April 6th, 2011

Key: x^2 = x squared; Vo = V nott (aka initial velocity); * = answer

Well, we have a quiz tomorrow and we reviewed for that. I just realized that I'm posting a little late for you to look at this and learn more from it. Woops. Not that I'm going to be much help anyways. Anywho, let's get to it.

We basically went over a worksheet that looks strikingly similar to this one that I found on google images on the first try:
a. So we've gone over this twenty million times, but in case your memory is a bit foggy, the marbles motion is parabolic because the ball's horizontal velocity, due to negligible friction and inertia, will be constant. However, after it leaves the table, since the only vertical force acting on it is gravity, it will accelerate towards the floor at 9.8m/s/s.

b.
While it is on the rail, we know that the ball is moving horizontally at a constant velocity of 10 m/s. This means no unbalanced forces, so the force diagram will have two arrows: an up arrow for the normal force the rail is exerting on the ball, and a down arrow for gravity. These two arrows are equal in length because I'm pretty sure the ball isn't accelerating upwards or downwards anytime soon.
When it leaves the table, since we are ignoring air resistance and inertia is not a force, there will only be one down arrow on your force diagram for gravity. (This computer doesn't have paint or anything similar to it, so no pretty drawings today :( ) It's horizontal movement is still constant because an outside force hasn't prevented the ball's inertia to keep it moving forward. As for vertical, it is accelerating downward.

c. Once the ball leaves the table, calculate how long it will take for the ball to hit the floor.
Okay, so here we use the equation: change in position = 1/2at^2+Vot
We are only talking about vertical movement. We know that the table is 1.5m tall, it's acceleration is 9.8 (gravity), and that it's initial velocity is 0, because the ball wasn't moving vertically at all before it left the table. So we plug this information in.

• 1.5m = 1/2(9.8m/s/s)t^2 + 0t (Then divide by 4.9 to get t^2 alone.)
• t^2 = 3.06 (Find the square root of both sides)
  
• t = .55s*

d. In the time you have calculated in part c, how far will the ball travel horizontally before hitting the floor?
Alright, so now we're talking about horizontal movement, but now we have t, which we found in the last problem. We also know that initial velocity is 10m/s, because that was the velocity left the table, and will always be the horizontal velocity until an outside force acts on it. Also, since velocity is constant, we know that acceleration is 0. So plug that in to our equation again:

• changeX=1/2(0)(3.06) + 10*.55s
  
• X = 5.5m*

e. Suppose the table was doubled in height to 3.0m.
i. Determine the horizontal range of the marble as it falls to the floor.
ii. What effect does doubling the height have on the range of the marble?
iii. What other factors affect the range of the sphere?

i. Okay, this is basically combining c and d, except we have a new vertical height. We need to find how long it takes for the marble to hit the floor so that we can plug that time into another equation to see where, in regards to horizontal direction, the marble will be.
So, once again, thinking about vertical movement, we plug in our facts into the equation. (See c for more explanation if necessary.)

• 3m = 1/2(9.8)t^2 = 0t
• t^2 = .612
• t = .782s

So take that time and plug it into our horizontal direction equation

• changeX = 1/2(0)t^2 + 10*.782
• x = 7.82m*

ii. As we can see, doubling the height increased the horizontal range.
iii. Other factors that could affect the range of the sphere:

• Horizontal velocity, because, given the same amount of time it takes to hit the floor, the marble would move faster to go farther or slower to come up short.
• Increased friction or air resistance, since it would slow the ball down and change its range
• *** One thing that would NOT change it's horizontal range would be the weight or mass of the ball, as long as the velocity was still the same because, as we've seen and tested ourselves, if you drop two objects varying in weight at the same height, they will always land at the same time, meaning none of of the variables have changed, producing the same horizontal range.

So yep. That's all. I would do more, like go to the back of the worksheet, but I'm lazy sick.

-Norman


SCIENCE!!!!

Angeline's Review, April 5

Hey guys! So today we continued discussing horizontally launched projectiles. Remember that if a projectile is horizontally launched at the same time an object is dropped, these two objects will hit the ground at the same time. This is because no matter what, gravity is still acting on the objects with an acceleration of 9.8 m/s/s. Even though the projectile is also moving forward, gravity is still acting on it. I'll post the answers to the problems we did in class later. I thought that this was at least a start :). Sorry the print above is really small. If you can't read it, the answer is .63 m.

Wednesday, March 23, 2011

Today, we reviewed. That's it.

Am I done? Great. I've got other things to do.

We went over the, ahem...

Unbalanced force Particle Model:
Newton's Second Law Review Problems

...worksheet.
Using a new, groundbreaking technique where the students put the answers on the blackboard, we answered problems 3, 4, and 5 in class:

3.

So, our baseball player can throw at 100 miles per hour (mi/hr) by extending his arm a distance of 1.5 meters (m). If his ball weighs 0.15 kilograms (kg), what is the unbalanced Force he must exert on the ball?

We will use this equation to find unbalanced Forces:



I hope you enjoy my childish scrawl.

M meaning our mass (which in this case is 0.15 kg) and A meaning our acceleration.

But, we need to find our A, which we measure in m/s/s.

However, you'll notice that we have an issue with measurements. We'll use measurements in m and seconds (s), but the question starts out with a measurement in mi/hr.

Homey don't play that.


None of you will really get this reference.

Which brings me to why I left Chemistry class: Dimensional Analysis.
This horrid mathematical device was the general tool in converting the terrible mi/hr to our darling meters per second (m/s).


Just in case you've forgotten how to do this, I'll try to explain this in a way that won't make me have another panic attack.
So, we set up 100 mi/hr in the upper left. According to our sheet, 1 mi is about 1600 m. So, we set the ratio (1600 m/1 mi) up next to it. Mi goes on the bottom because it's sort of like multiplying fractions, and the old measurements are supposed to cancel. We need to get rid of that measurement to convert our given.
So, we can successfully convert our mi to m by multiplying 100 and 1600. But, now we need s.
Because hr is below on the left, it has be be above on the right, again so it can cancel out. below it goes 60 minutes (min). The ratio is (1 hr/60 min). But, we want s, so we add another ratio (1 min/60 s), so we can convert our measurement to seconds.

I might have made this more confusing.



Multiply 100 by 1600. Then divide it by 60 twice. You about 44.4 m/s.

We'll use that to find the acceleration with this equation:


Velocity (V) squared equals Initial Velocity (Vo) squared plus 2 times A times change in position (ΔX).
Our V is 44.4 m/s. Our Vo is assumed to be zero. Our ΔX is 1.5 m minus zero m, which is 1.5 m.


So, with our acceleration, now we go back to our first equation and plug in our numbers:


98.6 Newtons (N) are exerted on the ball to throw it at 100 mi/hr.

4.

After staring from rest, a 0.50 kg ball falls 30 m. It has contact with the ground for 2 milliseconds (.002 s) before bouncing back up and reaching 20 m. We need to find the force exerted on the ball as it hit the ground.

Because we have a mass again, we'll use our Funbal = MA equation.
We need to find the acceleration as it hit the ground, which would be between our two velocities.


Oh my God, NO!!

Yes, two velocities. There is when it is falling initially and after it bounces on the ground. Luckily, we can use the same equation from before:



Or, at least I hope it is.

So, I put 9.8 in for A.

"But then, we have acceleration! Why do we need to figure out what it is if we have it?"

Well, very bold student, we have acceleration, while the ball is in the air, while the acceleration due to the force of gravity acts upon it, not when it hits the ground.

So, we get 24.25 m/s from the first one and 19.8 m/s from the second. We take these two and stick them into another equation:


Good ol' change in Velocity (ΔX) over change in Time (ΔT).
If we consider the bounce upwards to be "moving forward," then the ball dropping would be considered "moving backwards." That's why the velocity is negative now.
Doing this equation leads us to the ungodly acceleration of 22,025 m/s/s.

Plugging our numbers back into our Funbal equation:


According to maths, our answer is 11,012.5 N.

5.

Another sled problem. That's all I should have to say.

There's a pair of kids in a sled. One of them is having too much fun. With the sled, their mass is 100 kg. A man with no muscle mass is pulling the sled by way of a rope at a 25° angle. The tension on the rope is 120 N. The friction force is 15 N. We have to find the acceleration.

We can use the same formula we have been using, but this time we have to change it to solve for A instead:


We have a mass (100 kg), so we just need our unbalanced Force. We make a Force diagram:


X is our force forward, which we need to find for the unbalanced force.
The reason the normal force doesn't equal the force of gravity is because of that 50 N over on the right. The rope is pulling the sled to the right and up, to the ground isn't pushing up against the sled as much as it would is it were at rest.

To find X, we use Trigonometry:


We subtract 15 N from 108.76 N and get 93.76 N, our unbalanced force.
We plug it into our adapted Funbal equation:


The acceleration is .94 m/s/s, the child was having way too much fun, and this entry is finished.

IM FINALLY BLOGGING!

okay so sorry its taken so long. Anyways..

3/16

Mr. Dickie put multiple practice problems on the board that I payed very close attention to just for you guys (and my grade:/)

#1 the guy up there^ pulling the car (picture it with wheels). mass of car= 1.5kg; person pulling with a force of 10N. find acceleration.

a= Fun/mass SO a=10N/1.5kg and you get 6.67 m/s^2 as the acceleration

#2 okedoke so the picture with the finger pushing the block is next. needed the acceleration. Given: finger pushing with 10N; mass of block=1.5kg. oh and theres M (mew thing)=0.15

first we need to find friction. so we do Force of Friction= (Mlittlek)(Normal Force)

=(0.15)(15N) to get frictional force of 2.25 N

Next we need the acceleration so we take the Force of the pushing (10N) and subtract the force of friction (2.25) to get 7.75N. Then divide that by the mass (1.5kg) and we get 5.17m/s^2 as the acceleration.

#3 so there is a pulley system. the car on top is 1 kg and the weight pulling it is 0.2kg. to find the acceleration we take the Fun/mass.

a=2N/1.2kg equalssssss 1.67 m/s^2

#4 ELEVATOR PROBLEMMM

So Mr. Dickie, with a mass of 78kg is in an elevator that accelerates at 3m/s^2 when it starts moving up. **this time we are given the acceleration and want the Fun!

Fun= (a)(m)

Fun=(3m/s^2)(78kg) = 234N

You may think you are done but wait! you have to add 234 and 780 to get your grand total of 1014N as your normal force !

#5 Mr Dickie is in an elevator moving up at 2.7m/s^2. the elevator slows down and a rate of 3m/s^2. oh and he still has the same mass of 78kg.

So to do this problem we dont even need the 2.7 m/s/s part. ignore that. We just find the force unbalanced again which is still 234N and this time instead of adding that to 780 we SUBTRACT it from 780 to get 546 as the normal force

3/21

So we basically just worked on review sheet for our TEST ON THURSDAYY:O

that is all.

Jon Sheppard's Class Review for Tuesday March 22, 3011

Okay, so today be basically did the first 2 problems on the "Newton's Second Law Review Problems.

#1. An 80 kg water skier is being pulled by a boat with a force of 220 N causing the skier to accelerate at 1.8 meters per second per second. Find the drag force on the skier.

so the normal force and the gravitational forces are each 800 N(up and down arrows), and the force of tension of the rope on the skier is 220 N. To find the drag force, use the funbal=(a)(m)-1.8 X 80 = 144=funbal. So the drag force is 220-144=76 N.

#2. A 2000 kg car is slowed down uniformly from 20m/s to 5m/s in 4 seconds. Determine the average unbalanced force on the car during this time, and how far the car traveled while slowing down.

Well, to find the Funbal, you first have to find the acceleration, which is V=at+vo-5=a(4)+20=-3.75m/s/s. Then funbal=(a)(m)-(3.75)(2000)=-7000 N. Then the distance covered is Chang in X=1/2(at)^2 - -3.75 X 4= 15^2= 225m.

Thank you for your time and cooperation, i really appreciate it ;)

Love,

yours truly, Jon Sheppard

Accelerometer Project

Paulina and I didn't realize that we had to put this on the blog, hence why it's about 43789462839 days late. Our sincerest apologies.

Tuesday, March 15

Today in physics we reviewed unbalanced forces. We started with this problem:Using this information, we had to find the acceleration. First, we made a force diagram for both the weight and the cart. The force diagram for the cart looked like this:

and the force diagram for the weight looked like this:

To find the acceleration we used the formula acceleration=Funbalanced/mass. The unbalanced force is 1 N, and the mass is 0.6 kg, so the equation would look like: acceleration=1 N/0.6 kg. When you solve the equation, the acceleration is 1.67 m/s/s.

Monday, March 14

Today we white boarded the problem we were given on Friday of a block with a mass of 0.18kg on a slide with an angle of 44°. To find the F_f you take cos44°=F_f /1.8N and you get 1.29N. Then to find the F_N you take sin44°=F_N/1.8N and you get 1.25N. Then Mr. Dickie gave us a new equation, F_f =µ(F_N), to find the coefficient of friction. Using this equation you get 1.25N=µ(1.29N), then you get µ=0.97N.

Friday! March 11

Okie Dokie, so here is what we did in our lovely Physics class today:

After whiteboarding, Mr. Dickie gave us the definitions of kinetic and static friction:

• 
  kinetic friction ~ The force exerted on one surface by a second surface when the two forces rub against one another because one or both of the surfaces are moving (def from the book if anyone is wondering)


• static friction ~ The force exerted on one surface by second surface when there is no motion between the two forces

He also made the point that these would have different coefficients of friction when on the same surface.

Next we did a problem to find the friction force between the friction block and the board:

We also knew that the coffiecient of (static) friction was 0.45 and that the mass of the block was 0.18 kg. Using this information, here is the force diagram:

Using the tip-to-tail method we made a triangle to help us solve for friction:

To solve for the frictional force between the block and the board"

sin28= x/1.8

Ff= 0.845 N

To solve for the normal force between the block and the board:

tan28= 0.845/x

Fn= 1.6 N

Homework :

Find the frictional force using the same picture and the same given except that the angle of the board is 44.

Hopefully this helped a tiny bit, maybe?

Anyways, have a fantabulous weekend!

in response to my darling Losi: Pasqualino <3

3/10

Hello kiddies, I apologize in advance if this doesn't make much sense.

At the beginning of class those of us who originally did yesterday's lab on the relationship between mass and friction had to convert mass to normal force. We then whiteboarded our results and discussed them. Paulina and I, being the brilliant people we are, somehow measured the mass of the block to be 124 kg, so I spent a lot of the discussion trying to figure that little mess out. However, what I did get out of the conversation was that we explained the relationship between the normal force and the frictional force using the following equation:

Ff=uFn

(u=coefficient of friction)

Now for the second part of class. We did another lab slightly resembling yesterday's, except today we used the more rough side of the board, and we found the force when the block was stationary (as much force as it had right before it began to move) as well as the force as it was moving at a constant speed. We should be comparing our results tomorrow.

Special note to my dearest paulie: Luca<3>

Friction Force part 2

Today in class we white boarded our labs from yesterday. We did a lab on friction force. Basically, Mr. Dickie had us pick between three choices: Surface area, normal force, and mass and we had to find out the relationship with a friction force.

So when we got in a circle to discuss, most people had tested mass. We didnt conclude whether or not mass affects friction or not, mr. dickie tried explaining but we sort of skipped it and we ran out of time, so we will find out tomorrow. Surface area, however, does not affect friction. We found out that friction opposes surface movement. Normal force also was not officially concluded either because it ties in with mass.

Mr dickie also gave us an example to think about on the board. He put a magnet on the board and asked why it doesnt slide down. We stated all the forces acting on the magnet. First, there is a magnetic force towards the board. Second, normal forces of the board are pushing towards the magnet. Third, gravity is pulling a downward force, but the force of friction is pulling up so it does not slide down.

Thats basically what we did for the day, not too much since it was a short schedule.

I hope this somewhat made sense...

Random fact of the day: The king of hearts is the only king without a mustache.

The Force of Friction

Ok everybody, this blog is going to be the dummed down version of class today. Sorry to all the really smart people. Today Mr. Dickie set up a ramp and a cart on his desk. He then pushed the cart down the ramp. Then, he took a wooden block and placed it on the ramp. It didn't move. After that, Mr. Dickie preceded to ask us several questions about why the cart slid down the ramp, and why the block didn't. I'm pretty sure the cart slid down because it had these four really cool things on the botton called wheels, yeah, I know, interesting stuff. Because the cart had wheels, the force of friction wasn't really a factor. And because, shocker, the wooden block didn't have wheels, it couldn't overcome the force of friction. When then did a lab to test how mass affects friction. We took a forceometer, a block, several weights, and a board to test this(no Mr. Dickie we weren't screwing around the entire time). The block was then attached to the forceometer and pulled five times. Probably more than that but all you needed was five. A different weight was placed on the block every time it was pulled, just so nobody ended up with five of the same data points. After this, we graphed our data, used our handy dandy calculators to find and equation, and then wrote up a conclusion. "And for those still listening, I'm Ron Burgundy, you stay classy San Diego."

Blog for 3/7/11- the day of Jake Plona

Well idk what to put on this blog, but what did you expect. This is the blog for yesterday btw. Uh the seniors weren't here cause they sold more December dreaming tix than us, so it was a slow day.

First we got into class and chilled for a bit and talked about our mid winter breaks. It was also the birthday of mr Jake Atnip, so a shout out to him.

Alright so now it's time to get down to business. We tried to find the ratio of the earth to a basketball and the moon to a tennis ball. We found out by using this magic google thing that the diameter of the earth is 8000 mi and the diameter of the moon is 2000 mi. So clearly, the earth is 4x the size of the moon. The diameter of the basketball is 9 in and the diameter of the tennis ball is 2.5 in, also 4x the size. This keeps the correct ratio of 3.6.

Using this ratio we discussed newton's third law and which applies the larger force, the earth or the moon. And of course I was correct in saying that they apply the equal force, for every action there is an equal and opposite reaction, for every force there is an equal and opposite force.

After we determined all of this we chilled some more and BSed. Then this guy came in and had 6 of us carry these tubes so that was the end of our honors physics class for the day.

I hope you all enjoy this post because I poured my heart and soul into it. Also, I spent most of physics class today writing it, so I'll be lost with what we did today (not that i really pay attention, jk mr Dickie, i always pay attention). And a special thank you to Jon Sheppard for letting me use his iPad to type this.

This is a day late cause i wasnt invited to this blog so just lettin you know.
I believe that's all I have to say, much love to you honors physics. Deuces.

Accelerometer

David Cicala, Andrew Rauh, Brad Mueller

Angle = 78 degrees
Mass of Stopper = .00708kg
Funbal = ma



Sin 78 = .0708N/t
t=.07238N

Funbal^2 + .0708N^2 = .07238N^2
Funbal = .01505 N

.01505 N = (.00708kg)(a)
a = 2.13m/s^2

Extra Credit

I didn't have a video camera so I am going to use the pictures and solve for acceleration.

When the car accelerated it created a 70 degree angle on the accelerometer.

We can draw a force diagram and then solve from there.

To solve for tension just use the formula: sin70=5/t

t=6.5N

We can now solve for Funbal by using the Pythagorean theorem

Funbal^2 + 5^2 = 6.5^2

F^2 = 17.25

F = 4.15N

Now we can find the acceleration using the formula: a=Funbal/M

a=4.15N/.5kg

a=8.3m/s^2

Extra Credit

I converted all of the degrees on the protractor to their respective acceleration.

The formula is:

Tan X=10/Acceleration --> Acceleration=10*/tan *the mass of the weight.

So 90 degrees (straight down) is O m/s^2

80 degrees is 1.8m/s^2 and so on

I wrote the acceleration down on the board. So when the weight moves you can just read the acceleration off of the board.

Honors Physics 2/25/11--BREAK!!!

Hello everybody!!

So today in physics class, we did the accelerating truck problem.We know that the the mass of the weight is 10kg, the angle between the roof and the rope is 55°, and that the truck is accelerating. So for this problem, we needed to draw a quantitative force diagram for the weight to determine all the forces acting upon the weight.

First, we draw a system boundary around the weight and find out what contact forces are acting on the weight...We know that the rope is pulling up on the weight causing a Tension force and that Gravity is pulling down on the weight. The Tension force is unknown while the Gravitational force is 100N because 10kg X 10 N/kg=100N.
Our next step is to find the Tension force by making a triangle with what we know. So we draw a perpendicular dashed line from the roof to weight which is the Gravitational force (100N). With this, we made a right triangle and we can now use trig!! So, we take the sin55=100N/Force of T to find the Tension force (122N).

What!?!? The forces aren't balanced!!! But...but...the picture looks like it is not accelerating. What the deal?!? Well, what's actually happening is that since the weight is in the accelerating truck, the weight is also accelerating at a constant acceleration. So, we need to find the acceleration. But how??

We need to use the equation Acceleration (A)=Force unbalanced (Fun)/Mass (m). There is just one little problem...we don't know the force unbalanced!! Here's how we find it.
To find the force unbalanced, we need to use vector addition (which means we line up our arrows tip to tail).
The gray dashed line represents the force unbalanced. Now all we need to do is find what the actual value is. Just plug in your numbers (122N for T and 100N for G). Use the Pythagorean Theorem to find the force unbalanced.
Now that we have the force unbalanced, we can find the acceleration!!

A=Fun/m

***IMPORTANT***
Remember that Fun is the force unbalanced (NOT THE TENSION FORCE OR THE GRAVITATIONAL FORCE!!!)

So the weight is accelerating at 7 m/s^2 with the truck. We DO NOT put an arrow for the acceleration on the force diagram because acceleration is NOT a force!!


We also did a lab where we only changed the mass of the weight. We were to find the angle between the roof and the rope. We used 1kg, 5kg, 15kg, 20kg, 50kg, and 100kg. No matter what the mass was, the angle was 55 degrees just like the orginal problem. This is true because the acceleration stayed the same. We did not change the acceleration, therefore, the angle did not change!

Mr. Dickie assigned a Extra Credit assignment as well. Over break, we are to make an accelerateometer (of some sort) to find the acceleration of our cars. We were to make an accelerateometer, drive with it in the car (however it may work), take a picture of the angle of the accelerateometer, and then measure the angle. Once you have the angle, find the acceleration...GOOD LUCK!!! Also...please be safe and have somebody do this with you. Thank you!!

If you have any questions, comments, corrections, etc. please leave a comment!! = )

Have a FUN break!!!!!!!

Honors Physics 2/24/11

Today we went over the second half of our worksheet. #3 and #4.

#3 asked about our previous "elevator" problem. It stated that a person that was in an elevator was applying a force down, and his weight increased as the elevator went up, and decreased when he reached the top. This is due to the acceleration of both the elevator and the person. On the way up, the acceleration is 2.5 m/s/s. When the man reaches the top, the acceleration is -3.0 m/s/s. He is less wegith because the unbalanced forces are pulling down at the top causing an opposite acceleration and causing him to only feel the normal force at that instant from the elevator, not gravity.

#4 was about how much an elevator can hold. It can hold 30000 n of force without the cable snapping. It can hold 20 people of 75kg. The elevator weighs 500kg. The cable pulls at 30000n and the force of the people and the elevator pulled down by gravity is 20000 n. The total unbalanced force is 10,000n. The wieght of the people plus the elevator is 2000kg, so you take Funbal/mass=A. You do this and find that the elevator can accelerate at a maximum of 5m/s/s without the cable snapping.

Then we started some truck problem where harry potter was making a ball float. It doesnt make sense. So don't count on me!

Today in physics we worked on the first two problems from the 'net force particle model worksheet 2'.

1. For the first problem, the down arrow exerts 46,000 N since the helicopter weighs 4,600 kg. The down arrow is the force of gravity from the earth on the helicopter. Next the acceleration needs to be calculated.

(2.0m/s2)X(4600kg)=9200 N

Now this unbalanced forced needs to be added to 46,000 N since this is the amount of lift force the helicopter would need to hover in midair.

46,000N + 9200N= 55,200 N of lift force from the propellers on the helicopter.

2. Since the bag weighs 20kg the force of gravity from earth on the bag is 200N. Now the unbalanced force from the acceleration of the lift needs to be calculated.

(5.0m/s2)X(20kg)=100 N

In order for the hand to hold the bag in midair, the hand form the person applies 200N of force on the bag. This force is added to the unbalanced for to find the force the hand pulls up with.

200N +100N = 300N

Since the forces are unbalanced and the force from the pull of the hand exceeds 250N, the bag breaks.

Movie Day!! :D

MythBusters video today about the myth that if one jumps right before your elevator plummets to its doom at the bottom of the shaft, they would live. Unfortunately it was busted and the dummy in the experiment gave a definitive splatter at the end of his ride. Using the equation V^2 = 2a * change in X (since the dummy's initial velocity is 0). The drop from the elevator is 92ft and gravity is an acceleration of 32ft/s^2, which would make the drop velocity 76.7ft/s. Unfortunately the average jump of a person is 8ft/s (using the same equation with a 1ft high jump), while this will slow your fall, it's only to a still bone crushing 68.7ft/s. The myth was busted.

Notes, February 17, 2011

Notes: February 17, 2011From the Force Diagrams and Net Force worksheet:

1. An elevator is moving up at a constant velocity of 2.5 m/s. The passenger has a mass of 85 kg.
a) Force diagram of the passenger:

b) Calculate the force the floor exerts on the passenger: (85 kg) x (10N / 1 kg) = 850 N

2. The elevator accelerates upwards at 2.0m/s^2
a) Force diagram of the passenger:
b) Equation for the vertical force on the passenger: Force unbalanced = (acceleration) x (mass)
c) Calculate the force the floor exerts on the passenger:Force unbalanced = (2.0 m/s ^2) x (85 kg)Force unbalanced = 170 NForce unbalanced + Normal force = vertical forceVertical force = 170 N + 850 NVertical force = 1020 N

3. The elevator accelerates downward at 3.0m/s^2
a) Force diagram of the passenger:

b) Write the equation for the vertical forces on the passenger: Force unbalanced = (acceleration) x (mass)
c) Calculate the force the floor exerts on the passenger:( 85 kg) x (10 N / 1 kg) = 850 N

4. While descending in the elevator, the cable breaks. How big is the force on the passenger by the floor?
a) At the moment that the cable breaks, the passenger and elevator accelerate downward at 9.8 m/s ^2.
b) Afterwards, the force of the floor on the passenger would be zero.
c) Relative to the elevator, the passenger would be weightless.

5. A 70 kg skydiver jumps out of an airplane. Immediately after jumping, how large is the skydiver’s acceleration?
a) Acceleration due to earth’s gravity = 9.8 m/s ^2
b) Upon reaching a downwards velocity of 100 miles per hour, 300 N of drag resist the diver’s motion. Draw a force diagram and calculate how large is the skydiver’s acceleration.
Force unbalanced = (acceleration) x (mass)
Acceleration = (Force unbalanced) / (mass)
Acceleration = (700 N – 300 N) / 70 kg
Acceleration = 400 N / 70 kg
Acceleration = 5.71 m/s ^2

6. A 900 kg car that exerts 5000 N of traction force on a level road while being opposed by 1000 N of friction and drag forces (combined)
a) Force diagram of car:
b) Write a net force equation for the car:
Force unbalanced = (acceleration) x (mass)
c) Calculate the acceleration of the car: Acceleration = (Force unbalanced) / (mass)
Acceleration = (5000 N – 1000 N) / (900 kg)
Acceleration = 4000 N / 900 kg
Acceleration = 4.44 m/s ^2