Where does the energy go?

...after you drop an "object"?

  • You lift a book (GravE=$mgh$).
  • You drop the book.
  • As it's dropping, it's height, $h$ is going down,
  • so it's GravE is going down.
  • What happened to that energy on the way down?
  • Is that energy still available in some form that it could change the environment?

    A heavy enough book resting on a tack has enough gravitational energy to do work on the tack and push it in to the floor.

    But there are other ways to do work, for example, throwing a book at a tack...

    Isn't the book in this case also "doing work on the tack"?

    What measurable characteristics of the moving book is this related to if it's an energy??

    Energy of motion

    Also worth fearing for its destructive power!

    According to the hypothesis of energy conservation, the book's gravitational energy is not lost as it drops, but is converted into energy of motion: "kinetic energy".

    As one kind of energy (gravitational) was reduced, another kind (kinetic) increased.

    Getting quantitative about kinetic energy

    In your experiments with dropping balls, you have already gathered enough data to figure out how kinetic energy on velocity if we assume that energy conservation is true:

    Here are some candidate formulas for kinetic energy ($k$ is a proportionality constant):

    • K.E. = $k\cdot m\cdot v$
    • K.E. = $k\cdot m \cdot v^2$
    • K.E. = $k\cdot m \cdot \sqrt{v}$
    • K.E. = $k\cdot m / v$
    • K.E. = $k\cdot m / \sqrt{v}$
    • K.E. = $k\cdot m / v^2$

    If K.E. increases as an object drops,

  • Let's measure $y$ as the distance below the point at which an object was dropped, so that it's a positive number. But remember, because $y$ is pointed down, the ball is actually *losing* energy as $y$ increases.
  • Measure $v$ to be positive in the downward direction.

So energy conservation says: $$\begineq \text{GravE lost} &=& \text{Kinetic E gained}\\ mgy &=&KE(y) \endeq $$

We need to think about how we expect the function $v(y)$ to behave. We'll consider the 3 possibilities:

  1. K.E.=$kmv^2$
  2. K.E.=$kmv$
  3. K.E.=$km\sqrt v$

Now go ahead...

  1. For each possibility, set KE=mgy
  2. Use algebra to solve for v
  3. Make a sketch of roughly how each $v(y)$ should behave (you can use Wolfram Alpha to try out different functions).

Expectations and outcomes

So if the gravitational energy that the dropping book loses as it drops a distance $y$ below its starting point is $mgy$,

...and that energy shows up as a gain in kinetic energy, $$\begineq \text{KE gained} &=& \text{GravE lost}\\ \text{KE gained} &=& mgy \endeq $$

Then here are our 3 scenarios for different K.E. expressions:

KE=$kmv^2$KE=$kmv$KE=$km\sqrt{v}$
$\ \ \ kmv = mgy$
solving for $v$:
$\ \ \ v= \frac{g}{k}y$
Plots of $v(y)$ vs $y$...

Here's a spreadsheet with data from a video of a dropping ball which we'll use to decide...