
  2. jcsd
  3. Oct 29, 2006 #2
    Well, start by finding his velocity when he hits the mattresses:

    V^2 = 0^2 + 2(9.8)(26)
    V^2 = 509.6
    V ~ 22.57

    Now calculate his rate of decceleration from 22.57 m/s to 0 m/s in .95 m:

    0^2 = 22.57^2 + 2(a)(.95)
    -509.6 = 1.9a
    a ~ -268.21 m/s/s

    Now apply Newton's first law given his mass and the acceleration you just solved for to find the force exterted on impact:

    F = 75(-268.21)
    F = -20115.79 N

    Newton's 3rd law of action/reaction states that if x force exerted on an object, -x force will be exerted back. Thus, the mattress exerted 20115.79 N of force on the stuntman.
  4. Oct 29, 2006 #3
    I think it's more helpful not to give the answer. In any case, I think this isn't quite right. What is desired is the average force, and BunDa4th seems to be learning about work/energy.

    If you want to solve this using the energy approach, you don't really need the velocity. You can select the points at which the velocity is zero (that's hint number 1)...

    There are two forces, the gravity and the force of the mattress. Remembering that W = F*d, what is the 'd' in for each force (that's a big hint). Then apply the kinetic work-energy theorem.

    Last edited: Oct 29, 2006
  5. Oct 29, 2006 #4
    pyroman also that answer seem to be incorrect.

    Dorothy is correct i am learning the work/energy and inelastic and elastic collision.

    I am not sure what you mean by selecting a point where velocity would be zero. also would i be using this

    KE_i + PE_i = KE_f + PE_f?

    0 + (75)(9.8)(26) = 0 +.....okay i am not sure how to do it.
  6. Oct 29, 2006 #5
    Hi BunDa4th,

    I'm just learning about this myself, so maybe someone more knowledgeable should step in. I don't think you need momentum here, though (but I might be wrong). The kinetic energy at the start and end of the motion is zero, which makes the net work zero. So it seems to me that W_fireman - W_mattress = Delta K = 0. And then you can just solve for the force of the mattress. Try this, and see if it gives you the right answer. I'm curious to know if this works.

  7. Oct 29, 2006 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Don't use kinetic energy.

    First question: What is the TOTAL DISTANCE he falls?
  8. Oct 29, 2006 #7
    the total distance he fall is 26.95 m. since he is jumping 26m then he compress .95 m more when he land on mattress which total to be 26.95.
  9. Oct 29, 2006 #8
    I think Office_Shredder is suggesting the same idea to you.

    So, what is the work of gravity, and the work of the mattress?
  10. Oct 29, 2006 #9
    work of gravity is 75(9.8)(26.95) and work of mattress 75(9.8)(.95)?
  11. Oct 29, 2006 #10
    Almost. You don't know the force exerted by the mattress on the man, that's what you are trying to find out.

    Anyway, now you are almost there. Just use the Work-Kinetic energy theorem, and solve for this unknown force.
  12. Oct 29, 2006 #11
    then i would set the equation as 75(9.8)(26.95) = m(9.8)(.95)?
  13. Oct 29, 2006 #12
    Are you looking for the mass, or the force?
  14. Oct 29, 2006 #13
    im looking for the force. I thought maybe i have to find the mass of the mattress to find the force.
  15. Oct 29, 2006 #14
    it seem the more i work on this problem the further away i am getting the answer. This problem is just too hard for me.

    but i think this is what im suppose to do

    stuntman before impact: find KE and PE
    stuntman at rest: Find KE and PE

    now rest - impact = workdone by mattress (non-conservative force)

    problem is im now sure what number to use to find KE and PE before impact and KE and PE to find rest.

    i also know that KE = 1/2mv^2 and PE = mgh but one problem i seem to have is how would i find velocity before impact since at rest velocity is zero. also would it be KE - PE to find the total work before impact and at rest?
    Last edited: Oct 29, 2006
  16. Oct 29, 2006 #15
    (75 kg)(9.8 m/s^2)(26.95 m) = F(0.95 m).

    The F should be the force exerted by the mattress.
  17. Oct 29, 2006 #16
    why isnt gravity a part of the mattress? also how did you come up with that?
  18. Oct 29, 2006 #17
    We define the bottom of the fireman's motion as our zero point, so the potential energy there is zero, by definition. If you want a more physical picture, I guess it would be that the mattress is in equilibrium. Whether it is a spring mattress or a waterbed, or air mattress, the forces that are keeping the mattress at the level are in equilbrium with gravity, so the only force that is pushing it lower is the force of the fireman. The force of the mattress opposes that, and gravity has nothing to do with that, as far as I can see.

    As for how I got it, well, we got it together, really. These are the steps we went through:

    W_fireman - Work_mattress = Total Work.

    But by the Work-kinetic energy theorem:

    Total Work = K.E. at start - K.E. at end = 0.

    So Work_fireman = Work_mattress

    Then just use W = F*d.

    Last edited: Oct 30, 2006
  19. Oct 30, 2006 #18
    ooooh i understand what you did now. thanks for the help (sorry i took a bit long to say thanks) i was busy with a few projects after i solve that problem.
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