Calculating Kinetic Energy...

[Dustin Kesselberg]

Let's say we have a 5lb metal ball. Traveling at 500mph.

K = ½mv2

How would i do that?


K=1/2 of 2500 2
K=1250 2
K=1 953 125 000 ???


Would the answer be 1953125000 joules?

I would get a different answer using Kilometers per hour though,So that can't be right.

What is the correct way of doing this equation?

 

[Wudang]

In the correct units - kilograms, metres and seconds.

 

[DrDave]

Remember that the precendence of the operators counts here, and of course units.

So you should square the velocity first. Then multiply by half the mass, this will give you the numerical part of the answer.

Do the same to the units, then you will have your units.

If you want your answer to be in joules, I would advise converting all the initial data into SI units first.

Hope this helps!

Dave

 

[rwguinn]

Let's say we have a 5lb metal ball. Traveling at 500mph.

K = ½mv2

How would i do that?


K=1/2 of 2500 2
K=1250 2
K=1 953 125 000 ???


Would the answer be 1953125000 joules?

I would get a different answer using Kilometers per hour though,So that can't be right.

What is the correct way of doing this equation?

500MPH=733ft/sec
v^2=537777.8 ft^2/sec^2
5lb=.1554 lb-sec^2/ft (5/32.2=w/g=m)
KE=(.155*537777.8)/2=41745 ft-lb
=56653 J

edited to add--get the units right, and cosistent. If you have to convert, either convert everything first, or convert the answer at the end.

 

[Brown]

This simple problem illustrates one nice feature of the metric system: if your inputs are in SI units, the output units "take care of themselves." This is not the case with English units.

 

[ktesibios]

Re: Re: Calculating Kinetic Energy...

500MPH=733ft/sec
v^2=537777.8 ft^2/sec^2
5lb=.1554 lb-sec^2/ft (5/32.2=w/g=m)
KE=(.155*537777.8)/2=41745 ft-lb
=56653 J

edited to add--get the units right, and cosistent. If you have to convert, either convert everything first, or convert the answer at the end.

Umm, just thought it worth mentioning that by dividing weight in pounds by the acceleration of gravity in feet/sec^2 you get mass in slugs- the "English" system unit of mass.

The way you show it does have the virtue of making it easier to keep track of the units and to see that the end result is in ft*lb.

 

[rwguinn]

This simple problem illustrates one nice feature of the metric system: if your inputs are in SI units, the output units "take care of themselves." This is not the case with English units.

Nonsense. It never obvious that a kg-m is a "Joule", nor that a Newton/m^2 should be a "Pascal", (what wrong with N/m^2 ?) or that an ElectronVolt, Calorie and Newton-meter should all have the same unit ("Joule"), just as it is unobvious that foot lbf/hour should be a "horsepower"
As long as you stick with a system of units, and know it, you don't have to worry. Just do your conversions all at once.
And its a falicy that SUI is all "10"s. when g=9.8 m/sec^2 (Why is that easier that 386.1 in/sec^2)--it all goes out the window...
Pick a system, and stick with it.

 

[rwguinn]

Re: Re: Re: Calculating Kinetic Energy...

Umm, just thought it worth mentioning that by dividing weight in pounds by the acceleration of gravity in feet/sec^2 you get mass in slugs- the "English" system unit of mass.

The way you show it does have the virtue of making it easier to keep track of the units and to see that the end result is in ft*lb.

That is because "Weight" is a force, which is m*g back here on Earth.
I know that kg is sup[posed to be mass, but the way everybody screws around with the systems, it ends up being listed as weight, which is not mass, and I get sorely confused as to which I am given.
I like hanging on the the original dimesnions throughout the analysis, because as you said, it shows you how the units come out. You can do the same thing in SUI--just don't convert to J, Pa, N, whatever untill you are done...It ain't at all obvious what a Pa*m actually is when you encounter it...

 

[roger]

And its a falicy that SUI is all "10"s. when g=9.8 m/sec^2 (Why is that easier that 386.1 in/sec^2)--it all goes out the window...
Pick a system, and stick with it.

A fallacy? Really? Tell me then, what is g in, say, deci-meters/sec^2?

 

[Smike]

...or that an ElectronVolt...snip... all have the same unit ("Joule")

An Electron Volt is not a Joule. It is approximately 1.6*10^-19 Joules. It is a seperate unit that measures the same thing (energy). Often used for measuring the energy carried by a photon.

Ahem.

 

[Ziggurat]

An Electron Volt is not a Joule.

Yes indeed. It's a Coulomb Volt that's equal to a Joule.

 

[Ziggurat]

And its a falicy that SUI is all "10"s. when g=9.8 m/sec^2 (Why is that easier that 386.1 in/sec^2)--it all goes out the window...

g is not a unit, it is a physical constant that is peculiar to earth. There's a difference.

 

[Soapy Sam]

Oh, no it's the SI versus traditional debate. Quick, throw several litres of water at 0 deg C over them.

SI and the metric system ain't the same thing. Both are designed to allow mental arithmetic to be done wrongly, faster and more easily than traditional systems which require carrying the potrzebie and generally can only be done on napkins.

The impact of SI on rocket scientists has been significant , though it's higher in mph than in kips. (Unless I divided instead of multiplying...)

 

[Dustin Kesselberg]

Im trying to calculate how much kinetic energy a ball with 100lb mass traveling at 100mph would have.



100lb=45kg

100mph=44m/s


KE = ½mv²
KE = ½ × 45 × 44²
KE = 22.5 × 1936
KE = 43560 J


Does this look about right?

Im getting 43560 Joules..However Looking on google Im seeing that bullets do not even usually have 1000 joules.
So is there something wrong with my calculations?

Help me



There's another problem...
My book says

K.E.=Mass X Velocity 2
---------------------
2

Mass times Velocity squared over 2.

When doing that I multiply The mass times the velocity. THen I square that and then divide it by two. However when I do this I get a completly different answer than when doing it the other way.

 

[Vim Razz]

There's another problem...
My book says

K.E.=Mass X Velocity 2
---------------------
2

Mass times Velocity squared over 2.

When doing that I multiply The mass times the velocity. THen I square that and then divide it by two. However when I do this I get a completly different answer than when doing it the other way.

Make sure you square the velocity before multiplying it by the mass -- this may be a problem of unclear notation in your text. You did it right when you did it the other way.

on the earlier calculation
Im getting 43560 Joules..However Looking on google Im seeing that bullets do not even usually have 1000 joules.

Bullets also dont usually weigh 100lbs.

Your numbers look fine there.

 

[rwguinn]

An Electron Volt is not a Joule. It is approximately 1.6*10^-19 Joules. It is a seperate unit that measures the same thing (energy). Often used for measuring the energy carried by a photon.
Originally posted by rwguinn
...or that an ElectronVolt...snip... all have the same unit ("Joule")

Ahem.

Ahem, youself, idiot.
Take a (rule 8) reading course.
how stupid do you have to be to understand that when somebody says "all have the same unit" he is not equating the values--just the unit.
The unit is in joules. I never said it was equal to a J. All of those different things have as units the Joule. Some are more than 1 J, some are less than 1J. they still have the unit of Joule.

 

[rwguinn]

g is not a unit, it is a physical constant that is peculiar to earth. There's a difference.

Ok, so where do you live and do all your work?

 

[TjW]

Ok, so where do you live and do all your work?

Oddly enough, it's on a planet where pi is not exactly equal to 3.

 

[pmurray]

Let's say we have a 5lb metal ball. Traveling at 500mph.

K = ½mv2

625,000 miles pound miles per hour per hour. Enough energy to give a one pound mass an accelleration of 1 mile per hour per hour over a distance of 625,000 miles. Or, enough energy to accellerate a 625,000 pound mass by one mile per hour per hour over a distance of one mile (at the end of which it will be travelling at .707 mph).

 

[c4ts]

Oh, no it's the SI versus traditional debate. Quick, throw several litres of water at 0 deg C over them.

Well, I'll just retaliate by flinging a half-cup of water at 373.15 K and... crap.