How to Calculate Average Force in a Falling Man Scenario [SOLVED]

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To calculate the average force exerted by a 90kg man who falls 10m and buckles his legs, the initial force due to gravity is 882N. The average force calculation must consider the deceleration as he comes to a stop over an additional 23cm. The discussion highlights that simply using F=ma yields only the man's weight, not the average force during impact. The conservation of energy approach is suggested to account for the work done in stopping. Ultimately, understanding the dynamics of the fall and subsequent deceleration is crucial for an accurate calculation.
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[SOLVED] A falling man

Homework Statement


a 90kg man falls (freefalls) 10m, when he lands, his legs buckle and he goes down another 23cm. What was the average force he exerted on the ground?


Homework Equations



M = 90k
a = 9.8 m/s^2


The Attempt at a Solution



from freefall to ground

F(g) = 90kg (9.8m/s^2)
F(g) = 882N

buckling

F(g) = 90kg (9.8m/s^2)
F(g) = 882

(882 + 882)/2 = 882

average force = 882?

Im confused, does the distance have anything to do with the question? Because F = ma, so as long as its accelerating but the distance does not matter.

but this seems a bit too easy, so I am pretty sure i did something wrong =/
 
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Have you tried to solve this by using the law of conservation of energy? Gravitational potential energy can be transferred to work. You know that W=Fs. F is the force that is stopping the man and he exerts the same force on the ground (third Newton's law).
 
we haven't learned that yet...

is it possible to only solve this problem with F=ma and the kinematic equations
 
What you calculated as the average force is actually just his weight in Newtons.
How fast is he falling when his feet hit the ground?
Then he accelerates back to zero m/s over a distance of 23cm.
 
ahh, ic. Thank you montoyas and Dr. Jekyll. I understand it now.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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