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
To calculate the Average Force exerted, you need to divide the total force exerted by the object by the time it took for the force to be exerted. This will give you the average force exerted over a certain period of time.
The units for Average Force are typically in Newtons (N). However, depending on the situation, other units such as pound-force (lbf) or kilogram-force (kgf) may also be used.
Yes, you can use multiple forces to calculate the Average Force. Simply add up all the forces and then divide the total by the time it took for all the forces to be exerted.
If you do not have the time, you will not be able to accurately calculate the Average Force. It is important to have both the force and time measurements to calculate this value.
No, Average Force and Net Force are two different concepts. Net Force refers to the sum of all the forces acting on an object, while Average Force is the force applied over a certain period of time.