How can I calculate the force required to compress a spring in a given distance?

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Brunno
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Homework Statement



There's a question that i tried to solve in a "Newtonian way",but i could only solve it using the concept of energy:

supposing that the human bone only tolarates a force of 48.000 N without braking.Calculate the maximum height H from where a man could jump and the bone wouldn't break,suposing that when his feet touchs the ground your knees bend itself and he decends 0.6meters.
Is given the man' weight:80Kg.





Homework Equations



W= F.d = 48000.06=28800J

W = W' --> 28800 = mgH

28800 = 80.10.H

H = 36

H' = 36 - 0,6 = 35,4m


Thanks in advance!
 
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Hmm, I agree that energy is the easier way here.
To solve this "Newtonically" you will probably have to work in the falling and impact times somewhere, and then get rid of them again :)

I did get to the answer, I will just outline the steps:

Force is change of momentum, right? Technically, F = dp/dt, but over a time interval t with average momentum pavg, F = pavg / t.
So
[tex]F = \frac{\tfrac{1}{2} m v}{t}[/tex] (*)
where v is the velocity of impact (final velocity is 0) and t is the compression time of the backbone.
You can find this velocity
[tex]v = h / \tau[/tex], (**)
where [itex]\tau[/itex] (the time from start of fall until the beginning of impact) can be eliminated in favo(u)r of h and g using
[tex]h = \tfrac{1}{2} g \tau^2[/tex] (***)

If you plug (***) into both (**) and (*) and cancel, you will obtain the expression
[tex]F = \frac{m g h}{d},[/tex]
which is of course precisely the energy equation W = F s. Solving for the height h is now a piece of cake.
 
Hi,
Pluging the way was said what i found was this:

v = g.T/2F = (1/2.m.v)/t = m.g.T/4/t = m.g.T/4.t and not :F = mgh/d :(Certainly there's something that i couldn't get right.What would it be?
 
Sorry I was away for a few days.
If you rewrite (***) to [itex]\tau = \sqrt{2h / g}[/itex] then (**) becomes
[tex]v = \frac{h}{\sqrt{2h/g}} = \frac{\sqrt{h g}}{\sqrt{2}}[/tex]

Therefore
[tex]F = \frac{m \sqrt{g h}}{2\sqrt{2} t}[/tex]
and of course for the compression time,
[tex]t = \frac{d}{v} = \frac{d \sqrt{2}}{\sqrt{h g}}[/tex]
so you get
[tex]F = \frac{m g h}{4 d}[/tex]

I will leave it up to you to get the factors of 2 right (look carefully which quantities are averages between an initial value and 0 or between 0 and a final value).