The problem involves calculating the average force exerted on a man who jumps from a height of 1.0m and lands on a cushion of 0.50cm in his feet. The subject area includes concepts from mechanics, specifically relating to forces, energy, and kinematics.
Participants are actively engaging with each other's reasoning, exploring different interpretations of the problem. Some guidance has been offered regarding the application of the work-energy theorem and the need to account for all forces acting on the man during his fall.
There is some confusion regarding the units of force and work, as well as the correct interpretation of the work done by the man versus the work done on him by the ground. Participants are also navigating the implications of gravitational force in their calculations.
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 said: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 said: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.
When forces besides gravity are acting, Wground + PEi + KEi = PEf + KEfAenion said:Yes I believe I am are you talking about PEi + KEi = PEf + KEf ?
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 Wgravity + Wground = KEfinal - KEinitial, where the KE terms are zero at the beginning and end of the motion.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