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