
#19
Jun2507, 03:03 PM

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P: 6,561

But consider this when the mass is subject to only deceleration. For a reduction in speed from v0 to v1 over an arbitrarily short time interval the change in momentum is m(v1v0) and the average speed is arbitrarily close to (v1+v0)/2. If you combine n of these intervals: [tex]E = \sum_{j=0}^n m(v_{j+1}  v_j)(v_{j+1} + v_j)/2 = \frac{m}{2}\sum_{j=0}^n (v_j^2  v_{j+1}^2) = = \frac{m}{2}\left((v_0^2  v_{1}^2) + (v_1^2  v_2^2).... + (v_{n2}^2  v_{n1}^2) + (v_{n1}^2  v_n^2)\right)[/tex] But the intermediate terms all cancel out and you are left with: [tex]E = m(v_0^2v_{n}^2)/2 = m(v_n  v_0)(v_n + v0)/2 = \Delta p(v_0+v_n)/2[/tex] If v_n = 0 then: [tex]W = m(v_0)(v_0/2) = \Delta p(v_0)/2[/tex] So, to be perfectly correct, one should say that the work done is the change in momentum multiplied by the average of the beginning and end speeds ([itex]W = \Delta p (v_i + v_f)/2[/itex]). AM 



#20
Jun2607, 04:01 PM

P: 63

Scientists came up with the idea of work after noticing that the quantity Fd (force times distance) is conserved. This led them to call that quantity work, W.
They then sought to manipulate the equation to see what they can extract from it and they did this: W=ΔKE=Fd=mad= m* (VfV0)/t * (Vf+V0)/2 * t=(m/2) (Vf^2V0^2)=KE(final)KE(initial) 



#21
Jun2607, 05:31 PM

P: 555





#22
Jun2607, 06:10 PM

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P: 4,107

Of course, the phrase "not generally true" doesn't mean NEVER TRUE. The point of using this phrase is to suggest that it is NOT ALWAYS TRUE and relevant conditions when it is true should spelled out (especially ones that are likely to lead to misconceptions.. e.g., the above passage I first quoted). (If it's not clear by now, the common misconception is that "average velocity" is [always] (v_{i}+v_{f})/2. It's not always true... it is true in certain cases (as finally pointed out above)... and there are textbook situations when it isn't true. "Averagevelocity" as used in kinematics means (as nrqed points out) timeaveragedvelocity: [tex]v_{avg} \equiv \displaystyle\frac{\int v\ dt}{\int dt}= \frac{\Delta x}{\Delta t} [/tex]. Most textbooks unfortunately leave out the center expression [or at least a discretized algebraic version of it], which would probably help clear up the misconception.) 



#23
Dec909, 01:17 PM

P: 19

I think that by "not generally true" it was just meant to suggest that there is a more general expression which always holds which does not depend on qualifying statements, like requiring that the acceleration be constant or that the force be constant, for example. When a discussion begins with a certain context and carries on until some of the joining parties loose track of these qualifiers because they are not explicitly stated every time, confusion can creep in.
Here's my question. When is it correct to use the formula, i.e. kinetic energy is one half the mass times velocity squared, and when should one revert back to the integral of the velocity with respect to the velocity, etc. in order to get the correct result? 



#24
Dec1009, 10:01 AM

P: 44

Here’s my version of the derivation of the kinetic energy formula (first post using latex, so go easy)
Kinetic energy is force times change in distance. [tex] KE = F \Delta x [/tex] Force is mass times acceleration, which is a change in velocity over time. We can write this as [tex] F= m \frac{\Delta v}{\Delta t} [/tex] Now the change in distance, [tex] \Delta x [/tex] can be expressed as velocity times time [tex] \Delta x = \bar{v} t [/tex] Notice that the velocity is is average velocity, where we have a constant change in velocity, or acceleration. Average velocity can be written as the sum of the initial and final velocities divided by two. [tex] \bar v = \frac{v_0 + v_f}{2} [/tex] Now let's look at final velocity. We can write this as the initial velocity plus the change in velocity over time multiplied by time [tex] v_f = v_0 + \frac{\Delta v}{\Delta t} \Delta t [/tex] which can also be written as [tex] v_f = v_0 + at [/tex] Plugging this into the aveage velocity formula gives [tex] \bar v = \frac{v_0 + \left[ v_0 + \frac{\Delta v}{\Delta t} \Delta t \right]}{2} [/tex] which simplifies to [tex] \bar v = v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t [/tex] Now, we plug this into the change in distance formula [tex] \Delta x = \left[v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \right] \Delta t [/tex] [tex] = v_0 \Delta t + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t^2 [/tex] Now, let's assume the beginning velocity is zero. Then we take the product of the above and force [tex] F\Delta x = \left( m\frac{\Delta v}{\Delta t} \right) \left( \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \Delta t \right) [/tex] Collecting terms we arrive at [tex] KE = m \frac{1}{2} \Delta v \Delta v [/tex] Which of course is [tex] KE = \frac{1}{2} mv^2 [/tex] 



#25
Dec1009, 10:34 AM

P: 63

Looking at the history of an idea/concept can sometimes help to get a feel for it. Wikipedia has a small bit on the history of energy: http://en.wikipedia.org/wiki/Energy and http://en.wikipedia.org/wiki/Vis_viva




#26
Feb2010, 08:55 AM

P: 19

To be specific, my real question is this: since the first derivative squared and the second derivative are generally not equal, why is it that we assert in undergrad phys that mechanical work done on a particle is equal to the kinetic energy gained?
Work = $\int \frac{d\; \vec{p}}{dt} \bullet d \vec{x} = m \int \frac{d^2 \vec{x}}{dt^2} \bullet d \vec{x}$. But the kinetic energy is $\int \vec{p} \bullet d \vec{v} = m \int \frac{d \vec{x}}{dt} \bullet dfrac{d \vec{x}}{dt}$ It seems as though this is saying that the first derivative squared is equal to the second derivative of x with respect to t. The same thing happens in the Schrodinger equation, when p is squared to set up the kinetic energy. The momentum operator in coordinate space is squared, and this is taken to be equivalent to the second derivative with respect to x, which is true when the function is the plane wave, but not in general. But, since all wave functions can be expanded in a fourier series or transform, we can get away with this seeming slip of the pen. What is it about energy that, regardless of the functional dependence of the position with respect to time, we can unequivocally say that the first derivative squared is equal to the second derivative? 



#27
Feb2010, 03:44 PM

P: 2,050

[tex]W =\int F\ dx =m \int \frac{d v}{dt}dx =m \int \frac{d x}{dt}\frac{d v}{dx}dx =m \int \frac{d \frac 1 2 v^2 }{dv}\frac{d v}{dx}dx =m \int d (\frac 1 2 v^2) =\big[ \frac 1 2 m v^2 \big] =\Delta KE [/tex] 


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