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Question, this is engineering and physics (2 Viewers)

Twickel

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Yep Work is equal to negative kinetic energy because the final kinetic energy is always zero. Work = K2-K1.

I calculated my net force using mgSin@-0.2xmxgxCos@ For some reason in my experiment kinetic energy is equal to that for example in trial one.

Net force = 0.35x9.8xSin30 - 0.2x0.35x9.8xCos30=1.12, final velocity = 2.54 K=0.5x0.35x2.54-1.12 See? Why is that happening.
 

Twickel

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Does anyone know why in my experiment the net force is equal to the kinetic energy?
 

Mumma

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Kinetic energy is mv^2/2

Also why are you using mgSin@-0.2mgCos@ for your collision force?
 

Twickel

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I have to make calculations with everything in motion, work kinetic energy force et.c My net force is eqaul to my kinetic energy for some reason why?.
 

Twickel

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Is it wrong because it does not take into consideration the rotation friction and hence its just as in accurate as a=gSinx? Well the teacher asked me to take into consideration the friction of the wood and I am for her by using mgSin@-0.2mgCos@

I used mg sin @ - 0.2 x mgcos@ to calculate net force. so thats F, I then rearragned F=ma a=f/m to calulate acceleration.
 
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Twickel

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What would the acceleration be of then? and how come you provided me with the other formula for acceleration aswell this one gsin@ - 0.2gcos@
EDIT: THE ACCELERATION IM FINDING IS WHEN THE CAR IS SLIDING WITHOUT THE WHEELS? NOT WHEN IT HAS ITS WHEELS ON?
 
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Twickel

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BUt isnt that excepted because in year 11 we can not calculate the rotational kinetic energy. look.
 
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Twickel

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ok, had I let the car slide down with no wheels down my ramp it would have been accurate right?
 

Twickel

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Good guess lol the web does say its 0.2. The car is smooth, its a cheap model lol. No detail. This just shows the general trend you know what I mean. The teacher wants the friction of would I gave it to her. She never mentioned anything about rotational kin energy, but I put that in my discussion how it effects the reliability.
 

Twickel

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Plus I do not have any you know proper equipment ticker timers etc. And the method you gave me is for avg velocity, I needed final. Thanks for you help. This is accurateas I can make it.
 

Mumma

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3unitz said:
energy has to go into deforming the object structurally, heat, sound etc. so not all the kinetic energy is transferred in displacing the crumple zone through the collision distance. if you repeated the collision using a different material for the crumple zone (eg. feathers) it would displace much more because it would take less energy to deform structurally
Yeah, part of the assumption is that all energy transfer is through work done and not heat. Fair assumption? Maybe not, but then I doubt there is much else choice if you really want to approximate collision forces (without access to a highspeed camera to measure time). Dunno.

edit: ignore this post, don't think what i said here is correct. Most energy transfer out of the system during collision would be through heat but its mostly irrelevant when considering the initial impact force, i think.
 
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Forbidden.

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Mumma said:
Yeah, part of the assumption is that all energy transfer is through work done and not heat. Fair assumption? Maybe not, but then I doubt there is much else choice if you really want to approximate collision forces (without access to a highspeed camera to measure time). Dunno.
But then there is the First Law of Thermodynamics you may consider.
 

Mumma

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Well thinking about it, I suppose the equation F.s = deltaKE still makes sense. If we assume KE2 = 0 then s would just have to be really really large for like say a crumple zone made out of feathers (otherwise the car would bounce off and KE2 != 0).
 
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Mumma

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Its just the work energy principle.
In fact I found a source for what I was thinking http://hyperphysics.phy-astr.gsu.edu/HBASE/Impulse.html

The force during a collision would be variable. Fs would give you the average impact force. Yes, with a long feather, (if its long enough to stop the car, since we are assuming all kinetic energy is lost), the impact force will be small. If its metal and the deformation is small, then yes, the impact force would be large. Which is bad.

I attached my free body diagram.
 

Twickel

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Hey guys my final velocity ( impact velocity) is lower then my acceleration is that ok?
 

Mumma

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3unitz said:
Fs does not equal the cars kinetic energy - s is displacement and varies among different materials. if youre saying Fs = K then youre implying you can calculate K from nothing but the cars impact force and displacement of the material. since different materials have different strengths, they will have different displacements for the same impact force.

if i hit a sheet of titanium with a hammer its displacement will be less than if i hit a sheet of aluminum with the same impact force. in other words Fs is largely dependant on the material and does not equal kinetic energy.

i know that the shorter the time of collision the larger the force experienced: F = dp/dt, thats the whole point of crumple zones, airbags, seatbelts etc. i just disagree with Fs = K (notice that here F is the impact force).
I didn't say it was equal to its kinetic energy, I said it was equal to the change in its kinetic energy (in the end its equal to its initial kinetic energy simply because we are assuming the crumple zone is large enough to completely stop the car, ie. final KE is 0) . Also the impact force will be different since the aluminum will not be able to provide the same average impact force during collision as titanium due to it displacing more. Thats the point.

If the impact force was the same between the two materials, then the momentum method would not work either since the change in momentum is constant. That means time would also be constant (Ft = delta(mv) = constant)

All I used was the work-energy principle. You are basically saying this principle is wrong, or I'm applying it incorrectly. If I am applying it incorrectly I would be interested in knowing how I would apply it correctly in this case. That would mean I'm missing some other important force acting on the system.

Also, go to the hyperphysics link i posted and read the second frame, where they say the same thing I did.
 
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