Monday, October 27, 2014

Unit 2 Summary

Unit 2 Summary:

The second unit was about Newton's Second Law:

Newton's Second Law is:
 acceleration is directly proportional to force and inversely proportional to mass.
or:
a=F(1/m)

In the Newton's Second Law lab, we set out to prove this. 
We did so by changing mass and Force to see how it would affect acceleration.

In the first experiment:
-we added mass and found that when we added mass and measured the acceleration, the acceleration decreased. 
-Then we graphed the data and compared the equation of the line to Newton's Second Law. 

y=mx+b
a=F(1/m)

The slope of the line is always the factor we kept constant during the experiment. 
in this case, it was force(we kept force constant by keeping the weight of the hanger constant).

When we compared the slope(theoretical force), to the actual force we measured(the wight of the hanger), we found that they were often within 10 percent of each other, depending on which group you asked. If it was within 10 percent, we can say that it suggests Newton's Second Law is true.

In the second expierement, we repeated the same procedure, but kept mass constant and increased force. We did so by transferring the mass from the cart to the hanger. We found that when we increaced the force, the acceleration also increased.

Then we learned about Free Fall, that is moving without air resistance. 

In free fall, the only force acting on the object is gravity. The force of gravity in free fall is 10m/s^2. 
-When a ball falls, the speed increases by 10 m/s every second. 

-When a ball is thrown up, it decreases every second by 10m/s. 
-at the top of it's path, the acceleration is 0 and it is at equilibrium. 
-When it falls back down, It increases every second again. 
*when calculation for a ball that has been thrown up and has come back down, you can only use the formulas if you measure the path it took starting at 0m/s.

For instance, to find the total time in the air, you must use d=1/2at^2 only for the path from equilibrium(at the top) to the ground. Then you can double the time to get the answer. 

Throwing things up at an angle:

When you throw things up at an angle, you must use the vector as the hypotenuse of a triangle with the other sides being vertical velocity and horizontal velocity. You treat the vertical velocity the same as you would throwing straight up as it increases by 10m/s when falling and decreases by 10m/s when rising each second. The hypotenuse of the triangle is used to determine the actual velocity at each second. 
*the angle may help you determine of the triangle is a special triangle

Projectile Motion:

When measuring projectile motion, you must take into account the vertical and horizontal properties. You must use different equations for each one: 

vertical:
d=1/2at^2
v=at

horizontal:
v=d/t
d=vt

Use each of these to find whatever part of the problem you are asked for, but make sure you are using the right equation. Hight determines time so the time will be the same for both of them.
To find the actual velocity at any given second, you create a triangle with the speed on each side, both horizontal and vertical, and the hypotenuse represents the actual velocity.

Falling Through the Air(Falling with Air resistance):
When falling through the air, you can measure the net force of the object with this equation:

Fnet=Fweight- Fair. 
Two things can affect the force of air resistance
-speed
-surface area

When you first jump off, your speed begins to increase, this increases the force of air resistance and you acceleration starts to decrease, as does your Fnet(see equation with increasing Fair). This does not mean velocity starts to decrease as well, in fact it continues increasing. When the object reaches terminal velocity, it is at equilibrium.

But, if the object were say a person skydiving, and that person reached equilibrium and then opened the parachute, the acceleration and Force would both decrease because the Fair increases because the surface area increased. Then the skydiver will reach a second terminal velocity, however, it is going at a slower speed than the first.
*The force of air resistance is the same for both terminal velocities.



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