Work, Energy, Force of Pile Driver and Pile

[Lebombo]

Work, energy, force of P & D

Hi, I'd like to better understand the physics of an object's net force when another object collides with it, such as a pile driver applying a force to drive a pile into the ground.

Questions:

A pile driver drops from 500 meters (falls for 10 seconds at 10m/s^2 constant acceleration) and hits pile. The speed of the driver just as it hits the pile is 100m/s.

If a pile driver strikes a pile into the dirt, a collision occurs. From the instant of impact, if the pile driver and pile remain in contact until the pile is driven into the ground some known distance, then does this mean that the pile must have (or could have) reached the speed of the pile driver at some point at the beginning of impact?

In other words, how does the acceleration of the pile occur? Does the pile instantly begin traveling at a rate of 100m/s and then slow down to 0m/s. Or does the pile accelerate up to 100m/s in some very small fraction of a second, and then slow down to 0m/s. Or does the pile never accelerate up to 100m/s?
Two scenarios.

Scenario 1)

10kg driver drops from 500 meters (falls for 10 seconds at 10m/s^2 constant acceleration) and hits pile. Driver remains in contact with pile. Pile is driven .2 meters into the ground. (All energy is transferred into work)

Xf = ½(a)(t^2) + vt + r = ½(10)(t^2) , therefore time in free fall = 10 seconds

Vf = at + v = (10)(10) + 0 = 100m/s, velocity at time of impact

ΔKE = ½(m)(v^2) – ½(m)(v^2) = ½(10)(100^2) – ½(10)(0) = 50,000J

ΔKE = work = force(distance) => 50,000J = F(.2) => 50,000J/.2m = 250,000NCan you make sure my wording is correct as to what is being applied to what in the scenario:

The gravity is applying a force to the driver.

The driver is building kinetic energy as it falls.

The driver transfers all of its kinetic energy to the pile.

The driver applies force to the pile.

The driver applies a force to the pile over a distance of .2 meters.

The driver applies a force to the pile of 250,000N only at the instant upon impact.Scenario 2)

10 kg driver drops from 500 meters (falls for 10 seconds at 10m/s^2 constant acceleration) and hits pile. Driver bounces off pile, remaining in contact for only .01 seconds. Pile is driven .2 meters into the ground.Xf = ½(a)(t^2) + vt + r = ½(10)(t^2) , therefore time in free fall = 10 seconds

Vf = at + v = (10)(10) + 0 = 100m/s, velocity at time of impact

ΔKE = ½(m)(v^2) – ½(m)(v^2) = ½(10)(100^2) – ½(10)(0) = 50,000JI'm a bit curious as to what is happening here as well..

If a collision is elastic, such as the driver hitting the pile for .01 seconds, then is it at all possible to calculate the force that the driver applies to the pile when the distance that the pile is driven into the ground is given?

In other words, when the driver strikes the pile and transfers all of its energy to the pile, and the pile is driven .2 meters into the ground, will the force applied to the pile be 250,000N regardless of the duration of contact?Thank you.P.S. This is a question of my own. Not homework or any other kind of assigned work. I am not looking for any specific numerical answer, but more of a conceptual understanding. Thank you.

 

[Lebombo]

In thinking about this more, it seems like the force of the ground or dirt on the pile is a major factor in the motion of pile. If the ground is so hard that the pile does not move into the ground at all, then the work that the driver does on the pile is zero. And the work that the pile does on the driver is also zero because the change in velocity occurred over zero distance. However, a change in velocity still occurred, so kinetic energy of the driver must have gone somewhere, especially if we are assuming all energy transferred into work. It seems possible to assume that the driver could have done work on the entire earth. I guess it can be supposed that either the pile was driven into the ground .0000000001 meters or that the entire Earth moved .0000000001 meters so that the force on the driver on the pile is 50,000J/.0000000001m = 5x10^14 Newtons. So I guess it really depends on how much the entire Earth moves. But suppose the Earth doesn't move at all and the pile doesn't move at all in space, if zero energy from the driver is able to transfer into work, then it MUST transfer into some other form of energy like heat or something.

Okay if that makes sense then if the ground is kind of soft, then the driver will be able to do some work on the pile. Then since the pile is originally at rest before being hit by the driver and ends up at rest some distance into the ground, than can it be said that if all the kinetic energy of the driver transferred into work, then the driver does equal amount of work on the pile as the dirt in the ground, just in opposite directions?

If this seems correct, then how come it doesn't seem like the dirt is applying the same amount of force on the pile as the driver at each instant in time? If the ground applied the same amount of force on the pile as the driver on the pile at the same time, then it seems like there would be a net force of zero on the pile and the pile would not move. So I'm left to believe that different magnitudes of force are acting on the pile at different times during the movement of the pile into the ground. But in what order are the forces being applied to the pile?

Thank you