F = MA 2012 Exam # 19, 20 (Using Graphs)

[SignaturePF]

Woops, it's the 2010 exam -- my bad.

Homework Statement

They have diagrams that I'm not sure how to display; so find them at the following link:
https://aapt.org/physicsteam/2010/upload/2010_Fma.pdf
These are #'s 19 and 20 and they refer to the PE graph in 18.

Homework Equations

Not sure, perhaps:
E_mec = K + U

The Attempt at a Solution

I'm not at all sure how to use the PE graph and a position vs time graph to find total energy or describe the motion. Also, I'm not sure how a potential energy vs position could give rise to position vs time. If anyone could direct me to the concept or link that I'm missing -- that'd be great.

 

[Simon Bridge]

Use the graphs separately.
What is the motion of the particle in the potential?
(You did Q18 OK didn't you? So you know the forces?)

i.e. Will the particle have a constant velocity or just hold it's position, or accelerate or what?

Then look at the position time-graphs. What kind of motion do each describe?
Which one matches up?

You can do it the other way around ... look at the first graph: what is the particle doing? What happens to it's position?
Now look at the potential vs position graph - does that make sense? (This is probably easier.)

For Q20 - it is pretty much the same ... what is the motion of the particle? Describe it in words - is it sitting still, moving at a constant velocity, oscillating between limits, what? At what kinetic energy would the particle have t have in order to have that motion?

 

[SignaturePF]

So for numeral I, it makes sense since net force is 0 at 15m so acceleration is zero and this can be represented by a constant position. TRUE
For numeral II, there should be a positive acceleration but the position vs time graph is constant; clearly FALSE.
Numeral III, the velocity is constant - thus the acceleration is zero. This agrees with the graph. TRUE
So 19 is both I and III. Thanks

Here's a thought for 20:
E_mec = K + U
E_mec = 1/2mv^2 + U
Let's make it easy on ourselves by finding where v = 0 (where dx/dt = slope = 0)
Two points of this are: x = 5 m and x = -5m
Here K = 0 so
E_mec = U
Tracking these points on the graph both yield:
-5J

Wow, those questions were actually kind of easy; thanks for your insight. I guess I should really pay more attention to what I am given.

 

[Simon Bridge]

Reading graphs is a skill that many people resist learning because it looks harder than it is. It is actually a good shortcut to solutions.

i.e. #20 the graph shows the particle oscillating sinusoidally between -5 and +5 ... to find the kinetic energy that does that, draw a horizontal line through the potential energy graph and see where it intersects.

All the difficult part is in knowing which lines to draw.
Anyway - well done.