I cant find the Average Force exerted, help please?

[Aenion]

Homework Statement

A 75Kg man jumps from a window 1.0m above the side walk.

if the man jumps with his knees and ankles locked, the only cushion for his fall is approximately 0.50cm in the pads of his feet. Calculate the average force exerted on him by the ground in this situation.

I've tried many ways to get the answer but i seem to get quite there. I know the answer is 1.5 X 10^5

 

[PhanthomJay]

Which ways have you tried? Please show one of them, using energy methods or Newtons's laws and the kinematic equations. Welcome to PF!

 

[Aenion]

Well I thought maybe at fist i could use the work-Kinetic energy thm. and solve for the force but the force i was solving by using F(delta X) = 1/2MVf^2 was the force of the man on the ground (726N). Then i looked at it if the man was standing stationary on the ground he would then be currently at an equilibrium. So I concluded that the ground must have exerted a force of 726N plus the amt. of force exerted on the .005m of cushion on his feet.

now I just don't know how to add in the amt. of force used on the cushion.

and thanks it seems like a wonderful site ^^

 

[PhanthomJay]

Well I thought maybe at fist i could use the work-Kinetic energy thm. and solve for the force but the force i was solving by using F(delta X) = 1/2MVf^2 was the force of the man on the ground (726N). Then i looked at it if the man was standing stationary on the ground he would then be currently at an equilibrium. So I concluded that the ground must have exerted a force of 726N plus the amt. of force exerted on the .005m of cushion on his feet.

now I just don't know how to add in the amt. of force used on the cushion.

and thanks it seems like a wonderful site ^^

You can use the work energy theorem if you apply it correctly, but are you familiar with the more general conservation of total energy principle? When using the work energy theorem, you need to consider the work done by all forces, including gravity and the ground contact forces, starting at the beginning when the man jumps, and ending after he comes to a stop. When using conservation of energy, with the same beginning and end points, then the work done by gravity is already built into the equation in PE terms, so you just need to calculate work done by the ground contact force.

 

[Aenion]

You can use the work energy theorem if you apply it correctly, but are you familiar with the more general conservation of total energy principle? When using the work energy theorem, you need to consider the work done by all forces, including gravity and the ground contact forces, starting at the beginning when the man jumps, and ending after he comes to a stop. When using conservation of energy, with the same beginning and end points, then the work done by gravity is already built into the equation in PE terms, so you just need to calculate work done by the ground contact force.

Yes I believe I am are you talking about PEi + KEi = PEf + KEf ?
but i do see what you are saying and how i didnt account for the work done by the ground.
In order to find the force exerted by the ground though isn't it possible for me to set the man's force of 726N equal to the work done by the ground (726N = Fgr(delta X))

726N = Fgr(.005m)
726/.005 = Fgr
145200N = Fgr​

which you can round to 1.5 X 10^5

 

[PhanthomJay]

Yes I believe I am are you talking about PEi + KEi = PEf + KEf ?

When forces besides gravity are acting, W_ground + PEi + KEi = PEf + KEf

but i do see what you are saying and how i didnt account for the work done by the ground.
In order to find the force exerted by the ground though isn't it possible for me to set the man's force of 726N equal to the work done by the ground (726N = Fgr(delta X))

726N = Fgr(.005m)
726/.005 = Fgr
145200N = Fgr​

which you can round to 1.5 X 10^5

I don't know where your 726 N comes from, and you can't set force equal to work, they are different units. And you are not looking for the work done by the man, you are looking for the work done on the man. If you use the Work-Energy theorem, then W_gravity + W_ground = KE_final - KE_initial, where the KE terms are zero at the beginning and end of the motion.

 

[Aenion]

No, I'm sorry your right i meant Wman = WGround

Which would then still be

Wman = Wground
726N(1m) = Fground(.005m)
726/.005 = Fground
145200 = Fground

and that can be rounded to 1.5 X 10^5

Which after i Re read your comment i saw that you had already stated that the work of the man is equal to the work of the ground

Thank you very much for your help on this ^^

 

[PhanthomJay]

OK, but the work done by gravity on the man is 75(9.8)(1.005) = 739 joules, making the average force of the ground on him 1.478 X 10^5 J, which rounds to the correct answer.