Hooke's Law

Today in physics we reviewed what we did on Friday which was total energy and how the amount of kinetic energy and the amount of potential energy are always equal to the starting total energy.

We then moved on to look a springs, more specifically, Hooke's Law which contains stuff about the spring constant and how far from equilibrium the sping is depressed or stretched.

Ex: A compressed spring that obeys Hooke's law has a potential energy of 18 J. If the spring constant is 400 N/m, find the distance by which the spring is compressed.

Ep = 18 J

k = 400N/m

x = ?

Ep=1/2kx^2

=to find x we manipulate the formula to get the root of 2ep/k

= the root of 36/400

=3/10 or .30m the spring compressed.

We also looked at finding the total energy and the maximum height of an object, in this case, a golf ball.

Ex: Determind the total mechanical enerygy of a 48g golf ball if it has a velocity of 25 m/s when it leaves the club face.

m = .048 kg

vi = 25 m/s

hi = 0 m

et = 1/2mv^2

=1/2(.048)(25^2)

=15J

b) If the golf ball goes in an arc and has a speed of 15 m/s at its maximum height, what will the maximum height be?

h = ?

Et = 15 J

m = .048 kg

v = 15m/s

Et=mgh+1/2mv^2

15J=(.048kg)(9.8m/s^2)(h)+1/2(.048)(15^2)

And to find h we manipulate the formula and eventually get that the height is 20m.

Landon is next.

Roller Coasters!

You will be completing an assignment that uses the Physics of Energy to explain how a roller coaster works.

For Monday I need you to look over the following list of possible websites to use or to select a video that shows a roller coaster in action.  Choose a video or a program to use.
https://docs.google.com/document/d/1Q6nmNhSqJ6Nn-JXg7HFrjm7m_Gdja076XPdRKqFSu3U/edit

If you are using an online game/animation for your roller coaster, on Monday you will be given time on the laptop in the Physics Lab to make a video of your roller coaster in action.

On Tuesday we will go to the computer lab for you to view your video and come up with the explanation that you will present to the class.

On Wednesday, we will present them!

The general idea behind the assignment is:

Create a roller coaster. Use Smart Recorder to record it or save it if you can.

Present your video to the class. Pause the video as you go to explain the physics behind the motion:

Draw on the screen with the SMARTBoard

• key ideas:
• height
• velocity
• kinetic energy
• gravitational potential energy
• gravity
• use given or arbitrary values for the height to explain why the roller coaster works

Potential Energy and the Law of Conservation of Energy

Today in a heated and somewhat violent Physics class, we learned about potential energy and the law of conservation of energy. We started off by finishing the questions on kinetic energy we did the day before on page 290 #1-4

Then we read pages 291 and 292 in the textbook and answered questions about it.

From the reading we learned some new definitions:

Potential Energy- Energy that is stored and capable of being transformed into other types of energy.

Law of Conservation of Energy- Energy is not created or destroyed in any interaction, but is merely transformed from one type of energy into another.

Mr.Banow used the textbook as per usual to show us that when he holds the book above the floor and the book is not moving it has potential energy.

After the reading we moved on to learn about the most common type of potential energy and the one we will use most often in Physics 30 and that is Gravitational Potential Energy.

Gravitational Potential Energy- the energy stored as a result of the vertical position (height) of an object.

The formula to calculate Gravitational Potential Energy is Eg=mgh where:

Eg is Gravitational potential energy (J)

m is mass

g is acceleration due to gravity

h is change in height relative to the reference point

Ex.1

A 25.0kg box is lifted from the floor to a ridiculously large desk that is a whopping 1.87m above the floor.

a) What is the gravitational potential energy relative to the floor?

m=25kg

h=1.87m

g=9.8m/s^2

Eg=?

Eg=mgh

=(25kg)(9.8m/s^2)(1.87m)

=458J

Jeren (soon to be four eyes) Tuchscherererererererererererer is next

Review of Collisions

We reviewed yesterday's lesson again. It was on collisions. So, either elastic collision or inelastic collision.

Elastic collision- there is no change in kinetic energy after the collision has occured.

Inelastic collision- where some energy is 'lost' when colliding objects are in contact.

Mr. Banow showed us a whicked example of inelastic collision using two steel balls and a piece of paper.

First, he put one of the balls in the middle of a roll of tape and put the piece of paper on top of the ball. Next, he took the second ball and dropped it so it would hit the ball below the piece of paper.

What happened you may ask?

Well, a little hole was burnt into the piece of paper as well as a 'clunk' sound when the balls hit was made.

The reason why the collision of the two steel balls was inelastic because their was energy lost in the form of heat and/or sound. Not all of the energy was being conserved as kinetic energy.

We then did questions 1-4 on page 290.

Ryan Fly is up next.

Kinetic Energy

For some reason by pictures are up there ^^^^ but their relevance will be explained later. We started off class by reviewing the questions we started yesterday on power. These questions also reviewed calculating work quite nicely as we discovered that work needs to be calculated in order to solve for power.

Next, we expanded on what we had previously learned about energy. We we introduced to some new concepts:

1. Kinetic energy- the energy of motion. Ex: a ball rolling. Formula: $Ek=1/2mv^2$ Kinetic energy is a scalar quantity :)
2. Potential energy- the energy of rest. Ex: a ball being held in the air.
3. Change in energy= (final kinetic energy- initial kinetic energy) or $1/2(mvf^2-mvi^2)$

We learned that work= change in energy. This concept is important because it can make problem solving a lot easier. For example, we were given a problem that gave us the variables mass, initial velocity, and final velocity and were asked to solve for work. Instead of calculating force, displacement, and acceleration to eventually find work, we were able to simply calculate change in energy using $1/2(mvf^2-mvi^2).

Finally, we reviewed and expanded on our knowledge of collisions. We learned how energy fit into each type and were introduced to one new type:

• elastic collision- the two bodies don't stick together after the collision. There is no change in kinetic energy. Ex: the bat hitting the ball
• inelastic collision- the two bodies stick together after collision, some energy is lost. The energy lost is usually in the form of sound, heat, or light. Ex: the cars stuck together
• completely inelastic collision- the two bodies stick together after the collision but no energy is lost. Ex: the spaceship stuck in the planet.

Paige is next, I think.

Energy and Power

Monday we reviewed a bit of what we did on wednesday, we covered calculating work from a graph and when work is done by pulling something on a angle. We went over how work is produced: the force must be in the same direction as displacement to have work done.

What we learned on Monday was that when work is done, energy is transferred from one object to another. If the energy of an object increases, the work done on it will be positive. If the energy of an objects decreases, (ex. slowly bringing an object to the floor.) the work done on it will be negative.

Time has no effect when calculating work, but it does matter when we calculate power.

Power is the rate at which work is done and the rate at which energy is used. We calculate power in watts (W).... 1 watt is 1 J/s. Since watts are really small, we use kilowatts too (kW). To describe engines, the term horsepower is used. 1 horsepower equals 746 watts.

Things to Remember:

• P=W/t
• work is the product of force and the objects displacement (same direction)
• the less amount of time doing work, the more power being used (same displacement)
• to calculate kilowatt hours, multiply number of kilowatts by the number of hours used
• work is measured in joules (J)

Ex. 2. (in notes)

How much power is developed in lifting 82 kg of concrete to a height of 21 meters in 12 seconds?

m= 82kg
h= 21m (distance)
t= 12s
P=?
W=?
F=?

F= mg
F= 82kg x 9.80 m/s²
F= 803.6 N

W= Fd
W= 803.6N x 21m
W= 16 875.6J

P= W/t
P= 16 875.6J/12s
P= 1400W

Conservation of Momentum

The class began with the perpetual note-taking on the subject of Momentum; included in the aforementioned topic was the extended subtopic of the Conservation of Momentum. Various types of collision and explosion are included in the conservation of momentum—they are:

Explosion from Rest- One object splitting into multiple; given by the formula M1V1=-M2V2( and various manipulations of it).

Explosion from Motion-An object in motion, therefore with a initial velocity NOT of zero, splits into multiple objects; given by the formula (M1+M2)Vi=M1Vf1+M2Vf2

Elastic Collision-Two separate objects (with two separate masses and velocities) collide and rebound; given by the formula M1V1i+M2V2i=M1V1f+M2V2f

Inelastic Collision- Two objects colliding and becoming one; given by the formula M1V1i+M2V2i=(M1+M2)Vf

NOTE: All of these formulas can be derived remembering the Law of Conservation of Momentum, that is, initial momentum of an isolated system is equal to its final momentum, or the total momentum of an isolated system does not change.

Ex.)

A 95kg Unger collides while running 5.0m/s into a static Boss with a mass of 76kg. If Unger rebounds with a velocity of 3.0m/s, what is the velocity of the object he collided with?

M1V1i+M2V2i=M1V1f+M2V2f

M1=95kg

M2=76kg

V1i=5.0m/s

V2i=0.0m/s

V1f=3.0m/s

V2f=?

((95kg)(5m/s)-(95kg)(3.0m/s))/76kg=V2f=2.5m/s

We then continued with textbook questions on page 149, #1-5.

Whoever is left is next.