Terminal Velocity and Free Fall!

On Friday's class we did our fun Friday quiz, and then proceeded to learn about free fall and terminal velocity.

Everything is subjected to the force of gravity, which is the force that pulls down objects during free fall!
During free fall, objects experience a force that slows them down known as air friction.
Free falling objects in air reach a state where they no longer can accelerate, or increase speed, from the force of gravity and the force of air resistance equalling each other. This is known as terminal velocity! As something is falling, for example a baseball falling from a pop fly, air resistane against the ball has a direction upwards and becomes so great that it equals the force of gravity pulling the baseball down. Once the two forces equal each other, the baseball cannot fall any faster. Mass, shape, size, and surface texture are all different factors that can change the value of terminal velocity for certain objects.




We discussed an example of a sky diver. They are free falling until they hit terminal velocity, and to slow them down they pull to let their parachute out. The air resistance then acts on the parachute, allowing them to slow down.

Continuation with Force and Friction!

For the last few days we have been doing experiments to learn about force and friction.
From these experiments we have learned:

·         Increasing the mass of the cart increases the friction and force.

·         The rougher the surface the greater the friction and force.

·         When an object is pulled up a slant there is a greater amount of force due to gravity pulling the object down.

·         When an object is pulled down a slant there is less force and friction because gravity is pulling the object down the slant.

·         Force acts through the centre of the object. Meaning that no matter what the surface area is there will be the same amount of friction.  

After talking as a class about our results with the experiments we did an example with force.

A 4.0 kg block of wood is sliding on a horizontal table top. What force must be applied to the mass to give it an acceleration of 0.50m/s²? The coefficient of kinetic friction is 0.30.

m= 4.0kg                F_net=ma                                                    F_n=mg
a= 0.50m/s²                              =(4.0kg)(0.50m/s²)                              = (4.0kg)(9.8m/s²)
μ_k= 0.30                        = 2.0N                                                    = 39.2N
F_a=?
F_net=?                                                                  F_f= μF_n                                                                 
F_f=?                                                                                                    = (0.30)(39.2N)                            
F_n=?                                                                        =11.76N 

                                               F_net= F_a + F_f
                           2.0N[forward]= F_a + (-11.76 [forward])

                                           F_a= 2.0N[forward] + 11.76N [forward]

                                             F_a= 14N [forward]

After the example we did questions from the text book on page 169, #1-4. Lastly, Mr. Banow also mentioned about a Friday quiz.

Kenzie is next!

Finding the Force of Friction

In today's class we went over yesterday's blog post and then proceeded to come up with various experiments that we could do to measure the limiting static friction of a cart. Limiting static friction is the maximum value of the force of friction which occurs just before the object begins to move. We came up with various different experiments to find the limiting static friction:
- Using different masses on the cart.
- Testing the cart on different surfaces.
- Pulling the cart up an incline.
- Pulling the wooden block down an incline.
We also tested the frictional force of two separate sides of a box to determine whether surface area affects the force of friction. In addition, we were to determine the coefficient of static friction for the green cart and the black cart and compare the two values. The values of limiting static friction were very low for the cart, increasing as weight or angle increased, and also increasing on rougher surfaces.

The sled would have low limiting static friction due to the smooth texture of the sled and it's light weight.
Casey is next!

Friction Forces Unite!

Today (A.K.A. yesterday) we started off class by checking over Friday's quiz, and we reviewed the lesson from Friday.

We reviewed the formula:  FF = Fn*µ , that tells us that the force of friction is equal to the normal force (represented by the object's weight) multiplied by the friction coefficient.

We also reviewed that the coefficient for static friction is greater than the coefficient for kinetic friction.

We then took some notes, and learned that friction only exists when a force is being applied We also learned that forces act through the center of the object, so friction has nothing to do with the surface area.

Mr. Banow then did a demo where he attached a force sensor to a wooden object, to show us the idea of limiting static friction. The force peaked just before the object moved, then leveled out as it moved.

The last thing we did was some examples of the Friction Force Formula, fortunately forementioned for frequent and fervorous figuring of Friction Forces (holy alliteration, Batman!). One of these examples was the following question:

The total weight of Physics Boy (on skis) is 500N. If it takes 35N for Calculus Man to start Physics Boy moving (so he can help somebody find the slope of a tangent on a specific point), determine the coefficient of static friction.

The formula is: FF = Fn*µ

• We are trying to find µ.
• FF is 35N, because this is the Limiting Static Friction 
• Fn is 500N, because the normal force is the same as the object's weight.

So we substitute and solve:

                           

35N = 500N*µ

35N/500N = µ

µ = 0.07

The friction coefficient between Physics Boy's skis and the snow is 0.07

There are no units, because  Newtons/Newtons = 1, so they cancel.

Nicole, you're next!

Friday Quiz + Friction

On Friday's class Mr. Banow did some quick review with the class before handing out our quiz on GRAVITATIONAL FORCES.... After we had finished that the class proceeded with notes on friction.

We learned that there are many different types of friction such as:

Static Friction- The friction that tends to prevent a stationary object from moving

Limiting Static Friction- The maximum value of the force of friction before the stationary object moves

Kinetic Friction- the force of friction that opposes the motion.


We also wrote down the Force of  Friction Formula

Fn= Normal force
Ff= Force of friction
μ= Coefficient of friction


Next up will be Brady if he shows up.

Law of Universal Gravitation

We started Thursday's class by finishing correcting assignment six.  We then reviewed Wednesday's class by going over the Law of Universal Gravitation by taking notes on the key points. These notes included the following:

• The forces of attraction are inversely proportional to the square of the distance between the two objects. I.e. if the distance doubles, the force will be one quarter the original amount.
• No value of distance would eliminate the force fully.
• Remember 'g' is an extremely small number. 

Later, we went through examples followed by the assignment #1-4 on page 225.

Ex. 1 Calculate the force of attraction between two bodies that are placed 6.0cm apart and have a mass of 345kg and 430kg respectively.

M1= 345kg               F=G(M1)(M2)           F= 6.67x10^-11  Nm^2 [345kg(430kg)]

M2= 430kg                          d^2                                               Kg^2        0.06m^2
d= 6.0cm ( 0.06m)
F=?                                                                 F= 2.7x10^-3 N


Next is Brady!

The Exciting World of Gravitational Forces

In yesterdays class we started learning several topics including; Gravitational Field Strength, Mass and Weight, and the Law of Universal Gravitation.

Gravitational Field Strenght (GFS)

-the force of gravity is not the same everywhere and is defined as the amount of force acting on a mass of 1Kg. The units used is N/kg

-Fg=mg; where g is the gravitational field strength.

-Example: The force of gravity on a 300.0kg spacecraft on the moon is 489N. What is the GFS on the moon?

F=489N                                      F=mg therefore; g=F/m

m=300.0kg                                                            g= 489N/300.0kg

g=?                                                                        g=1.63N/kg

Mass and Weight

-mass is defined as the amount of matter in an object and is measured in kilograms while weight is described as the force of gravity on an object and is measured in Newtons and is a vector quantity.

-We learned that the mass of an object is not affected by its location. An object with a mass of 50kg on earth would have the same mass in space.  Only the weight is affected by the location of the object.

Law of Universal Gravitation

-To finish off the class we started learning about Newton's Law of Universal Gravitation.  Newton found that every particle in the universe attracts every other object.

   

 m= mass

d= distance between center of objects

F= equal and opposite forces of gravity

Indianna is next.

More Newtons Third Law stuff

Today in class we went over part c of the Newtons Third Law example we began the other day. Afterwards we did similar questions on page 149, 1-3a in the text book.  Once we were done, we began another assignment due tomorrow, or I guess this present class.  We also got our quizzes back.


part c) find the force exerted by toboggan one on toboggan two,

Fnet=ma
       =(55 kg)(2.2 m/s squared [E])
       =121 N [E]

We take the force of 22 N [W] from b and plug  it into another equation to find the total force.

Fnet= Fa + Fa
 121= -22 + Fa
   Fa= 143 N [E]


Next is Kurtis

Continuation of the "Newton's Third Law" Lab

We started today's class by reviewing questions regarding the “Newton’s Third Law” lab. We completed this lab by doing the Extension Experiments. We then proceeded to watch videos on inertia and Newton’s Third Law.  We finished off the class by doing a word problem on Newton’s Third Law, as shown below.

ex. Three toboggans are connected by a rope. The first toboggan has a mass of 75 kg, the second has a mass of 55 kg, and the third has a mass of 10.0 kg. If the force exerted on the first toboggan is 310 N and the surfaces are considered to be frictionless. Find:

a. the acceleration of the toboggans

         F_net   = 310 N [E]                                                                       
         m     = 10 kg + 55 kg + 75 kg = 140 kg
         a      = ?

b. the force exerted by toboggan two on toboggan three

F_net3=m₃a₃

F_net3= 10.0 kg (2.2 m/s² [E])

F_net3 = 22 N [E]

 

 Krystle Marie Anne Hanrahan is next.

                 

Newton's Third Law: "To every reaction there is an equal and opposite reaction."

During today's  class we worked on the "Newton's third law" lab. We had to answer preliminary questions and make predictions before collecting our data. We measured the force of two objects moving away from each other with an applied force which was recorded on the Lab Quest. This showed us that if force is only acting on one of the objects it will act on both. We finished the analysis questions on the lab and were about to begin the Extension questions when the bell rang!
 
Next is Tessi

MORE NEWTONS LAWS!

        Yesterday we continued to look at Newtons Second Law, "  "An object accelerates in the direction of the force acting upon it." We furthered our learing from looking at simple forces acting upon an object such as ( up, down, forward, back) to more complex forces acting upon an object such as forward35°up. We looked at a few examples as a class, followed by text book questions #1,3, pg. 145.

Example:
                A boy pulls his wagon, a mass of 15kg, giving is a horizonal accelleration of 1.0m/s². If the wagons handle makes and angle at 35° with the ground as he is pulling it, there is a frictional force of 3.0 N opposing to the wagons motions, with that what is the force he is pulling the handle of the wagon?

M=15kg
A= 1.0m/s² ( forward)
Ff=3.0 N ( backward)
Fa= ?

Fnet= mass X Acceleration        (15.0kg){(1.0m/s² ( forward)}= 15 N Forward... + 3 N= 18 N Forward

Cos35 °= 18 N     =          18 N    =  22 N ( Forward35°Up)

                 Fa                   Cos35°

We then continued on with Newtons Third Law, " To every action there is an equal and opposite reaction." We began our new Lab. We started to get the Labquests ready by calibrating the force sensors. We also got our string and mass. Now we can begin our labs today effeciently.

A Continuation of Newton's Second Law

        Today in class, we continued working with Newton's Second Law. We did a couple practice problems on the board, followed by questions 1-4 on page 140. We then attempted to learn about the effect that vectors have on these problems, but were sadly interrupted by the bell.

• Newton's second law states "An object accelerates in the direction of the force acting on it."
• The net force acting on an object can be calculated by multiplying the mass of an object by its acceleration.

Ex:
              A car has a mass of 2300kg. When starting forward from rest, the motor can exert a force of 4100 N, which pushes the car forward. How long does it take the car to get up to a speed of 35m/s?

M= 2300kg                   Fnet =MA                           A= 4100 N
Fnet = 4100 N               Fnet =A                                    2300kg
A=?                                 M
Vf = 35 m/s                                                              A = 1.783 m/s/s
Vi = 0
T = ?                               T= Vf-Vi            T =     35m/s              T = 20. s
                                                 A                      1.783m/s/s


Dylann is up next

Newton's Second Law = Excitement at its finest

          On Friday, we continued our lab from the previous days, reviewing the project, finishing up the Analysis questions and doing the Extensions portion of the assignment.

          In the Analysis section, using brain power we were able to determine that the slope of a force vs. acceleration graph has the units kilograms and therefore represents the mass of the object. 

If m= mass x= acceleration y= force and b= 0 we can determine Newtons Second Law:

y= mx+b

Force= Mass(Acceleration) + 0

Force= Mass x Acceleration

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Newton's Second Law

          In the Extensions portion of the lab we attached weights with an unknown mass to our cart and  moved the cart back and forth to generate a force vs. acceleration graph, which we would use to determine the total mass of our object. After much observation and a mishap with a broken sensor we were able to deduce that finding the slope of a force vs. acceleration graph is an accurate way to determine the mass of an object.  

          After our lab was completed we clarified and began taking notes on the newly learned Newton's Second Law. We found that: 

• If an unbalanced force acts upon an object then the object accelerates in the direction of the force.
• The smaller the mass, the more acceleration will occur.
• The larger the mass, the less acceleration will occur.

We also looked at a Force, Mass, Acceleration triangle:

Which was then followed by some examples:

Ex. What is the acceleration of a 1600Kg race car if the net force acting upon the car was 4500N [forward]?

Fnet = 4500 N [forward]       Fnet = ma         a = 4500 N [forward] / 1600Kg

m = 1600Kg                          a = Fnet /m       a = 2.8 m/s squared [forward]

a = ?

Newtons Second Law in Action:

http://www.youtube.com/watch?v=MXgnIP4rMoI

Brittany is next :)