What is Energy? Split fro Can energy move faster than light?

[quickquestion]

First of all, I am trying to figure out what energy is, scientifically.

 

Because my science textbook says PE=height*gravity.

And KE=.5*m*sqr(v).

 

But if KE is an exponential function of v...how then can PE be simply linear height and gravity.

Edited April 23, 2017 by quickquestion

[Mordred]

The scientific definition is "The ability to perform work" Now apply that definition to the problem sets on potential and kinetic energy above.

Edited April 23, 2017 by Mordred

[quickquestion]

The scientific definition is "The ability to perform work" Now apply that definition to the problem sets on potential and kinetic energy above.

If that is the case, then I don't get the definition of PE.

Because it violates the equation of KE.

[Mordred]

Does it?

Potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.

 

So in order to calculate PE you have to determine its position from the center of gravity in the case of graviational potential energy.

Edited April 23, 2017 by Mordred

[quickquestion]

Does it?

Potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.

 

So in order to calculate PE you have to determine its position from the center of gravity in the case of graviational potential energy.

The science book i read said that the main thing you have to factor in is the distance from floor level. Not its Cog.

[Mordred]

that works too depends on the book in this go with your textbook. I take it this is homework? If so let us know so we can move this thread to the homework section

Edited April 23, 2017 by Mordred

[quickquestion]

that works too depends on the book in this go with your textbook. I take it yhis is homework?

No.

My praxis is that I deeply suspect that PE equation is borderline fictitious.

[Mordred]

No it isn't. Ok lets take an example a weight sits on a table. Calculate the potential energy of a 1 kg mass sitting 10 feet off the floor with g=9.8 m/s.

 

That weight can obviously perform work after all if you remove the table the weight will fall. Potential energy of course will convert to kinetic energy once the weight starts moving.

 

One thing you will discover both PE and KE are used also in more advanced physics including GR or the formulas for effective action. Its extremely important to properly understand

Edited April 23, 2017 by Mordred

[quickquestion]

No it isn't. Ok lets take an example a weight sits on a table. Calculate the potential energy of a 1 kg mass sitting 10 feet off the floor with g=9.8 m/s.

 

That weight can obviously perform work after all if you remove the table the weight will fall. Potential energy of course will convert to kinetic energy once the weight starts moving

And this is where the fictitious part comes in.

It says PE=height*gravity.

 

But KE=.5*m*sqr(v).

 

On impact it will return KE.

The KE will give a much higher value than the PE allots.

And the PE will not yield the same results as the KE equation.

[Mordred]

Then you may be doing something wrong in your calculations. Your Ke should return your orginal PE value. (here is a hint)

 

as it falls, its total energy (the sum of the KE and the PE) remains constant and equal to its initial PE.

 

Take the scenario above with the formulas you posted and show your calcs so we can find what your missing

 

(were now dealing specifically with the conservation of energy) involved in the scenario above.

Edited April 23, 2017 by Mordred

[quickquestion]

Then you may be doing something wrong in your calculations. Your Ke should return your orginal PE value. (here is a hint)

 

as it falls, its total energy (the sum of the KE and the PE) remains constant and equal to its initial PE.

 

Take the scenario above with the formulas you posted and show your calcs so we can find what your missing

Pe=mgh

bowling ball of mass 5 kg, height of 450 meters, gravity of 9.8 m/s.

 

 

Ke=0.5*m*sqr(v)

velocity after 450 meters is 54 m/s.

ke=7250 j.

pe=22050 j.

 

So its a bit suprising, I thought ke would be more than pe, but still they do not match.

[Mordred]

What is you velocity of the ball just prior to impact? (I noted I forgot to square seconds above.) should be 9.8 m/sec^2 Your velocity should be higher than 54 metres/sec after falling 450 metres

Edited April 23, 2017 by Mordred

[imatfaal]

Just in case this is a confusion not a typo

 

KE = 1/2 m v^2 Not KE = 1/2 m sqrt(v)

[swansont]

Pe=mgh

bowling ball of mass 5 kg, height of 450 meters, gravity of 9.8 m/s.

 

 

Ke=0.5*m*sqr(v)

velocity after 450 meters is 54 m/s.

ke=7250 j.

pe=22050 j.

 

So its a bit suprising, I thought ke would be more than pe, but still they do not match.

Let's see your calculations.

[Mordred]

One error for sure is the velocity after 450 metres. time to fall is 9.579 seconds velocity should be 93.95 m/s.

 

[latex] t=\sqrt{\frac{2h}{g}}[/latex]

[latex]v=\sqrt{2gh}[/latex]

Edited April 23, 2017 by Mordred

[Strange]

[latex]v=\sqrt{2gh}[/latex]

 

 

And that, by itself, shows that PE and KE are equal.

And (surprise!) using 93.95 m/s for the velocity gives the same as PE.

[quickquestion]

What is you velocity of the ball just prior to impact? (I noted I forgot to square seconds above.) should be 9.8 m/sec^2 Your velocity should be higher than 54 metres/sec after falling 450 metres

Google tells me terminal velocity is 54 m/s and height to reach terminal velocity is 450 m.

 

One error for sure is the velocity after 450 metres. time to fall is 9.579 seconds velocity should be 93.95 m/s.

 

[latex] t=\sqrt{\frac{2h}{g}}[/latex]

[latex]v=\sqrt{2gh}[/latex]

I dont know how you got 93.95. That is nearly double. I did not know air resistance could reduce the velocity by double like that.

Edited April 23, 2017 by quickquestion

[imatfaal]

Google tells me terminal velocity is 54 m/s and height to reach terminal velocity is 450 m.

 

I dont know how you got 93.95. That is nearly double. I did not know air resistance could reduce the velocity by double like that.

 

This is a mechanics thought experiment - air resistance,terminal velocity, friction etc do not come into the calcs.

 

This cartoon perfectly sums up physicists' ideas of experiments

 

[Strange]

Google tells me terminal velocity is 54 m/s and height to reach terminal velocity is 450 m.

 

I dont know how you got 93.95. That is nearly double. I did not know air resistance could reduce the velocity by double like that.

 

 

I wonder if you mean "Wikipedia tells me"?

 

Near the surface of the Earth, an object in free fall in a vacuum will accelerate at approximately 9.8 m/s², independent of its mass. With air resistance acting on an object that has been dropped, the object will eventually reach a terminal velocity, which is around 53 m/s (195 km/h or 122 mph^%5B1%5D) for a human skydiver. The terminal velocity depends on many factors including mass, drag coefficient, and relative surface area and will only be achieved if the fall is from sufficient altitude. A typical skydiver in a spread-eagle position will reach terminal velocity after about 12 seconds, during which time he will have fallen around 450 m (1,500 ft).^%5B1%5D

https://en.wikipedia.org/wiki/Free_fall#Examples

 

But if you are going to use terminal velocity, then you need to calculate the energy lost to air resistance, etc. At which point it all starts getting very complicated (but if you do it right, you will find that energy is, indeed, conserved).

[MigL]

Back to your original question as to the difference between the two energies...

 

For a conservative force such as gravity, where the work done is not path dependent, kinetic and potential energies are related by the fact that one type is exchanged for the other.

We can assign a potential energy to any height, but as the height is reduced, this potential is exchanged for kinetic energy ( speed )such that the total of the two is conserved.

[swansont]

Google tells me terminal velocity is 54 m/s and height to reach terminal velocity is 450 m.

 

Terminal velocity implies air resistance, which is not a condition of the problem.

[Sriman Dutta]

If initial velocity is zero, then [math]v^2=2gh = 2*9.8*450 =8820 [/math] . We are here neglecting aerodynamic drag.

 

So, [math]K=\frac{1}{2}mv^2 = \frac{1}{2}*5*8820= 22050J[/math]

And,

[math] P=mgh=5*9.8*450=22050J[/math]

 

So simple.

Edited April 24, 2017 by Sriman Dutta

[quickquestion]

Ok, so what about this though. Let's make it even simpler and give a situation with zero air resistance.

 

On earth an object will take 1 second to fall 9.8 meters.

PE equation is mgh.

Lets say mass is 4 kg, gravity is 9.8 m/s and height is 9.8 m.

So it will tell me

PE: 4*9.8*.9.8

 

KE=.5*m*sqr(v)

KE=2*9.8*9.8

 

These do not match.

So what am I doing wrong?

Or is the equation itself, wrong?

 

Since bullet experiments in water proved KE is correct equation, then I say PE is the wrong equation.

[imatfaal]

[latex]

PE_{at 9.8m} = mgh = 4 * 9.8 * 9.8 = 384.16J[/latex]

 

[latex]KE_{groundlevel}= \frac{mv^2}{2} = \frac{4 * v^2}{2}[/latex]

 

but what is v? look at suvat

 

[latex]v^2 = u^2 +2as[/latex]

 

u=0m/s (starts from a standstill) a=9.8m/s^2 s=9.8m

 

[latex]v^2 = 0+2 * 9.8 * 9.8 = 192.08[/latex]

so

 

[latex]KE_{groundlevel} = \frac{mv^2}{2} = \frac{4 * 192.08}{2} = 384.16J[/latex]

 

So your 384.16 Joules of PE at 9.8 metres is converted entirely to 384.16J of KE by the time it reaches the ground

[Strange]

KE=.5*m*sqr(v)

KE=2*9.8*9.8

 

These do not match.

So what am I doing wrong?

 

 

It is not clear where you get the numbers for the KE equation, but they are clearly wrong.