OPENING QUESTIONS: What part of the test stumped you most? Why?
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WORDS O' THE DAY:
- Work (Force through displacement)
- Dot Product ("scalar product")
- Work ("Newton meter (Nm)" or "Joule (J) ")
- Hooke's Law (F = -kx)
- Work done by or done on a spring = 1/2kix2 - 1/2kfx2
- Kinetic energy of an object in motion = 1/2mv2
- Potenial energy in a compressed spring = 1/2kx2
CALENDAR:
WORK O' THE DAY:
Let's take a look at your tests --
Notice that the first problem *can* be solved in multiple ways using:
- energy considerations
- force considerations
- a combination of the 2
Let's look at energy considerations first:
A solid QA of the situation yields the following:
The coiled spring has potential energy. When the spring is released, ALL of that potential energy is transferred to the block. The block then experiences 100% of that energy (no friction is present) as kinetic energy. As the block rises up the ramp, the kinetic energy goes to gravitational potential energy:
Uspring = KEblock = Ug
So...
1/2kx2 = mgh
h = (1/2)kx2/mg
h = (1/2)(1400 n/m)(.101m)2 /(.200 kg)(9.81 m/s2)
= 3.6395 meters..
HOLD THE BUS... You're not quite done since h is the HEIGHT not the distance up the ramp.
We know the angle = 60 degrees however, and we just found h so using trig we can quickly see that sinθ = opp/hyp
so hyp = opp/sinθ
3.6395m/sin60 = 4.20 meters
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Using Force Considerations:
We know that when the spring releases that it imparts a force of kx on the block. We also know that force is being resisted by a component of gravity pushing the block down the ramp.
Forces up the ramp: kx
Forces down the ramp: mgsinθ
So....
∑Framp = maramp
substituting in we get:
kx - mgsinθ = ma
solving for a we get:
a = (kx - mgsinθ)/m
a = [((1400 n/m)(.101m) - (.200 kg)(9.81 m/s2)(sinθ)]/(m)
a = 698.504 m/s2
using kinematics we know that:
v2i - v2f = 2ax
where vf = 0 so...
solving for a we get:
v = sqrt[(2)(698.504 m/s2)(.101m)
v = 8.4 meters (sig figs)
Interesting conundrum here, any ideas why v is different in this situation? (it's ok that it is, really)
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using a combination of both forces and energy:
We know that the energy of the spring is 100% converted into energy of motion of the block.. so
1/2kx2 = 1/2mv2
solve for v
then use the kinetics equation we used above to find out how far up the ramp it goes
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So... take a few moments to digest this beastie.
Try to set aside your frustrations.... I'm sorry to say that a fact of life in this class is dealing with frustration.
Write me a paragraph or so on:
1) Does the path you take to solve the problem impact how efficiently you CAN solve the problem?
2) What did you learn from solving the problem that you didn't know before.
The rest of the problems will be discussed more generally in class.
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Now let's take a gander at the homework...
I asked you to take a gentle look at section 8.1.... how did that go?
Now let's review the homework...
Let's get them all up on the board but I'm particularly interested in your take on problem 8.7
Let's take some time to talk about that one!
Now let's recollect our FOUNDATION conservation of energy equation:
∆KE + ∆U = - Elost
KEf + Uf = KEi + Ui - Elost
What does THAT mean?
Basically that formula shows us that energy is ALWAYS conserved... doesn't matter which, when, why or how. It also takes into account that sometimes energy is lost due to the work done by friction (for example see eq. 8.14)
Notice how elegantly this equation works with/replaces the myriad of equations the author throws at us:
For example: rewriting that equation using \\\\\
HOMEWORK (est minimum time: 20 minutes of class, 1 hr at home):