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msu94
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I am in the middle of physics debate with someone who has some rather, shall we say, unconventional views regarding classic mechanical physics. It started when he posted that he feels 9/11 was a conspiracy because a section of a building falling does not increase the "load" (as he called it) on the lower floors.
Our rebuttal was well do you think a brick resting on your head exerts the same force as one that falls and hits you in the head and he said YES.
I replied that yes, of course the force of impact is greater than when it is at rest and gave an example of a 5 kg bowling ball dropping 1 meter and fully stopping within .02m after impact, and I worked out the forces at rest.
If you put a 5 kilogram bowling ball on a scale, you will get 5 kilos. Its weight (which is a force) 5 * 9.8 = 49 Newtons of force. F = M G
Now let's take that 5 kg ball and drop it onto the scale from 1m high. Do you really think it will still only show 5kg upon impact? Of course not.
It will impact the scale at velocity of 4.43m/s2, with a Kinetic energy of 49 Joules. KE = 1/2 MV^2 (4.43^2 *5 /2 ) = 49 joules.
Here is one of the other concepts you are not understanding, its the Work/Energy principle, that the CHANGE in Kinetic energy is directly related to the amount of work done. You recognized a change in kinetic energy but disregarded it.
Here is something else you did not take into account. When the object is DECELERATED (acceleration in the opposite direction), the longer the deceleration, the less of an impact force. If you catch something fast, is there a greater impact on your hand if your hand does not move, or it you give with the impact and the hand moves with it?
So if we take that 49 Joules of kinetic energy, and it is equal to the work being done (work energy principle), so we can substitute KE for Work.
Force * Distance = Kinetic energy which is 49 joules. We say that 49 Joules of work was expended over a deceleration of .02M upon impact with the scale.
Force = 49N / .02m.
Force = 2450 N - average force during impact (when decelerated within .02m)
Compare that with 49N when at rest on scale.
Our rebuttal was well do you think a brick resting on your head exerts the same force as one that falls and hits you in the head and he said YES.
The brick that hits someone on the head after being dropped has no more force that the brick sitting stationary on someone's head. The variables for mass and acceleration do not change. I repeat, the brick's mass does not change. The acceleration constant of Earth's gravity does not change. Force remains the same. What changes is VELOCITY. What changes is momentum and kinetic energy. Upon impact we are measuring something called instantaneous velocity which sounds like and oxymoron, but really means the velocity the brick was traveling at the moment of impact with the object is was traveling relative to. We take that velocity and we insert it into the equation, 1/2 MV^2 and we get the Kinetic energy of the object. This is what changes...NOT force.
I replied that yes, of course the force of impact is greater than when it is at rest and gave an example of a 5 kg bowling ball dropping 1 meter and fully stopping within .02m after impact, and I worked out the forces at rest.
If you put a 5 kilogram bowling ball on a scale, you will get 5 kilos. Its weight (which is a force) 5 * 9.8 = 49 Newtons of force. F = M G
Now let's take that 5 kg ball and drop it onto the scale from 1m high. Do you really think it will still only show 5kg upon impact? Of course not.
It will impact the scale at velocity of 4.43m/s2, with a Kinetic energy of 49 Joules. KE = 1/2 MV^2 (4.43^2 *5 /2 ) = 49 joules.
Here is one of the other concepts you are not understanding, its the Work/Energy principle, that the CHANGE in Kinetic energy is directly related to the amount of work done. You recognized a change in kinetic energy but disregarded it.
Here is something else you did not take into account. When the object is DECELERATED (acceleration in the opposite direction), the longer the deceleration, the less of an impact force. If you catch something fast, is there a greater impact on your hand if your hand does not move, or it you give with the impact and the hand moves with it?
So if we take that 49 Joules of kinetic energy, and it is equal to the work being done (work energy principle), so we can substitute KE for Work.
Force * Distance = Kinetic energy which is 49 joules. We say that 49 Joules of work was expended over a deceleration of .02M upon impact with the scale.
Force = 49N / .02m.
Force = 2450 N - average force during impact (when decelerated within .02m)
Compare that with 49N when at rest on scale.
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