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Very simple problem: Derrive equation for KE.

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Very simple problem. Derive the equation 1/2mv^2 for kinetic energy.


    2. Relevant equations
    1/2mv2.


    3. The attempt at a solution
    ∫f(x) dx
    ∫ma dx
    ∫m (dv/dt) dx
    m∫(dv/dt) dx

    Now, here is where I get confused. Here is the supposed solution that I don't understand.
    m∫(dx/dt)dv
    m∫v dv
    m (1/2v2) +c
    =1/2mv2

    Yet, this begs two questions.
    1. Why am I able to move the derivative from (dv/dt)dx to (dx/dt)dv?
    2. I don't understand the concept. Why is the work done (well, after considering Kf=ki + w) equivalent to the intergral of a force with respect to time? I don't really understand this concept. Can anyone explain this to me?

    For example, the following equation for a spring.
    kx is the force of a spring, so to find the work done by a spring, we do:
    k ∫x dx
    k(1/2x2)
    = 1/2kx2 - 1/2kx2

    Of course, this is the familiar equation for work done by a spring. But wouldn't we need to multiply this equation by x (to get F*d), so actually 1/2kx3?
     
  2. jcsd
  3. Nov 19, 2011 #2

    rock.freak667

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    Homework Helper

    Well if we apply the chain rule to dv/dt, we will get dv/dt = (dx/dt)*(dv/dx).

    so the integral becomes: ∫(dx/dt)*(dv/dx) dx = ∫ v(dv/dx) dx = ∫v dv

    While it's not the correct way, you can think of (dv/dx)*dx as having the 'dx' part cancel out.


    By the definition of work done

    W = ∫F dx

    For the spring, F = kx so W = ∫ kx dx = 0.5kx2 (the limits are x and 0)
     
  4. Nov 20, 2011 #3

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    This is some important math that some physics textbooks gloss over, or explain by "abuse of notation" without proving it properly. We know that [tex] a = \frac{dv}{dt} [/tex] Using the chain rule, we can write this as [tex] a = \frac{dv}{dx}\frac{dx}{dt} [/tex]Notice, however, that dx/dt = v by definition. So the expression becomes [tex] a = v\frac{dv}{dx} [/tex]Now notice that this last expression looks like something that has been differentiated using the chain rule. In particular, if you define a function u(v) = ∫vdv, then the expression becomes [tex] a = \frac{du}{dv}\frac{dv}{dx} [/tex]It looks like the chain rule has been used to differentiate u(v) with respect to x. So we can REVERSE the chain rule and write[tex] a = \frac{d}{dx}[u(v)] = \frac{d}{dx}\left(\int v\,dv\right)[/tex] With this in mind, the derivation becomes:[tex] W = m\int a(x)\,dx[/tex][tex] = m\int v \frac{dv}{dx}\,dx [/tex] [tex] = m\int \frac{d}{dx}\left(\int v\,dv\right)\,dx [/tex] At this point the integration and differentiation with respect to x cancel each other out, and we are left with: [tex] W = m\int v\,dv [/tex]This idea of reversing the chain rule is the key thing that allowed us to derive the property of integrals that we used, namely that [tex] \int v\frac{dv}{dx}dx = \int v\,dv [/tex]Most physics textbooks just explain this by "abusing" the notation i.e. by saying that you can "cancel" the dx's," in spite of the fact that the dx's are not supposed to be regarded as numbers/quantities.
    I think you meant to ask, why is work equal to the integral of force with respect to distance, not time. In any case, yes I can explain this to you.But first, let's address this:
    NO. The thing you must realise and get used to is that work is NOT equal to force*distance in general. The most general definition of work (i.e. the equation that is always true in all (1D) situations) is [itex] W \equiv \int F(x)\,dx [/itex]. WHY is this true? Well, one could argue that it is just the definition of work. But I can give you some motivation for this definition, which I think is what you want. Let's say that an object moves in 1D from initial position x1 to final position x2. Let's say also that the force acting on this object varies continuously along the path. I.e. the force is not constant, but rather it is a function of position: F(x). It should be pretty clear to you that you cannot compute the work done in going from x1 to x2 simply by multiplying the applied force by the total distance travelled (x2 - x1). The problem is that the force varies continuously over this interval, so which value of the force would you use in the product? How can we solve this problem? Well, one thing we could do would be to consider a very small distance interval [itex]\Delta x_i = x_{i+1} - x_i [/itex] along the path (xi+1 and xi are two points along the path that are fairly close to each other). We'll break up the total distance into "N" such sub-intervals, so that the interval size [itex] \Delta x_i = (x_2 - x_1)/N [/itex]. Let's say this interval size is small enough that we can pretend that the force is roughly constant over this short sub-interval, and that it is just equal to [itex]F(x_i)[/itex] So the little bit of work done during each interval could be approximated as [tex]W_i = F(x_i) \Delta x_i [/tex]To get the total work done in going from x1 to x2 then, we'd just take the sum of all of these little bits of work that are done in each of the sub-intervals that we've broken the path up into:[tex] W = \sum_i W_i = \sum_i F(x_i)\Delta x_i [/tex]Now, obviously this is only an approximation to the true work done over the interval. It has some error, because we pretended that the force was constant over each of the little intervals Δxi, when in reality the force varies continuously with x. However, you can make the approximation better by making the intervals smaller (and having more of them). In fact, you can make the approximation arbitrarily close to the true value by making the size of the intervals arbitrarily small (and the number of intervals "N" arbitrarily large). In other words, the true value of the work done is equal to the limit of this sum as the interval size Δxi goes to 0:[tex] W = \lim_{\Delta x_i\to 0}~\sum_i F(x_i)\Delta x_i = \lim_{N \to \infty}~\sum_{i=0}^N F(x_i) \left(\frac{x_2 - x_1}{N}\right) [/tex] You should recognize this last expression. It's just the definition of the definite integral [tex]\int_{x_1}^{x_2} F(x)\,dx [/tex]So that's why work, in general, is equal to the integral of force with respect to distance. I should note that the physicist's sloppy "abuse of notation" way of explaining this is to say that the infinitesimal bit of work dW done over the infinitesimal distance interval dx is given by dW = F(x)dx, and hence the total work done over the path is just W = ∫dW = ∫F(x)dx. I think that the way I did it above is more rigourous, because the concept of a "limit" is something that is mathematically well-defined, whereas the concept of an "infinitesimally-small" quantity is not clearly-defined (at least not at the level of introductory calculus). Nevertheless, you'll encounter the abuse of notation all over the place, and it will give you the right answer.

    You'll note that I said that work is not equal to force*distance in general. But we can derive this result from the more general definition for the special case where the force is a constant i.e. F(x) = F = const.:[tex] W = \int_{x_i}^{x_f} F(x)\,dx = F\int_{x_i}^{x_f}\,dx = F(x_f - x_i) = F\Delta x[/tex]So that's where "work = force*distance" comes from. It's ONLY true in the special case where the force is constant, whereas the integral definition is true in general.
     
    Last edited: Nov 20, 2011
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