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Relationship between Potential Energy and Force

  1. May 1, 2013 #1
    Hello Everyone,
    I am having a very difficult time intuitively understanding the formula -dU/dx = Force(x). I don't want help deriving it, I'm simply looking for an intuitive understanding about why this might be true. An example with gravity would really help.

    My thoughts so far: I find working with -Δ(U)/Δ(x) = avg Force easier for understanding. So lets say a ball is falling from initial height xi and ends up at final height xf. Keeping this same displacement, if a ball loses 1,000,000,000,000 J of potential energy in that displacement, then the avg Force over that displacement must be very big........why? what force? net force? gravitational force? Instead, if the ball loses .000000001 J of potential energy in that exact displacement, then the force acting on the ball must have been.....small? why?

    Instead, let's say that over a displacement, the ball loses 100 J of potential energy. If we make the displacement particularly big, say 1,000,000,000 m, then, man, it took that much displacement for the ball to lose 100 J? so the force must be ... very small? Makes a little more sense... if the ball loses 100 J in .001m, then the force was very large? What does the force have to do with how much energy is lost?

    Lastly, why the negative sign? Bonus points for connecting this with ∫f(x)dx = -U

  2. jcsd
  3. May 1, 2013 #2
    The force and the distance dictate the work done, and the work done combined with the theory of conservation of energy dictate the loss of potential energy. The minus sign is there because the work done on an object by a force must equal the loss of potential energy due to a displacement of that object in the field.

    force of gravity * distance fallen = work done on ball by gravitational force.

    work done on ball by gravitational force + change in gravitational potential energy = 0 (conservation of energy)

    therefore change in gravitational potential energy = - force*distance,

    or as you put it,

    ΔU/Δx = - force

    A ball falling further OR due to a stronger gravitational field leads to a more negative change in gravitational potential energy. But from a conceptual point of view, it makes more sense to treat the force and distance as the independent variables here. The change in potential energy that results is merely a consequence of the laws of motion (and the definition of the potential energy itself).

    The integral form ΔU = -∫Fdx is an extension in the case that the force is not independent of the position.
  4. May 1, 2013 #3
    Hard for me to answer because it seems you have explained it yourself just fine. Change in energy is had by doing work (or having work done). Work is defined as your last equation, its the force applied over a distance.

    You can do the same amount of work by applying a large force over a small distance or a small force over a large distance. This is like earning money, you can earn the same amount of money if you make a high wage for a short time or a low wage for a long time.

    One practical place that you do this is when you jack your car up, right? You want to lift your car a few inches up, thats a small distance. Because the distance is small you have to pull really hard to lift your car, its more force than you can apply. But if you stretch out the force you are applying to a few feet rather than a few inches you can do the same amount of work with your arm's force. This is done using leverage (mechanical or otherwise) in your jack.

    The negative sign is there because forces act to minimize potential. If you want to raise your potential energy, you have to apply a force against the force field (gravity, or EM). To lower your potential energy, you apply your force in the direction of the field.
  5. May 1, 2013 #4
    Ahhhhh! I think I see it now!!! "I can see clearly now, the rain is gone!" lol.
    My problem what that I was thinking about it with respect to time instead of distance.
    Here's my current understanding:
    Let's say I have two positive point charges, one held in place. Let's say I'm far away from the held in place one: As I displace the other point charge small amount IN THE DIRECTION of the one held in place, the potential energy grows by a small amount, almost noticeable, because I'm so far away.

    Now let's say I'm really close to the point charge held in place. As I displace the positive point charge by the same small amount, the change of the potential energy is now VERY large. I've now added much more energy to the point charge than before, and now it really wants to shoot away. So the rate of change of potential energy with respect to displacement is a lot bigger very near the source field than very far away.

    However, the force on the movable point charge is ALSO largest near the source field, so when the rate of change of potential energy with respect to displacement is very large, so must the force be very large! And it takes more and more force just to be able to displace the point charge by the same amount! So whenever dU/dx is very large, you can count on the force also being large.

    Is this all correct?

    Thanks everyone.
  6. May 1, 2013 #5
    One last thing:
    Last time I checked, a derivative was just a scalar, so it can only be positive or negative. How can you derive any direction for the force from this?

    An example: dU/dX = 10J/m at a displacement -.5 m for a rock being loaded onto an elastic slingshot. Origin is at relaxed position. What is the force? I can give a magnitude, but how can I derive a mathematical direction from a scalar? Maybe a derivative isn't a scalar....
  7. May 1, 2013 #6

    It's not a scalar, have you heard of the gradient operator? Strictly, F = -grad(U), which is a vector. Simply put, grad(U) points in the direction of steepest incline of U, and has magnitude equal to the gradient of that incline. In 3D space (x,y,z) you calculate it as the vector [dU/dx, dU/dy, dU/dz].

    The implication is that the force on an object acts towards the steepest downward slope, just like a ball on a hill.

    In 1D space (in your example), it is simply dU/dx, and as long as you preserve the convention of +/- direction along the x-axis, you'll be fine. The resulting force is strictly a 1D vector (i.e. it has magnitude |F| and a direction (+/-)).
    Last edited: May 1, 2013
  8. May 2, 2013 #7
    U is scalar but it's derivative (gradient) is vector
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