Voltage

Hello everyone MrSiwWy here for today's physics scribe post, in which we actually covered quite a bit. The class began with Ms. K correcting the sheets that we had for homework over the weekend and also by reading pages 4 - 5 on the topics of Work and Kinetic Energy and Electric Potential in the handout booklet she gave to us sometime last week. This booklet's title on the cover is Grade 12 Physics: Gravitational Potential Energy.

For electric potential:

1. "Just as PE (potential energy) transforms to KE (kinetic energy) for a mass lifted against the gravitational field, the electric PE of an electric charge transforms to other forms of energy when it changes location in an electric field. When released, how does the KE acquired by each compare to the decrease in PE?"

The answer: equal (since both require work to move an object while still under the influence of the field whether it's gravitational or electric. Once work has been applied some potential energy has been transformed into kinetic energy in both cases.)

2. "Similarly, a force pushes the charge (called a test charge) closer to a charged sphere. The work done in moving the test charge is the product of the average force and the distance moved. W = FD. This work is equal to the PE of the test charge. If the test charge is released, it will be repelled and fly past the starting point. Its gain in KE at this point is equal to its decrease in PE." (note: terms in bold font are the answers since it was fill in the blanks.)

3. "complete the statements.
-Electric PE/charge has the special name Electric potential.
-Since it is measured in volts it is commonly called voltage."

4. "When a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of 2 J, its potential is = 1 V." (This is due to the fact that electric potential is the electric potential energy divided by the charge, as stated above in question 3)

5. "If a conductor connected to the terminal of a battery has a potential of 12 volts, then each coulomb of charge on the conductor has a PE of 12."

6. "If a charge of 1 C has a PE of 5000 J, its voltage is 5000 V."

7. "If a charge of 0.001 C has a PE of 5 J, its voltage is 5000 V."

8. "If a charge of 0.0001 C has a PE of 0.5 J, its voltage is 5000 V."

9. "If a rubber balloon is charged to 5000 V, and the quantity of charge on the balloon is 1 millionth coulomb (0.000001 C) then the PE of this charge is only 0.005 J.

10. "Some people get mixed up between force and pressure. Recall that pressure is force per area. Similarly, some people get mixed up between electric PE and voltage. According to this chapter, voltage is electric PE per unit charge."

Next, we moved on to a lab which consumed the majority of today's class time. The lab was essentially the assembly of a figure that embodied the main idea of increasing potential energy within an electric field (raising a charge AKA applying work to raise the charge against the electric field). The procedure and materials required for the lab are approximately similar to the following (since I don't have a green book with me right now):

Materials:

A Ruler
Clay
Steel balls (Any object can substitute for this, such as pennies in this case)
Masking tape

Procedure:

1. Insert a ruler within a molded ball of clay so that the ruler can stand upright without much independent mobility.

2. Take a 2 cm x 8 cm piece of tape and label it "3 V - 3 J/C" and repeat this process for three more equal sized pieces of tape with the labels "6 V - 6 J/C", "9 V - 9 J/C" and "12 V - 12 J/C."

3. Attach these pieces of tape to the ruler approximately at each 3 cm interval indicated on the ruler.

4. Once each piece of tape is attached to the ruler, attach the steel balls (or pennies) to each piece of tape, starting with 4 at the bottom on the 3V piece of tape, then with 3 on the 6V, and so on.

Results

Basically, we were given questions accompanying the lab as usual, and this was our overall analysis for the lab. Now essentially for the lab there was a structure which indicated that the amount of charges decreased as the voltage increased while you move farther up. If you think about the surface the structure is resting on in terms of the base of the electric field, you move the charges upwards and the voltage increases while the charges decrease. This also implicates an energy change, which can be calculated by multiplying the voltage with the amount of charges present at that level. So for this lab, the following data table was completed as the lab progressed:

Here are the corresponding analysis questions for the lab (accompanied by my answers):

"1. How much energy is required to lift each coulomb of charge from the tabletop to the 9-V level?"
The energy required to raise one coulomb of charge from the tabletop to the 9-V level is 9 J of energy. This is due to the fact that there are 2 charges at a voltage of 9 J/C accumulatively giving an energy value of 18 J, but since the question only inquires about the energy of one charge this 18 J is divided by 2 to give 9.

"2. What is the total potential energy stored in the 9-V level?"
(Note that this is basically explaining the calculations made in the table for each energy value)The total potential energy at a given location when you raise the object further up against the electric field can be found by using the electric potential of the object / charge. Since V = PE/q, and both q and V are known, solving for PE gives PE = V*q, PE = (2 C)*(9 J/C) = 18 J.

"3. The total energy of the charges in the 6-V level is not 6 J. Explain this"
The total energy of the charges in this level is not equivalent to 6 J since there are 3 charges present at a voltage of 6 J/C, rather than just one charge at a voltage of 6 J/C. Since there are 3 charges at a voltage of 6 J/C giving an overall energy for the level as 18 J.

"4. How much energy would be given off if the charges in the 9-V level fell to the 6-V level? Explain"
Now since the charges are dropping to a lower voltage at 6 J/C from 9 J/C, an energy equivalent to 3 joules would be emitted from the system since the charges are falling from 9 to 6. * not sure about this one since I can't remember what I actually put and am not in the mentality to reason out the answers momentarily.

Application question

"1. A 9-V battery is very small. A 12-V car battery is very big. Use your model to help explain why two 9-V batteries would not start your car."
Now if you think about it, the more charges that are present then the higher the energy value for that level. Now since these two cases involve voltages that are quite similar, there are obviously much more charges found in the large car battery than in the small 9-V battery, therefore allowing for quite a bit more energy available for use by applying the battery correctly.

Once all the labs were in, Ms. K went over the answers to the other sheet we were given on friday. The answers to the questions are as follows:

1. V = 31 V
2. a] PE = 3.6 x 10^-14 J
2. b] V = 180 V
3. 1.9 x 10 ⁷ m/s
4. 3.2 x 10 ^-9 J
5. a] w = 1.92 x 10 ^-18 J = 12 eV
5. b] E = 1.92 x 10 ^-18 J = 12 eV
5. c] V = 2.0 x 10 ⁶ m/s

Once she finished putting that up, we were let to work on our own on either finishing up the lab or to finish the sheets she put out in the front of class. Jeez, I wish I could've put more so that anyone who doesn't quite understand this unit yet or is yearning for some help with their tribulations concerning problems in this unit. Now as for our homework, the sheets that were up front for pick up are probably going to corrected tomorrowed, and by the way, the test is on friday. Good night everyone and the next scribe for tomorrow's class will be Anthony!

Scribe^2

Well, sorry class and sorry Ms. K for not posting yesterday, I rarely have access to the computer so it's difficult to scribe or sometimes to check the blog at all. Well, on with the scribe for monday, October 29, 2007.

Class began as Ms. K was setting up a movie on the television screened parked somewhat diagonal from the front desk. She also notified the class that the labs were to be handed in by the end of class, and to make sure that we all hand our labs in successfully. We then diverted our attention back towards the television set as Ms. K handed out a sheet containing several questions pertaining to the imminent movie. These questions were as follows:

1. Can an object move with constant sped and accelerate at the same time?
2. What is special about a vector originating from the center of a circle to a point on the circle?
3. State the two ways of finding the position of a point on a circle.
4. Briefly describe the Copernican universe.
5. Who revealed how the Copernican system worked?
6. Using the circle below, draw the acceleration and force vectors. Are both vectors in the same direction?
7. How is acceleration defined?
8. What is speed?
9. Can a body accelerate if speed is constant?
10. What does the rate of change of a vector mean?
11. Give three ways a vector can change.
12. Using the circles below, draw the radius and velocity vectors on circle 1, and then draw the velocity and acceleration vectors on circle 2. What is the relationship between both pairs?
13. What equation gives the relationship between acceleration, velocity, and radius?
14. If you superimpose vectors from circle 2 on the vectors from circle 1, what is the direction of the acceleration vector?
15. Is a force needed to accelerate a body in circular motion? What is the name of this force? What direction does it take?
16. If the moon is in orbit around the earth, what force is causing the moon to maintain its uniform circular motion? State the mathematical equation for this constant speed.

And then we continued on with the movie!

The movie was extremely fast paced in terms of the conveyed knowledge within each segment. When watching the movie many people in the class scrambled to assemble their answers to the questions she handed to us, and ipso facto Ms. K gave us an excess of 10 minutes to confer our results amongst our peers prior to reviewing the answers. The answers are as follows:

1. An object can in fact be constant in speed and accelerate simultaneously.
2. A vector inscribed within a circle drawn from the center to any point along the circle will always be equal in length as the point on the circle is altered. This is due to the fact that such vectors would be the radius of the circle.
3. They can be found by either:
- Cartesian co-ordinates (x,y)
- It's distance from the center of the cirlce and the angle from which this measurement is taken relative to the horizontal. (In other words, the position can be found by using trigonometry)
4. The coperinican theorem of planetary revolution consisted of the planets revolving the sun in a nearly uniform circular motion (each planet's moons, particularly Earth's, are described to have the same motion).
5. Nicolaus Copernicus' theory of planetary revolution was further explained by Sir Isaac Newton.
6.

The vectors are both directed towards the center of the circle ergo they share the same direction.
7. Acceleration is simply described as the change in speed or velocity.
8. Speed is the length or the magnitude of the velocity vector.
9. A body can indeed accelerate if it's speed is constant (as in the case of circular motion).
10. The rate of change of a vector indicates how fast a vector is changing.
11. Three ways that a vector can be altered are as follows:
- The size or magnitude of the vector.
- The direction of the vector.
- Both of the aforementioned methods.
12. The relationship between both pairs is that they are perpendicular to their respective vector quantities.
13. a = v² / R
14. The acceleration vector would be opposite the radius ,or rather towards the center of the circle.
15. Yes there is a force accompanying bodies in circular motion, i.e. Centripetal force, and this force acts towards the center of the circle.
16. The force responsible for such phenomena is known as gravity. The mathematical equation interrelating circular motion and gravity (though we are not required to know this formula quite yet) is given by:
V = √(Gm_e/R)

Once we concluded this portion of the class, Ms. K then showed some more short clips that pertained to circular motion, the first of which relied on a professor indicating that the magnitude of displacement is the radius of the circle, and a velocty vector consistent in magnitude is perpendicular to the displacement vector. He then projected the fact that acceleration is also constant in magnitude, but is directed towards the center of the circle. Then we watched some cool videos of children that were playing on a merry-go-round, displaying some of the indefinitely entertaining properties of circular motion. Once the movie sessions were done with for the class, Ms. K let us work amongst our peers on whatever we had to finish.

Some questions that we were given time to work on if not already finished over the weekend were the questions found on Centripetal Acceleration and Centripetal Force. And that concluded our class for monday. The scribe has already been determined for today, and that scribe is:
AICHELLE

dixi

Initiation of Review

Hi, I'm Chris (from third period physics) and I will be the scribe for today. I simply titled the post "initiation of review" since all we really covered today was a small review of some sample problems involving forces and free body diagrams. Throughout the class, Mrs. K attempted to detail each of the presented cases (found within a small booklet that was also given out today) sufficiently so that the entire class could comprehend the problems at hand. Though, I'm not so sure everyone in the class was quite as comfortable with these cases as she might have thought. In this post, I will outline each of the cases that were covered in today's class with hopefully proficient detail and explanation that the class might delve into the monstrous test on monday with some apt confidence and security.
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Case 1: Constant Acceleration Motion

"If an applied force of 100 N is opposed by friction μ = 0.15 on an object with a mass of 50 kg, calculate:
a) Ff
b) Fnet
c) a"

Now, this is probably one of the simplest dynamics question that we might encounter on the test on monday.
For part (a), the question asks for us to find Ff. Simple enough, we have been taught that to calculate Ff, the equation would be as follows:

Ff = μFn ; where Ff = friction, μ=coefficient of friction and Fn = normal force.

Now since we know that Fn = -Fg, since the normal force is equal and opposite to the weight of the object, we can then substitute Fg into the equation instead of Fn. We would do so because we know the mass of the object, we can then substitute m*g instead of Fg to determine the weight or normal force (since they are equal in linear dynamics), then we would just multiply it by the coefficient of friction to determine the frictional force. This can be shown algebraically as follows:

Ff = μFg = μmg = (0.15)(50 kg)(-9.81 N/kg) = -73.5 N (negative since it is moving left)

For part (b), we are looking for the net force in the free body diagram, and since we know that Fn and Fg are equal, we do not need to take them into consideration. We can also determine from the free body diagram, and also since there is acceleration occurring, is that the applied force is greater than the frictional force. So in order to determine the net force, we must add up the forces Fa and Ff. Since we know both, this portion becomes quite easy.

Fnet = Fa + Ff = (100 N) + (-73.5N) = 26.5 N (since it's positive, it's moving right)

For part (c), the accceleration can be determined from rearranging the equation Fnet = ma to solve for a:

a = Fnet / m = 26.5 N / 50kg = 0.53 m/s²
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Case Two: Force on Two Masses

"If the applied force is 200 N, m1 = 5 kg and m2 = 3 kg, and μ = 0.1, calculate
a) Fnet
b) a
c) F₂"

Now this question presented tribulations for the class as Mrs. K attempted to present how to approach this problem to the class.
For part (a), to calculate the net force on the objects, all we must do is add Ff and Fa together (since once again, Fn = Fg). Now, we already know Fa, but we then need to determine Ff. Since Ff = μmg, as I outlined earlier, we need to apply that once again but instead of having just one mass, we have two, so we must combine the two masses to get m. By this I mean that in this instant we treat the two masses as one mass by combining them into one. This will give us:

Fnet = Ff + Fa = μ(m₁+ m₂)g + Fa = (0.1)(5 kg + 3 kg)(-9.81 N/kg) + 200N = 192 N

Now for part (b), we are attempting to determine the acceleration of the two masses. All we must do for this is use the net force (since it is the total of all forces in the situation) and divide it by the total mass of the object, once again adding m₁ + m₂. Algebraically:

a = Fnet / (m₁+m₂) = 192 N / 8 kg = 24 m/s²

Part (c) requests that we now determine F₂, this usually presented trouble for some students so I will try my best to outline the process of calculating F₂. In order to calculate this force value, we must see that F₂ is caused by mass 2 and is actually driving m₂ forward, but is pushing back on m₁. This is a direct result of Newton's third law, that every action will have an equal and opposite reaction. We must also realize that this force is a ratio of the overall force being applied, in other words, it is a specific portion of the 200 N being applied to the overall object (m₁ and m₂ together). To determine this so-called ratio, we must divide the mass of m₂ by the total mass, since we're basically looking for how much of the total object is just m₂. Now, here's how the calculations should look like:

F2 = [m₂ / (m₁ + m₂)] F = (3/8) (200 N) = 75 N
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Case Three: Connected Masses

"If m₁ = 2 kg, m₂ = 4 kg, m₃ = 1 kg, and Fa = 10 N, calculate:
a) a
b) T₁
c) T₂"

To determine the acceleration, since there is no friction acting (and Fg = Fn, so they quantitatively cancel each other) Fnet = F. Also, in this case, we must incorporate all three masses into our calculations using m since F is indirectly acting (pulling) the entire connection of the masses together. Now, to calculate acceleration, we can use the formula F = ma rearranged to solve for acceleration:

a = F / (m₁ + m₂ + m₃) = 10N / 7 kg = 1.43 m/s²

To determine the first tension force (acting on m₁ and m₂), we can use the first free body diagram to carry out this calculation, seeing as it is more simple. Now, as we inspect the first diagram, we notice that the only force acting on the object IS T1, meaning that the net force in this case IS T₁. So, in order to calculate T₁, we can use the equation T₁ = m₁a.

T1 = m₁a = (2 kg)(1.43 N/kg) = 2.86 N

For T₂, we must use the third free body diagram to calculate it since we can use F as a constituent of the net force to determine T₂. This, in other words, means that we can use the net force (the total of the forces) to determine T₂ since F is known. The vector T₂ would be negative in this situation, since it is moving left, and F would be positive as it would be moving right. This doesn't always have to be the case, but I have designated left negative and right positive, or the force T₂ negative with respect to F. Since the net force is not known, but m₃ and a are both known, we can use those values in place of Fnet since Fnet = m₃a. This gives us:

T₂ --> F - T₂ = m3a --> T2 =-(F - m3a) --> T2 = -[10N - (1 kg)(1.43 N/kg)] = -8.57 N
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Cases Four and Five:Mass on an Incline

Now, for these cases (though we did not go over them in class, but were told to complete them) I will outline how to come about solving such questions. These inhabited a majority of the current unit and is quite likely to be found frequently throughout the upcoming test. Now, the following diagram will be used to describe both scenarios.

The normal force, Fn, will always act perpendicular to the surface. This meaning that if the mass is resting on a surface, the normal force exerted on the mass will form a 90^o with the surface. Cases which involve an object and an incline will always direct gravity downwards, regardless of the slope of the surface, the direction of the normal force, etc. Gravity is always pointed down (as shown in the diagram) in an inclined plane problem. Now, in order to conduct calculations, we must break the force of gravity, Fg, or weight (all the same thing), of the object into two components, a parallel component and a perpendicular component. These are basically just x and y components (or vertical and horizontal components) of the force of gravity, but since Fg is now acting as the hypotenuse pointed directly down, they are named according to their position in the diagram. By this I mean that these components are named relative to the surface, as with in the case of the normal force. These components are illustrated in the freebody diagram depcited above. To calculate these components, we visualize the triangle formed by breaking Fg into it's components as a right triangle, where Fg itself is the hypotenuse. The angle θ will be formed between Fg and F perpendicular, and from there on use basic trigonometric functions (particularly sinθ and cosθ) to determine either F perpendicular or F parallel. And if Fg is not given, but the mass is, then mg can always be used to substitute for Fg.

F parallel plays a specific role in an inclined plane. Fg does not act directly downward on the surface, but in two ways, thus why Fg was dissected and broken down into two components. F perpendicular simply acts equal and opposite to Fn, similar to how gravity does in linear scenarios. But F parallel acts as Fa, driving the object down the incline. The only force to oppose this force (with respect to what we have calculated as of yet) is friction. When friction comes into play, the process might seem more complex and intuitively complicated, though contrary to popular belief this might not be entirely true. As in the case of linear dynamics, Ff opposes Fa (in this case F parallel) and then determines what motion will occur. Here are three important tidbits to remember when regarding net force and Ff vs. Fa:

- If Fa is less than Ff, then the object is not moving.

- If Fa = Ff, then the object is moving with a constant velocity.

- If Fa > Ff, then the object is accelerating.

Continuing with the explanation of frictional force in an inclined plane is imminently complete. As I noted earlier, Ff opposes F parallel, but also, in the case of all situations regarding frictional forces there underlies a specific coefficient of friction (the ratio of friction over the normal force). Using either of these two facts leads to the calculation of the frictional force acting on an object on an inclined plane. To use these two ways, either subtract the F parallel force from the net force to get the frictional force, or manipulate/directly use (depending on for what purpose) the equation Ff = μFn.
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Now, I hope many of you, especially those who were absent from today's class, have acquired more of a formally but yet simplified (to an extent) explanation of some of the main concepts and processes within this unit.

As our class finished covering the above through a somewhat less elaborate volubility of explanation, we were given two sheets entitled "C. Using Concepts" and "Grade 12 Physics: Dynamics Problems" to complete. Though I am not sure whether they are for homework in the sense that we will correct them or hand them in or not, I strongly encourage many to not only attempt and complete the questions, but to shamelessly inquire as necessary. These questions will not only provide more practice and a more definitive grasping of the subject at hand, but also a stronger habitual behaviour to approach problems that we might encounter on the upcoming test. As for the test, it will be on Monday, though some argue (can't remember who else besides Vincent) that the test should be on Tuesday. That is all for my scribe post, I know it was lengthy, albeit an obvious resultant of my instructional endeavor for the rest of the class(es). Hope this post can help anyone at all in this unit, have a great night and don't forget to do homework!

Oh yes, the scribe for tomorrow will be Anthony!