Energy of motion?

Here were our 3 scenarios for different K.E. expressions:

Bring in balls on rolling track too.

KE=$kmv^2$KE=$kmv$KE=$km\sqrt{v}$
$$\begineq kmv^2 &=& mgy\\ v^2&=& gy/k\\ v&\propto&\sqrt{y}\endeq$$> $$\begineq kmv &=& mgy\\ v&=& \frac{g}{k}y\\ \endeq$$ $$\begineq km\sqrt{v} &=& mgy\\ \sqrt v&=& gy/k\\ v&\propto&y^2\endeq$$
Plots of $v(y)$ vs $y$...



So, the graph that matched our data was the one where $KE\propto mv^2$, and actually, the best fit was to...

$$\text{KineticE}=\frac 12 mv^2.$$

Solving problems with KineticE and GravE

Assuming that we don't have to worry about friction, there are a number of problems we can do by using the idea of energy conservation for a single object:
$$ \text{GravE lost}\\ \text{KE gained}$$

1.) A student on the top floor of the Ad building drops a book out of the window. How fast is it moving when it hits the sidewalk?

2.) A roller coaster that has no engine of its own is pulled up to a height of 100 m. If it travels on a (nearly) frictionless track, how fast is it moving when it has dropped 50 m below the place where it was released?

These problems would be much harder (particularly the roller coaster) if we had to (for each second of time of the motion...)

  1. Figure out the force $f$ at the starting position,
  2. Apply $a=f/m$ to find the acceleration,
  3. Use $\Delta v=a \Delta t$ to figure out how much its speed changed in the last second,
  4. Use $\Delta d = v \Delta t$ to figure out how far it moved in the last second,
  5. Now, at the new position, re-do the calculations for the next second...