Energy Transfer Challenge

Energy Transfer Challenge

Goal

Determine the relationship between the spring stretch distance and the force on the spring.  

Data collected

Graphical representation:

*the steeper slope had a stiffer spring, meaning that F and x have a larger relationship.

Stretchy Spring: 

Equation of the line: F = 8.98x + 0.0223

Stiff Spring:

Equation of the line: F = 80.841x + 1.0651

Post Lab Questions:

1. What's the relationship between the spring stretch distance and the force?
    The distance it stretches and the force exerted are identical.
2. How does the total energy in the spring compare when it is stretched more?
    The energy in the stretched spring is greater than an un-stretched spring.
3. When you look at an F v.s. x graph, what does the slope represent?
    The slope represents the stiffness in the spring (the spring constant).
4. How do the F v.s. x graphs show how much energy was present?
    The steeper the slope, the greater amount of energy present.  Also, the energy was shown through the area under the curve, which could be calculated using the Eel = 1/2bh formula.
5. What is the equation to calculate the force needed to stretch a spring a certain distance?
    F = Kx   (K is the constant energy)
6. What is the formula for the elastic potential energy in a spring?
    This can also be found as the area under a slope as Eel = 1/2bh.  Otherwise, the equation is Eel = 1/2Kx^2
7. Do a pie chart analysis for each of the two springs used for the carts showing them compressed and then released as the cart moved forward.

Stiff spring:

Stretchy spring:

Energy and Power Unit

 Energy

Different Types of Energy (Always measured in J)

Eel - Elastic Energy (usually used with a spring/rope)

      - Eel = 1/2Kx^2

Echem - Chemical Energy (usually used when a person is in the system)

            - Echem = 

Ek - Kinetic Energy (when a cart is moving at a speed and no other forces are adding energy into it)

     - Ek = 1/2mv^2

Eg - Gravitational Energy (has to do with how high the object is off of the ground)

     - Eg = mgh

Etherm - Thermal Energy/Dissipate (always caused by the presence of friction, makes the object slow down)

W - Work (when energy is being put into an object from an outside force)

     - W = F x (change in X)

Energy Pie Charts

Say a car is pushed with a spring up a slightly inclined hill.  As the car moves, up the hill, where does its energy go?

(Total circle contains Eel)    (Half Eel, 1/4 Ek, 1/4 Eg)     (Half Eg, Half Ek)

Graphs and Elastic Energy

When using a graph, the elastic energy can be determined by using the formula for the area of a triangle.  The area of a triangle is 1/2 base x height, and then you add the length x width to get the full elastic energy.

Bar Graph Key Points

(I'm sorry, it's not letting me upload pictures so I am unable to show bar graphs as examples)

- make sure you list all components in the system. If you miss out on one, you may mistake energy as work instead of chemical (for example).

- when friction is included, there is a very large chance that Ediss or Etherm is going to be used in the end.

- make sure you label how many boxes you have used for each graph, because the amount of energy you start with is exactly the same as the amount that you end with.

- if two back to back problems have the same idea, but one just asks for friction, use the same ratio of boxes concerning the height of the object if it is being thrown up/ moving up.

- you only really use Echem when the energy comes from a person, and if the person is part of the system. If the person is not part of the system, but is still putting energy into the object, you would call that Work.

Quick Problems!

1) How much kinetic energy does a 2000kg SUV traveling 50 mph have?

Ek = 1/2mv^2

Ek = 1/2 (2000) (50)^2

Ek = 1000 (2500)

Ek = 250,000J

2) How much energy is stored when a railroad cart spring is compressed to 10cm? (The spring requires about 10,000 newtons to be compressed 3 cm).

Eel  = 1/2Kx^2

Eel = 1/2 (333333) (0.1)^2

Eel - 1666.6 J

*Finding the spring constant given newtons and distance:

       Fel = Kx

       10,000 = K(0.03)

       333333 = K

3)Determine the final velocity of a rollercoaster with a mass of 40kg, assuming a 10% loss to friction, that is 5m above the ground.

Energy conservation equation -> Ek = Eg

Eg = mgh

Eg = (40) (10) (5)

Eg = 2000J

10% of 2000 is 200, and 2000-200 = 1800J

Ek = 1/2mv^2

1800 = 1/2 (40) (v^2)

1800 = 20v^2

90 = v^2

v = 9.5m/s

4) A student eats a breakfast before school that contains 350 calories. (One food calorie = 4186 joules).  While he is working on classwork during school, he radiates about 150 joules per second. How long would the student have to work on classwork to radiate all of the energy from breakfast?

350 x 4186 = 1,465,100

1,465,100 / 150 = 9767 seconds = 1.62 hours

Power

- power = energy transferred / time
- measured in watts
- the quicker you are at transferring energy, the more powerful you are

1) Bob pulls on a rope to lift a 12kg pail out of a deep, dark well.  She lifts the pail straight up for a total of 10 meters at constant speed.  How much power is required to complete this task in 5 seconds?
Eg = mgh
Eg = (12) (10) (10)
Eg = 1200J

power = energy transferred / time
power = 1200 / 5
power = 240 watts

MTM Challenge

Unknown Mass Challenge

What was the challenge?

For this challenge, Will, Robert and I had to figure out the weight of a sack of salt without measuring it.  We were given a cart, a sack of salt, and a motion sensor to collect the velocity.  Here are the steps we took:

1) The empty cart's mass is 0.5kg.  Find the velocity of the empty cart by using Capstone and the motion sensors.  This will be your Va.

2) Put the sack of salt on top of the cart after finding the empty cart's velocity.  Find the full cart's velocity and record it.

3)By using the mava + mbvb = (ma + mb) (Vab) formula, solve for the salt's mass.

Data, Picture and Calculations

The empty cart's average velocity was 0.5 m/s, so here's how we solved for the momentum:

pemptycart = mv

pemptycart = (0.5) (0.57)

pemptycart = 0.28 kgm/s

We found the average velocity of the full cart to be slower, traveling at .24 m/s.  Once we obtained this information, we were able to solve for the mass of the sack of salt.

ptotalbefore = ptotalafter

mava + mbvb = (ma + mb) (Vab)

0.28 kgm/s + 0 = (0.28 + mb) (0.24)

0.28/0.24 = 0.28 + mb

1.16 = 0.28 + mb

0.89 kg = mb

We were very close to the actual weight.  We calculated that the salt's weight was 0.81 kg, but it was actually 0.77.  My thought is that we were a bit off on the velocity data collection, and for the future I would suggest that we do more trials to solve for the velocity, and then just take an average for an accurate collection of data.  

Unit 6 - Momentum and Inertia

Momentum and Inertia 

Formulas you need to know:

(REMEMBER: P = Momentum and J = Inertia)

p = mass x velocity ---->  momentum is measured in kgm/s 

∆p = mass x ∆velocity

J = F x ∆time ----> inertia is measured in NS

J = p

mava + mbvb = mava + mbvb ------> finding the total momentum of a system

mava + mbvb = (ma + mb) (Vab) ----> total momentum when the objects stick together

ptotalinitial = ptotalfinal

Momentum and Inertia Questions that are frequently seen:

1) What is the momentum of an 8kg bowling ball rolling at 2m/s?

p = mv

p = 8 (2)

p = 16kgm/s 

2) What's the momentum of a 50kg cart sliding at 7 m/s across a tile floor?

p = mv

p = 50(7)

p = 350 kgm/s

3) Which has a greater momentum, a heavy truck at rest or Julian Martin on a skateboard rolling at 4 m/s?

Julian has a greater momentum, because of his velocity.  If you were to calculate this, you would see that the truck's velocity would be 0, thus making his momentum 0 as well.

4)Describe the change in momentum in this situation: A small bee hits a big mac truck's windshield.

The change in momentum of any action/reaction scenario stays the same; therefore, the bug and the mac truck lose the same momentum.  The truck has a big mass and a small acceleration, where the bug has a small mass and a big acceleration.  This concept is proven through Newton's 3rd Law, saying each action has an equal and opposite reaction.  See below:

Mava = mbVb

mac truck ∆momentum = bee ∆momentum

5) If you throw a ball horizontally while standing on a skate board, you will roll backwards.  If you go through the motions of throwing the ball, but hold onto it instead, will you still roll backwards?

You theoretically shouldn't and won't move backwards because there's no Newton's 3rd Law force pushing back on you.  The net force would have to be something other than 0, because in order to have that acceleration, there can't be a balanced force.

6) While being thrown, a net force of 100 Newtons acts on a baseball with a mass of 0.5 kg for a period of 1 second.  What's the change in momentum of the ball?

F∆t = J

(100N) (1) = 100NS

J = p

p = 100kgm/s

7) If a bat hit a baseball, how large would the ball's force be on the bat?

The forces, according to Newton's 3rd Law, would be equal and opposite.  

8) An astronaut with a mass of 60kg carries a cooler of CapriSun with a mass of 15kg in space.  By using the cooler away with a speed of 3 m/s, the astronaut floats in the opposite direction.  Solve for the total system's momentum and find the speed at which the astronaut moves away from the cooler.

ptotalinitial = ptotalfinal

(ma + mb) (Vab) = mava + mbvb

(75kg) (0) = 60 (va) + 15 (3)

-45 = 60va

Va = - 0.75 m/s

9) On an icy road, a 7000 kg mac truck hits a 1000 kg car that had been previously traveling at 15 m/s, causing the truck to slow from 20 m/s to 14 m/s and the car to speed up.  Find the total momentum and find the final velocity of the car.

ptotalinitial = ptotalfinal

mava + mbvb = (ma + mb) (Vab)

7000 (20) + 1000 (15) = (7000+1000) (Vab)

290000 = 8000 Vab

Vab = 36 m/s

Big Question!

Why would you bend your legs when jumping off a roof? (This question is very similar to the Egg Toss and padded dashboard scenarios). 

- You are going from some amount of velocity to completely stopped no matter if your legs are bent of not

- The mass and change in velocity are the same with or without bent legs

- ∆p = m∆v and the ∆p is the same with or without bent legs

- Since the ∆p is the same, and the ∆p = J, the J is the same too

- Since J is constant, bending legs will increase the time, which decreases the force (leading to a safer landing)

- Bad landing:  J = F∆t  (Big force on you with little time)

- Good landing: J = f∆T (Small force on you with greater time)

From my orange sheet:

- Mava = mbVb (Big momentum and little velocity equals little momentum and big velocity)

- ∆p = J

- When two objects hit, the momentum is equal and opposite.  So is the impulse.  However, the acceleration may be different for an object (for example, if a bug hit a car windshield, it's acceleration would be greater)

- F∆t = m∆v

- The ∆p is the same no matter if the ∆p is fast or slow 

- Remember to convert weight to kg before using it in a formula

- ptotalinitial = ptotalfinal

- Use SOHCAHTOA when finding the sides of an angle (for example, if two balls collided and bounced off at different angles).

- Individual objects can have a change in momentum, but when working with a system, the momentum of two objects has to be equal and opposite.

Rocket Challenge

Rocket Challenge

Background information on our challenge:

Our task was to move the target to predict where the rocket would land on the field hockey field.  I, along with Will Rankin and Will Casse, used a bicycle pump to launch our rocket into the air.  By using different wooden angles, we were able to launch our rocket at different height angles.  This allowed for us to gather a good amount of data to predict where to put the target when launching our test rocket.  Our test angle was 35 degrees.

Here's a video of us collecting data:

We first found the velocity by calculating how fast the rocket went for different angles:

We had to use various formulas and collect many pieces of data to make our prediction.  Once we found the velocity (by calculating distance over time), we tried solving for the x and y velocities. Since the object was at an angle, we had to use tangent to find these variables:

We discovered that the Viy = 14.12 m/s and the Vx = 9.89 m/s.  With this information, we were able to use the change in X in the y direction to solve for the time.  We also had to use the quadratic formula for this.  The time ended up being 2.82 seconds, saying that it took 2.82 seconds for the rocket to fly up and return to the ground.  Once we had solved for the time, we were able to plug it into the {Vx = (change in X in the X direction) / time} equation, thus solving for the change in X in the x direction.  It ended up being 27.9 meters.  

Were we correct?

Yes!!  Our rocket landed directly in the circle that we made out of rope on the field hockey field!!  

Unit 5 - Projectile Motion

Projectile Motion

Formulas:

From Miranda's yellow sheet:

• If an object is going down, the velocity and displacement are negative
• In the hypothetical situation of a ball being thrown into the air and caught in the same spot, the displacement is 0
• The acceleration is -10m/s^2 in the vertical direction for the object.  The horizontal direction does not have an acceleration, it has constant velocity.  
• Remember to differentiate the horizontal components from the vertical components (Vx and Viy)
• Horizontal velocity = Vx
• Vertical velocity = Viy
• The initial Viy is usually 0 (especially in the scenario of dropping something)
• When given the initial velocity with an angle, make sure to solve for the Vx and the Viy.  DO NOT use the velocity given to you.

To not get confused with your components, try writing out one of these charts on the side of your problems to differentiate where each piece of information goes.

This is what a motion map for an object falling off of a table would look like.  Notice the X and Y components along with the actual movement of the object (in red).

Practice problems!!

1. A ball is thrown downward with an initial speed of 20m/s.  What's the ball's acceleration?  Calculate the displacement during the first 4 seconds.  Then, calculate the time it takes to reach a speed of 50 m/s, and how long it would take to fall 300 meters.

a = -10m/s^2

First, solve for the displacement by plugging in 4 for time.

 Then, Use the Vf that you were given to solve for the time.  Remember that the final velocity is negative because the object is going downward.

To solve for the amount of time it would take to reach 300m, plug in -300 meters as the change in x, and transform the problem into an ax^2 + bx + c formula (to use the quadratic formula).  

2. A student finds that it takes 0.2 seconds for a ball to pass through photo gates placed 30 cm apart on a level ramp.  The end of the ramp is 92 cm above the floor.  Where can we place a coin to ensure that the ball will land on top of the coin?

 Solve for the time by plugging in -.92 as the displacement (remember, negative because it's going down)

 Then, use the information you know (displacement in the x direction and the time) to solve for the velocity in the x direction.

Once you have found the velocity in the x direction, use it and the time to solve for the displacement in the x direction.  This tells you how far away you should place the coin.

3. A kickoff sends a football with an initial velocity of 25m/s at an angle of 50 degrees above the horizontal.  Find the x and y components of the velocity, the time the ball is in the air, and the horizontal distance the ball travels before it hits the ground.  

First, solve for the X and Y components.  You don't want to use the velocity that the problem gave you.

Then, plug the Viy that you solved for into the displacement equation, using 0 as the displacement in the y direction.  You use 0 as the displacement in this case because the ball did not change it's position in the y direction, it landed at the exact same height that it started at.  Solve for the time with this information.

Solve for the displacement in the x direction by plugging in the x velocity you solved for, and the time you solved for in the previous step. 

4. A water balloon is launched at a building 24 meters away with an initial velocity of 18m/s at an angle of 50 degrees above the horizontal.  At what height will the balloon strike the building?

First, split the velocity that you are given into X and Y components. 

 You now have to find the time.  You are not given the displacement in the y direction, so you must use another formula.  You know that the displacement in the x direction is 24, and you just solved for the Vx, so plug those into the corresponding formula and solve for the time.

Once you have solved for time, you can solve for the displacement in the y direction.  

Ball Bouncing Challenge

Ball Bouncing

On Saturday, Christina and I took a stroll down to the gym to test the velocity and position of a Champion basketball after bouncing it.  

To further help you visualize our collection of data, here's a video of Christina bouncing the ball:

This is what the video looked like with the data points:

Here's a free body diagram of what the ball looked like in the air:

The reason why there are no other forces on the ball is because only the force of gravity is pulling the ball to the earth.

Here's our v v.s. t graph:

On our v v.s. t graph, we were observing two things: the ball's velocity in the vertical direction and the balls' velocity in the horizontal direction.  The data with the light blue border corresponds with the light purple line on the graph.  This line was a representation of the ball's velocity in the horizontal direction.  It is basically a straight line, only a bit of air resistance was slowing it down.  The data with the pink border corresponds with the green line, which depicts the ball's vertical velocity.  As you can see, when the ball hit the ground, it's velocity was at 0, but then increased in the negative direction as it continued on to another bounce.

Here's our x v.s. t graph:

In this graph, you can see how the ball's vertical movement gave us a parabola, showing that it had an acceleration.  The ball's horizontal line showed that the ball moved forward at constant velocity.

Analysis

The acceleration in the y direction was -9.6 m/s^2 (which would make sense, as the average force of gravity is 10N).  This was the slope of our green line in the v v.s. t graph.The acceleration in the x direction was -0.54 m/s^2.  This was the slope of the purple line in the v v.s. t graph.  

The initial velocity in the horizontal direction was about 6 m/s, and the initial velocity in the vertical direction was about 4.6 m/s (the graph is a little bit off, but our data is correct).

The ball's velocity at the top of it's path in the horizontal direction was constant, as no forces (except for minor air resistance) prevented it from moving forward.  The ball's velocity at the top of it's path in the vertical direction was also 0, and you know this because it had been slowing down until it had reached it's peak, where it stopped momentarily before accelerating to the floor.

The final velocity in the vertical direction was about -5 m/s, and the final velocity in the horizontal direction remained at a constant 6 m/s.

The ball's highest point was 2 meters, as represented on our graph of x v.s. t.  In order to solve for the distance travelled in the horizontal direction, you can use:

Δx= 1/2 a t^2 + vi * t

By plugging in -0.54 as the acceleration, 1.05 as the time, and 6 as the initial velocity, you end up with the Δx being 6.01 meters.

It took the ball 0.6 seconds to reach the peak of it's bounce.  The ball was in the air for 1.1 seconds, which is shown on out x v.s. t graph.  

Conclusions

The experiment was pretty impressive, as we studied the balls' motion not only in the forward direction, but in the upward direction as well.  The vertical acceleration was almost equal to the force of gravity, which is 9.8N (or 10N).  The horizontal acceleration however was irrelevant, because the ball bounced at a constant velocity of 6 m/s across the gym.  This justifies that the ball's upwards velocity was not constant, but it's forwards velocity was.  

We solved for the horizontal displacement by using the change in x formula.  By plugging in the acceleration, time, and initial velocity, we were able to conclude that the total horizontal displacement was 6.01 meters.  The ball's velocity at the top of it's path in the vertical direction was 0m/s, as it was changing from going up to going down.  It had been going in the upwards direction, slowing down in the process, until it vertically stops momentarily, and then continues accelerating in the downwards direction.  It's horizontal velocity however remained constant at 6 m/s, as the ball continued moving forward while it's vertical path had acceleration.  To solve for the vertical component, you would also use the 'change in x' formula, and plug in the total time.  However, to solve for the ball's height, you would use the 'change in x' formula, and plug in the amount of time it took to reach it's height. 

UFPM Challenge

UFPM Challenge

Group ~ Miranda, Will C, Robert

Our goal was to calculate the exact time it would take for the cart to reach the same spot of a hanging weight that was quickly dropping.  We found the total mass of the system, which was .639kg (rounded to .64kg), and used that to find the acceleration.  

Here's a diagram of what our setup looked like:

We solved for the acceleration of the weight, enabling for us to solve for time later on.

To make it easier to understand the information, we drew a free body diagram of the system.

  We measured that the distance the weight would have to travel to reach the ground, which was .74 meters, and we knew that the acceleration was .78 m/s2.  Thus, we used our information to solve for the time it would take for the weight to reach the floor.

We already solved for the acceleration of the weight, so then we had to solve for the acceleration of the cart.  We measured that it took 3.6 seconds for the cart to move 1 meter, meaning it had a velocity of .28 m/s.  Therefore, we had to find the distance it would travel in the same amount of time it would take for the weight to hit the floor. 

This information meant that we had to position the cart .38 of a meter away from where the weight would hit the floor.  By releasing the cart from this distance and the weight from its height at the same time, they would potentially collide.  

Our hypothesis was correct!  When we released the cart and the weight at the same time, the weight landed in the center of the cart.  We had no percent error, because our time was exactly what we had predicted.

If you want to see our success, go to https://youtu.be/P3dvTJU3kcU 

Unit 4 - Unbalanced Force Particle Model

Skateboard Challenge

Initial Trends and Patterns:

- The object accelerates when more force is applied

- The more force you apply, the faster it accelerates.  The less force you apply, the slower it accelerates

- The lighter the object is, the faster the acceleration (and vice versa pertaining to heavy objects)

- There's an inverse relationship between mass and acceleration

- There's a direct relationship between force and acceleration

Elevator Challenge Question

An elevator is moving up at constant velocity of 2.5 m/s.  The passenger has a mass of 85kg.

The elevator now accelerates upwards at 2m/s^2.

What's the equation of the vertical forces on the passenger?

Fnet = Fg + Fn

Newton's 2nd Law Key Points:

Situation - You have a 2kg box being pushed with a net force of 4N.

1) Calculate the box's acceleration:

a = Fnet/mass

a=4/2

a=2m/s^2

2) How would the acceleration compare if the mass of the box was cut in half?

The acceleration would increase/double.

3) How would the acceleration compare if the force was doubled?

The acceleration would increase/double.

4) How would the acceleration compare if the force was cut in half?

The acceleration would decrease/be cut in half.

5) How would the acceleration compare if the mass was doubled?

The acceleration would decrease due to no change in the force.

6) How would the acceleration compare if the mass and force was doubled?

The ratio would be higher between mass and force compared to the original, but essentially the object would be moving with the same acceleration.

7) How would the acceleration compare if the force and mass was cut in half?

The ratio would be lower between mass and force compared to the original, but the object would still move with the same acceleration.

8) How would the acceleration compare if the force was doubled and mass was cut in half?

The object would accelerate at a quick rate because a lot of force would be exerted on a light object.

9) How would the acceleration compare if the force was cut in half and the mass doubled?

The acceleration would decrease by a lot, because not a lot of force would be exerted on a heavier object.

Practice Problems!

1) A race car has a mass of 710kg.  It starts from rest and travels 40 meters in 3 seconds.  How big is the net force on the car?

First, solve for the acceleration.  Plug in all of the variables you have:

When you find the acceleration, solve for the Fnet.  This will be the total force acting on the car:

Here's what a force diagram would look like: 

2) A rollercoaster car has a mass of 300kg with passengers.  It accelerated down a 65 degree hill.  Draw a force diagram, determine the gravitational force parallel to the hill, and find the acceleration.

3) A sled weighing 300N is moved at constant velocity over a horizontal floor by a push force of 50N.  Determine the coefficient of kinetic friction between the sled and the floor.  Then, determine the sled's acceleration if the kinetic friction was 0.

Formulas to be familiar with:

vf = at + vi

Δx = 1/2 aΔt^2 + viΔt

vf^2 = vi^2 + 2aΔx

a = Δv/Δt

a = Fnet/mass

Ffriction = µ (Fnormal)