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Homework Help: Not Exactly Newton's Bucket U & K questions

  1. Jun 29, 2009 #1
    Hi I have a few questions on kinetic and potential energy, slightly mundane lol but still interesting.

    So technically, if i am carrying a bucket of water from upstairs to downstairs the water within the bucket heats up, even if ever so slightly. Is this true?

    Also, if we are all on the surface of the earth we are all actually carrying potential energy as we are constantly "pulled" by gravity towards the center of the earth? True or not?

    Finally, could you explain the integral within the equation of "work" in a horizontal direction, as in a box being pushed along the ground. The way i understand it to be is as follows (please let - (S) stand for the integral summa within this, and the way i am writing it is the way that what I am studying laid out, in successive parts)

    My question about the equation is, (i'm new to integrals so i wonder), where the (*) indicator is, why did it change from dv/dx x, to dx/dt dv?
    Where the (**) is, the (S) summa, does it move from each symbol successively, as in it had integrated the m (which is constant right) and moved on to the vdv.
    And to finish, how did 1/2v^2 come about, I've looked on the net and can't seem to understand, everywhere i check just shows it as if it is obvious.

    Amabo Te :):):)
  2. jcsd
  3. Jun 29, 2009 #2


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    Why would the water in the bucket heat up?
  4. Jun 29, 2009 #3
    Because it's potential energy is changing into Kinetic? Maybe only if you threw it off the stairs (with a cover on lol) or dropped it?
  5. Jun 29, 2009 #4
    U=mgh (Potential Energy = mass x acc due to gravity x height) so it's height above the ground (nvm it's height from the centre of the earth) has it stored within...
  6. Jun 29, 2009 #5


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    No, carrying it down the stairs just loses PE -- you are having to do work to walk down the stairs, and your legs warm up, but not the water.

    If you dropped the bucket, the loss in PE is matched by the gain in KE as the bucket gains velocity. Still no change in temperature.
  7. Jun 29, 2009 #6


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    Regarding the derivation of the formula for kinetic energy from the definition of work, I'll post a proper derivation that should answer your questions in a minute. However, you've said you don't know integral calculus. If so, it may be difficult for you to understand the explanation. It would be helpful to know what your background is. Do you at least know:

    - differential calculus (i.e. what a derivative is)

    - some calculus-based physics (i.e. the fact that velocity is the derivative of position with respect to time, and why)

    - the chain rule for differentiation (this is very important here)

    - the power rule for differentiation

    - the fundamental theorem of calculus (i.e. what the relationship is between differentiation and integration)

    If you don't know these things, then my derviation will not make any more sense than what you've already read, even though I've put more effort into explaining each mathematical step.
  8. Jun 29, 2009 #7
    Yep, that makes more sense lol, thanks you :) I figured out more, in the integral it does move from symbol to symbol individually integrating them, still unsure about the other two parts of the equation tho if anyone could help :)
  9. Jun 29, 2009 #8


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    In any case, here goes:

    IF a particle is confined to move in one dimension (along the x-coordinate axis) and the force acting on a particle is also constrained to point along that coordinate axis, but the magnitude of the force is not constant, but rather varies as a function of x, F(x), THEN the work done on that particle as it moves from point x1 to point x2 is defined by:

    [tex] W = \int_{x_1}^{x_2} F(x) dx [/tex] ​

    By Newton's second law, F = ma, and since the force is not constant (it varies with position), we can also conclude that the acceleration varies with position i.e. that the acceleration is a function of x, a(x):

    [tex] W = \int_{x_1}^{x_2} ma(x) dx [/tex] ​

    Now, by definition, the acceleration is the derivative of the velocity with respect to time.
    [tex] W = \int_{x_1}^{x_2} m \frac{dv(x)}{dt}dx [/tex] ​

    Now, we have that the velocity is a function of position v(x). Position is a function of time, x(t), since the particle is moving. Therefore, velocity is a function of time as well. In this situation, we can use the chain rule for differentiation. Do you know the chain rule? In this case it states that:

    [tex] \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} [/tex] ​

    So, *using the chain rule*, we can make the substitution for dv/dt in the integral:

    [tex] W = \int_{x_1}^{x_2} m \frac{dv}{dx}\frac{dx}{dt} dx [/tex] ​

    However, dx/dt is just the velocity, by defintion. v = dx/dt. Therefore:

    [tex] W = \int_{x_1}^{x_2} m \frac{dv}{dx}v dx [/tex] ​

    Now let's take a closer look at the integrand (the thing being integrated). If you look at it, it looks like something that has already been differentiated using the chain rule. In other words, it looks like there is some function h(v) such that, by the chain rule:

    [tex] \frac{dh}{dx} = \frac{dh}{dv}\frac{dv}{dx} [/tex] ​

    The thing on the right hand side is the same as the integrand IF dh/dv = v. In other words, h(v) is an antiderivative of v. Imposing that condition, we have:

    [tex] \frac{dh}{dx} = v\frac{dv}{dx} [/tex] ​

    Our work integral becomes:

    [tex] W = \int_{x_1}^{x_2} m \frac{dh}{dx}dx [/tex] ​

    Now, from the fundamental theorem of calculus, integration and differentiation are inverse processes, meaning that the integral of the derivative of the function is just the function itself h(x). More precisely:

    [tex] \int_{x_1}^{x_2} \frac{dh}{dx}dx = h(x_2) - h(x_1) [/tex] ​

    We have stated that h is function of v, and we can interpret h(x1) to mean h(v(x1)), because the former is, in words, "h(x) evaluated at x = x1", and the latter is "h(v) evaluated at (v evaluated at x1)"

    So the result becomes:

    [tex] \int_{x_1}^{x_2} \frac{dh}{dx}dx = h(v(x_2)) - h(v(x_1)) [/tex] ​

    For notational simplicity, we can write this as:

    [tex] \int_{x_1}^{x_2} \frac{dh}{dx}dx = h(v_2) - h(v_1) [/tex] ​

    Where v1 and v2 are the velocities of the particles at postions x1 and x2 respectively.

    Now, also from the fundamental theorem of calculus, it must be true that:

    [tex] \int_{v_1}^{v_2} \frac{dh}{dv}dv = h(v_2) - h(v_1) [/tex] ​

    If these last two integrals (one with respect to x, and one with respect to v) are both equal to the same thing, then they must be equal to each other:

    [tex] W = \int_{x_1}^{x_2} m \frac{dh}{dx}dx = \int_{v_1}^{v_2} m \frac{dh}{dv}dv [/tex] ​

    Substituting the expressions for dh/dx and dh/dv, we get:

    [tex] W = \int_{x_1}^{x_2} m v \frac{dv}{dx}dx = \int_{v_1}^{v_2} m vdv [/tex] ​

    THIS general result for integrals (which comes from working backwards from the chain rule for derivatives) is what was assumed (and therefore omitted) in the step you identified with an asterisk (*).

    Now we have:

    [tex] W = \int_{v_1}^{v_2} m vdv [/tex] ​

    How do we actually evaluate this integral? In other words, how does one integrate the function f(v) = v? For this, we use the power rule for integration, which states that


    [tex] f(y) = y^n [/tex] ​


    [tex] \int f(y)dy = \frac{1}{n+1}y^{n+1} [/tex] ​

    Here's another rule for you. The integral of a constant times a function is equal to the constant times the integral of the function:

    [tex] \int af(y)dy = a\int f(y) dy [/tex] ​

    That's why the m gets taken outside the integral sign.

    The variable y is being used as a generic variable here. I won't derive these results for you here. In any case, applying them to our work integral, (obviously with n = 1).

    [tex] W = \int_{v_1}^{v_2} m vdv = \left[\frac{1}{2}mv^2 \right ]_{v_1}^{v_2} [/tex]

    [tex] = \frac{1}{2}m(v_2^2 - v_1^2) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 [/tex]

    [tex] = K_2 - K_1 = \Delta K [/tex]

    And so we have derived the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy
    Last edited: Jun 29, 2009
  10. Jun 29, 2009 #9


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    Certainly not. Temperature is usually defined in terms of the average random kinetic energy associated with the molecules in a substance. While, by lifting the bucket we may give the bucket kinetic energy (and do work against the Earth's gravitational field), this kinetic energy certainly isn't random.

    Remember that we can only measure changes or differences in potential energy. For example, it's true that we (at the surface of the Earth) have graviational potential energy with respect to the center of the Earth. However, I can just as easily define a point on the surface of the Earth with zero GPE and make all my change in potential energy calculation from there. Note: This is why we use the equation GPE = (m)(g)(x) when an object is near the surface of the earth.

    Well, if you're not too picky about the semantics, the expression for the kinetic energy (the work-energy theorem) can be derived using only algebra:

    Suppose a block accelerates from rest (Vi = 0) to some final velocity Vf in a time interval Δt. Clearly, the average acceleration is given by the equation,

    aavg = ΔV/Δt = Vf /Δt

    The distance the block travels is defined to be the product of the average velocity and the time, so we have,

    x = (Vf Δt)/2

    Since the work done on an object is the product of the force acting on the object and the distance it travels, we have

    W = F * x

    And by Newton's Second Law,

    W = (m)(a) * x

    Substituting our expressions for a and x into this equation produces,

    W = m * (Vf /Δt) * (Vf Δt)/2

    Which, with some algebra reduces to,

    W = mVf2 /2

    Hope this helps!
  11. Jul 9, 2009 #10
    Seriously thank you so much, was a very detailed reply and i am very grateful. I understoood half of it lol so i got myself a proper calculus book, i am with you until you bring in "dh" but i'll get into my book and come back here and look back again. thanks
  12. Jul 10, 2009 #11


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    You're welcome!

    h(v) is just the arbitrary name I chose for the function that is the antiderivative of the function f(v) = v with respect to v. That's why I said, "it looks like there is *some* function h," emphasizing that I have simply defined it that way, it has no specific physical significance. The problem with my derivation was that the actual physics was interrupted by some math that I had to go into and explain, which broke up the flow and made it more convoluted.

    Good luck with the calculus studying!
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