Go Back   Physics Help Forum > College/University Physics Help > General Physics

General Physics General Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Nov 9th 2013, 08:53 AM   #1
Senior Member
 
Join Date: Jun 2010
Location: NC
Posts: 418
Smile What is Energy?

Something curious to me. For a BODY as system,
a "mass equation" is a simple scalar idea.

1. The 2nd Law might be called a "momentum equation." This equation is applied and solved to yield, as answer, some momentum of an event.

2. Multiplication of the momentum equation by a vector differential displacement (attention given to the math) yields an "energy equation."
For cases, the energy equation is solved to tell the BODY energy for some event.

I see the logic of solutions by way of "1", i.e., use of the 2nd Law alone and only. And solutions via "2", use of energy equation only - make sense.

My "ill at ease" arrives when the equations: 1) momentum and 2) energy are used together. After all, 2) is not independent of 1). 2) is a vector scalar product of 1). The equation 5[(x + 5y) = 4] has the same information as (x + 5y) = 4.

So what is going on? How can momentum get multiplied by displacement and a new independent equation happens? I suppose I'll feel foolish when the answer comes in.

Beware, I have a physical example to explain next.

TSH
THERMO Spoken Here is offline   Reply With Quote
Old Nov 11th 2013, 07:49 AM   #2
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,352
If I understand your question, it's how does going from:

1.

to:

2.

introduce a new independent equation? I would say it doesn't. Absent some other effect not covered by Newton's Laws - such as friction (evident by heat) the work-energy equation is simply an extension of the impulse-momentum equation.

Last edited by ChipB; Nov 11th 2013 at 01:30 PM.
ChipB is offline   Reply With Quote
Old Nov 11th 2013, 12:34 PM   #3
Senior Member
 
Join Date: Jun 2010
Location: NC
Posts: 418
Constrained Free-Fall

Hi Chip...
Thanks for thinking on this. Maybe KE is a variant of momentum?

I was thinking momentum was a property and then energy, another property, IS obtained by multiplication.
That makes me wonder.

For a physical situation I was thinking about free-fall (see my figure) The 2nd law solves (by itself) for free-fall of the block - Case 1 with no constraint and - Case 2 with a linear constraint.

Case 3) It seems to me N's 2nd Law is insufficient (alone) to tell where the block leaves the cylinder in this free-fall 2nd-order constrained. I looked at the equations and I didn't see a solution. Others get the answer but they use v = sqrt(2gh). I don't want to use use v = sqrt(2gh).

It seems the 2nd Law ALONE should be enough.

Not important I suppose.

Thanks, Jim
Attached Thumbnails
What is Energy?-three_slide.png  
THERMO Spoken Here is offline   Reply With Quote
Old Nov 11th 2013, 02:02 PM   #4
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,352
I would argue that v=sqrt(2gh) does come from Newton's 2nd law:

Fdt = mdv
F(dt/dx)dx = mdv
F/v dx = mdv
Fdx = mvdv. Call this equation (1). For F=constant this integrates to give:
Fh = (1/2)mv^2

F=mg, so
mgh = (1/2)mv^2
v= sqrt(2gh).

But what if F is not constant? In your problem F = component of gravitational force acting in the direction tangent to the cylinder, which changes as the blocks slides:

F=mg sin a, where angle 'a' is measured from the vertical, and dx =R da. So we have:

int(F dx) = int(mgR sin a )da = mgR(1- cos a). Set this equal to (1/2) mv^2, and it yields: v = sqrt(2gR(1-cos a)). R(1- cos a) is the same as 'h' - the vertical distance the block has slid.

So it doesn't matter that the ramp is curved and F is not constant - the velocity of the block depends only on how for it has slid vertically. For convenience we call this idea "energy principals" - change of KE equals negative change in PE. But it all derives from the 2nd law.

Last edited by ChipB; Nov 11th 2013 at 02:10 PM.
ChipB is offline   Reply With Quote
Old Nov 11th 2013, 02:34 PM   #5
Forum Admin
 
topsquark's Avatar
 
Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,814
You do realize that this question (or the basic definition of just about anything in Physics) is going to drive you crazy. Basic quantities such as distance, speed, energy, etc. can be more or less defined in terms of their units. The problem is that the unit for distance, meters, cannot be defined without recourse to a concept of length. Circular definition.

Some quantities are more useful than others. Energy and momentum are two of the biggies. (In fact any quantity that is conserved tends to be in this category.) Something I've been working on over the last decade is how to change the system of units to be such that the three base unit quantities are energy, momentum, and action. The problem is how to interpret a quantity with the units J/(kg m/s). No real progress there. But it's an interesting hobby. (Yes, the above unit can be broken down, but I'm talking about in reference to the new system of base units.

As far as the rest I'm only going to make the comment:
Originally Posted by THERMO Spoken Here View Post
Maybe KE is a variant of momentum?
In Relativity we have the energy-momentum 4-vector:

so energy and momentum have a fundamental connection.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.
topsquark is offline   Reply With Quote
Old Nov 12th 2013, 08:03 AM   #6
Senior Member
 
Join Date: Jun 2010
Location: NC
Posts: 418
Thanks All!

Thanks. This is just something I've thought about. The wife and I are travelling to her parent's for Thanksgiving. I'll take the problem with me and go over it thoroughly again.

Jim
THERMO Spoken Here is offline   Reply With Quote
Old Jan 3rd 2014, 03:15 PM   #7
Pmb
Physics Team
 
Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,576
Originally Posted by topsquark View Post
In Relativity we have the energy-momentum 4-vector:

so energy and momentum have a fundamental connection.

-Dan
It turns out that this expression is not an identity but an equality that holds only in SR where one uses inertial frames. If you chose to use a non-inertial frame and thus the coordinates were those corresponding to such a frame then, if the components of the metric tensor are independant of time then energy is conserved. However the energy is not equal to E = P^0 but E = P_0 which is quite different. In all cases E = P_0 since this is a definition of E whereas E = P^0 only holds in special coordinate systems.

However in all coordinate sustems mc = P^0 is a definition where m = relativistic mass.
Pmb is offline   Reply With Quote
Reply

  Physics Help Forum > College/University Physics Help > General Physics

Tags
energy



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Solar energy to mechanical energy transformation thinhnghiem Light and Optics 1 Aug 30th 2016 09:51 AM
ionistaion energy of H atom and energy released ling233 Nuclear and Particle Physics 3 Oct 20th 2014 03:27 PM
Binding energy + fission energy help please fishkeeper Nuclear and Particle Physics 4 Mar 1st 2011 11:37 PM
Is the increase in kinetic energy equal to the decrease in potential energy? s3a Energy and Work 1 May 28th 2009 06:02 PM


Facebook Twitter Google+ RSS Feed