energy/ force new question!

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Offline ramachandran.akshay

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energy/ force new question!
« on: 30/11/2012 08:13:50 »
I have a question rattling my head for some time. Say an object (100kg mass) drops from a height of one meter, the fall according to laws of motion would make its impact velocity around 4.4721m/s. At this velocity, the object just before impact, would have acquired 1KJ of Kinetic energy.
Now since work= force x displacement, how would you relate this kinetic energy to the "deformation" caused onto the object. Please help me out with this problem!

Thank you!
« Last Edit: 30/11/2012 08:16:02 by ramachandran.akshay »


Offline teragram

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Re: energy/ force new question!
« Reply #1 on: 02/12/2012 16:44:10 »
Im not sure I understand your question, but here are my thoughts:-
A moving object (such as your 100kg) striking something (ie. Floor) obviously comes to a stop, by a process of deceleration (negative acceleration). If the object stopped instantly, the force required would be infinite (F=ma), which of course could not happen. What happens, and I think this is at the root of your question, is that the object, and the thing it is striking, deforms, to allow a less than instant deceleration. This is what happens in a car crash.
My mathematics is poor, but I think you will find that if you make an assumption for how much the object deforms, say 1 centimeter, this being the distance over which the object comes to rest, you will be able to work out the deceleration from the velocity you stated. You can then calculate the force (F=ma) experienced by the object (and the floor) in order to bring the object to rest . A major assumption here is that the object deforms at a linear rate. Also of course in real life the floor shares in the deformation.
My calculations indicate that if the 100kg mass deforms in 1 centimeter (i.e. the distance for deceleration)  during the impact, the force will be about 98,000Newtons.
I stand ready to be corrected by any experts out there!
« Last Edit: 02/12/2012 17:00:02 by teragram »


Offline evan_au

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Re: energy/ force new question!
« Reply #2 on: 08/12/2012 17:33:01 »
Real materials behave in an elastic manner for small deformations (stress is proportional to strain), the material springs back to its original shape after teh impact, and physics can handle this relatively simply.
You do need to know a few details about the materials involved, such as Young's Modulus, which you can get from a data sheet for the material (

Note that even for "simple" materials like pure copper, these properties can vary enormously depending on how the material was heated, cooled and stressed during manufacture. More complex alloys like steel have a large menu of formulations, with a correspondingly wide range of properties.

Subjected to high forces, real materials behave in an inelastic manner, and are permanently deformed after the collision. To model this you need a more detailed knowledge of the material properties ( You need to understand how forces propagate through the material, which is affected by the shape of the object (and a similar level of detail about the object with which it is colliding).

This gets even more complex if the objects consist of multiple materials or alloys.

Engineers often use the Finite Element Method to model an object as small blocks, and work out the forces on each small block, which then lets them work out the forces on the adjacent blocks. This requires a supercomputer, which initially limited it to applications like simulation of the weather. As prices of computing came down, the car industry adopted it for crash-proofing vehicles. Now even home computers have considerable processing power, and can use this method (if a little slowly).
« Last Edit: 08/12/2012 20:26:53 by evan_au »


Offline saspinski

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Re: energy/ force new question!
« Reply #3 on: 22/11/2015 23:38:41 »
Suppose your object is a rod of steel, keeping its axis perpendicular to the ground during the fall.

After the impact, during a brief period of time, its length shortens due to the compressive force. This force = EAε, E = 2,06 x 10 N/m is the modulus of elasticity of the steel, A the cross area of the rod and ε the elastic strain.

This force starts from zero until reaches a maximum when the rod speed is zero. The work done is ∫EAεdx = ∫EAεLdε, where L is the length of the rod. W = EALε / 2.

For a steel object of 100kg, the volume is 100 / 7850 = 0.01274 m. If the ground is hard enough, like a big hardened anvil, the kinetic energy transforms to an elastic energy of 1000J = 2,06 x 10 x 0.01274 x ε / 2 => ε = 0,000873. The tension σ = Eε = 179830724 N/m = 18,3kg/mm, what is below the yield point for a normal carbon steel, so the elastic assumption is OK.

In order to know how much (mm) the rod would deform, it depends on its length. For a length of 1m for example, it would shrank by 1000 x 0,000873 = 0.8mm. In this case, the force = σA = 179830724 x 0.01274 = 2289828 N or 233 tons.