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AP Physics “B”
San Dieguito Academy2012 – 2013
Book 1Collisions in Two Dimensions
Center of MassSprings
Density - PressureFluid Dynamics
Flight
1
Book 1Table of contents
Chapter Topic Contents Pages
1 Introduction and Review 1 – 12
2 Momentum in one dimension 14 – 23 in two dimensions 24 - 28Center of mass 29 - 34
3 Springs Hook’s Law 35 - 37Energy in a spring 38 - 46
4 Density Volume – Density 47 - 56
5 Fluid dynamics Pressure and Force 57 - 65Pressure and Energy 66 - 76 Bernoulli’ Principle 77 - 87Buoyancy 88 - 92
Chapter 1 – General Understandings from first year physics
Basic terms, concepts, and problem solving techniques that you should have learned include:
Inertial mass (m) is an object’s resistance to change in motion, or conversely, its desire to continue in its current motion. Inertia is a scalar. It has magnitude but not direction. In the MKS system it is measured in Kilograms
The basic law of inertia as stated by Newton: Objects in uniform motion want to stay in uniform motion forever; unless acted upon by a net unbalanced force.
Gravitational mass (m) an object’s ability to attract other matter and its ability to curve space. The gravitational attraction created by mass is experienced as gravitational force. This force is equal on both masses but opposite in direction. G is a constant 6.67 E-11 in the MKS system
Fg = G m1 m2 / r2
Note: It has never been proved that inertial mass and gravitational mass are equivalent. On the positive side, there has never been a case where the two were not equivalent.
Force (F) The amount of push or pull exerted on matter. It is important to note that force is a vector. It has both magnitude and direction.
The basic law of force as stated by Newton: Force causes objects to accelerate.
Fnet = ma Ffrict = Fnor = m g (cos
Fslope = m g sin
Fg = m g
Fg = G m1 m2 / r2
Kinematics: the study of how things move.
2
Acceleration (a) is a measure of how fast an object speeds up, slows down, or changes direction. It is measured in meters/sec2. In general it is given by
a = v / tFor circular motion it is called acceleration centripetal and is given by the equations
ac = v2 / r
ac = 4 2 R / T2
ac = 4 2 R f2
Basic Kinematics equations (note that both Vickie and Dickie require constant acceleration)
Vickie vf2 = vo
2 + 2 a (d)
Dickie df = ½ a t2 + vo t + do
Arthur a = v / t
Newtonian reference frame: For Newton motion in the universe is all relative. There is no prime location in the universe. Each frame of reference works as well as all others. The only thing that really mattered is that for Newton’s equations to work out you needed to be in a non-accelerated frame of reference.
dobserved = din reference frame + dof reference frame
vobserved = vin reference frame + vof reference frame
Notice: if you are in an inertial (non-accelerated) frame of reference, all of the kinematics and dynamics equations will work and give working answers in that frame and in any other observing inertial frame of reference.
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Momentum and Force: First Universal method of studying the world.
Momentum () which is an inertial measure of an object’s current motion = mv is the basic equation for the determination of momentum. Unlike inertia, momentum is a vector. When working with momentum you must include direction. In a closed system momentum is always conserved. Total momentum in a closed system remains constant.
Momentum before = Momentum after
1o + 2o + . . . = 1f + 2f + . . .
= mv so
m1v1o + m2v2o + . . . = m1v1f + m2v2f + . . .
Impulse (J) is the measure of the strength of a mechanical interaction between objects. It is seen in the momentum change in objects that the interaction produces J = (mv) It is also expressed in terms of the force of the interaction J = F t
Impulse = Impulse
(mv) = F t
It is important to remember that momentum and impulse are vector quantities. You have to keep track of positive an negative directions.
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Energy and Force: the second universal way of studying the world
Energy (E) is an alternative way of measuring the mechanical world. Energy comes in many forms and it is measured by its ability to do work with the assumption being that in a closed system the total energy will be constant. The forms of energy that we used were
Kinetic energy KE = ½ m v2 Gravitational Potential energy PEg = mgh
Conservation of Energy
Energy Before = Energy After
Work (W) is a measure of how much energy has changed form or has entered or left a system. Mechanically, Work is found in terms of force W = F d. Most commonly when energy leaves a system it is called Wout and it is usually in terms of heat generated by friction.
W = F d
Power (P) is the rate at which energy changes form, enters, or leaves a systemIt is measured in Watts.
P = E / t
In situations where velocity is constant we can use
P = F v
The English units for power are Horsepower. One horsepower is equal to 745 Watts.
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Review problems. Show your work
1. A 240 kg block initially stationary is acted upon by a 400 Newton force find the following after 6 seconds has passed:
A. acceleration of the block.
B. displacement of the block.
C. velocity of the block.
D. KE of the block.
E. momentum of the block.
F. impulse on the block.
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2. A 24 kg block moving to the right with a speed of 12 m/s collides with and sticks to a 16 kg block moving to the left with a speed of 30 m/sec.
A. What is the total kinetic energy of the objects in the system before the collision?
B. What is the net momentum of the blocks in the system before the collision?
C. What is the final speed of the 24 kg after the collision?
D. What is the total kinetic energy of the objects in the system after the collision?
E. What is the net momentum of the objects in the system after the collision?
F. What is the impulse on the 24 kg object due to the collision?
G. What is the impulse on the 16 kg object due to the collision?
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3. A 1200 kg car is driving up a 4 degree slope at a speed of 30 meters/second. Answer the following:
A. What is the minimum coefficient of friction needed to maintain this speed?
B. How much power must the engine produce to maintain this speed?
C. What average force must the engine be producing to maintain this speed?
4. A 2 kg rock is thrown straight upwards from the top of a 50 meter cliff with an initial speed of 20 m/sec. The rock ends up at the ground at the bottom of the cliff, answer the following:
A. The rock’s initial KE is.
B. The rock’s maximum elevation above the bottom of the cliff is.
C. The rock’s impact velocity at the bottom of the cliff.
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D. The rock was in the air for how many seconds.
E. When the rock hits the ground, find the magnitude of the impulse it experiences.
F. If the final collision lasts 0.05 seconds, what is the magnitude of the average force of impact with the ground?
5. A 70 kg coasting bicycle enters a loop the loop with an initial speed of 14 m/sec. The loop the loop has a radius of 3 meters. Find the following:
A. The initial kinetic energy of the bicycle
B. The initial centripetal acceleration of the bicyclist.
C. The true weight of the bicycle
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D. The apparent weight of the bicycle at the bottom of the loop
E. The speed of the bicycle at the top of the loop
F. The centripetal acceleration of the bicycle at the top of the loop
G. The apparent weight of the bicycle at the top of the loop.
6. A 5 kg block is resting on a horizontal surface. The coefficient of friction between the block and the surface is 0.7
A. Find the maximum force of friction the surface can exert on the block
B. If you push down on the block with a force of 35 Newtons, now what is the maximum force the surface can exert?
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Simple concept problems
1. You are fired upwards with an initial speed of 10 m/sec, what will your acceleration be at the highest point of your projectile motion?
2. You are fired at an angle of 60 degrees with respect to the horizontal with an initial speed of 10 m/sec. What will your velocity be at the highest point of your projectile motion?
3. Which exerts a greater force on the other. The Earth’s gravity on the Moon or the Moon’s gravity on the Earth?
4. A bee flies between two bicyclists as shown below. How many Kilometers did the bee travel
before the two bicyclists collide?
5. In which case above, will block A move with a greater acceleration?
6. In the picture above, the blocks have a weight of 100 Newtons each. What is the tension in the rope connecting them together?
7. In the problem above, Block A is initially moving downward and Block B is initially moving upwards with a speed of 5 m/sec. How does this effect the tension in the rope?
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Define in layman’s terms the following:1. Difference between displacement and distance
2. Difference between velocity and speed
3. Difference between work and power
4. Difference between KE and PEg
5. Describe two ways in which a person can have negative acceleration
6. Give two instances where an object can have zero velocity and still have acceleration.
7. Fg is
8. Mass can be defined in terms of “inertia” explain
9. Mass can also be defined in terms of “gravity” explain
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Commitment of Understanding
Do not let anybody fool you. AP Physics is tough. It requires commitment, hard work, and teamwork. Let me make this perfectly clear, AP Physics is a team sport. Look around you for a moment. Your obligation here is not just to learn, but to teach physics to each and every individual in this classroom. Holding yourself back holds the whole class back. For the sake of the rest of us, if you are not willing to do the work and go the distance, drop this class now!
This is taken from another challenge but it applies here.
I am here to work hard, to work with the best, to teach the best, and to take the AP Exam with the best of intentions.
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Signed: _________________________________________________
Chapter 2 MomentumCh 7 in Giancoli
One of Newton’s greatest contributions to physics is the concept of momentum and the argument that in closed systems momentum is always conserved. For him an object’s mass and velocity share equal importance when discussing the impact that objects will have on other objects when interactions occur.
Momentum p = m v
Momentum is a vector and we have to include direction.
Lets look at some simple review problems where objects collide and stick together.
1. A 10 kg cat runs with a speed of 8 m/sec into a stationary 20 kg bale of hay. Assume that the cat and hay are on a frictionless surface. Find the resultant speed of the two.
Conservation of momentum
mcat vcat = _______________
vfinal = _______________
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2. A 10 kg cat comes up from behind with a speed of 8 m/sec and runs into a 20 kg bale of hay. The hay was already going 2 meters per second in the same direction Assume that the cat and hay are on a frictionless surface. And they stick upon hitting. Find the resultant speed of the two.
3. A 10 kg cat runs with a speed of 8 m/sec into a 20 kg bale of hay. The hay was going in the opposite direction with a speed of 8 meters per second. Assume that the cat and hay are on a frictionless surface and they stick when they collide. Find the resultant speed of the two.
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From a global view (closed system) momentum is always conserved. From an individual particle’s point of view that particle either gains or loses momentum depending on the magnitude of the interaction. He called this change in momentum Impulse.
Impulse on an object = (m v)
He also discovered that impulse on an object was equal to the combination of force and time of interaction.
Impulse on an object = F t
Lets look at examples of impulse.
4. Return to problem 1 in this section and find the magnitude of the impulse on the cat.
5. Return to problem 1 in this section and find the magnitude of the impulse on the bale of hay.
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6. If the collision in problem 1 lasted one tenth of a second, how much force did the cat experience?
7. Return to problem 3 in this section and find the impulse on the cat.
8. If the collision lasted one tenth of a second how much force did the cat experience.
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Momentum and Impulse apply to all sorts of interactions between objects. An explosion is another sort of interaction.
9. Assume an explosion deep in space (no mentionable gravity) blows apart two masses. Mass A is 6 kg and goes upwards with a speed of 10 m/sec. How fast and what direction is the 2 kg Mass B going?
10. What is the impulse on mass B? (Show your work)
11. Which would you rather be hit by: the fastest pitched baseball (40 m/sec) or the fastest served tennis ball? To answer this question, determine the following quantities …the momentum of each ball the impulse delivered if the baseball stops dead on impact but the tennis ball rebounds at nearly the same speed.
Write your answers in the table below.
Quantity Baseball Tennis BallMass 145 grams 57 grams
VelocityMomentum
Impulse
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% 12. A problem for Americans — because Americans still think non-metric units are fine. Analyze this …Now, the bears I live with average, the males, eight to twelve hundred pounds. They're the largest bears in the world …. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach. Timothy Treadwell author of Among Grizzlies. The Late Show with David Letterman. NBC. 20 February 2001.
A. Mr. Treadwell states two different top speeds for the grizzly bear — one directly and one indirectly. Are they effectively the same? Explain.
%B. Pick whichever speed you wish and compute the momentum of a Grizzly bear
(in SI units) using the average mass quoted by Mr. Treadwell.
C. How fast would a 200 lb man have to run to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)
D. How fast would a 2000 lb car have to drive to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)
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% Read this passage about one of the difficulties of interstellar travel., the faster we go, the more difficult it is to avoid collisions with small objects and the more damage such a collision will wreak. Even if we are fortunate enough to miss all sizable objects, we can scarcely expect to miss the dust and individual atoms that are scattered throughout space. At two-tenths of the speed of light, dust and atoms might not do significant damage even in a voyage of 40 years, but the faster you go, the worse it is — space begins to become abrasive. When you begin to approach the speed of light, each hydrogen atom becomes a cosmic ray particle and will fry the crew. (A hydrogen atom or its nucleus striking the ship at nearly the speed of light is a cosmic ray particle, and there is no difference if the ship strikes the hydrogen atom or nucleus at nearly the speed of light. As Sancho Panza: "Whether the stone strikes the pitcher, or the pitcher strikes the stone, it is bad for the pitcher.") So 60,000 kilometers per second may be the practical speed limit for space travel.
% Isaac Asimov. "Sail On! Sail On!" The Magazine of Fantasy & Science Fiction, Vol. 72 (February 1987):
%% 4. The density of the interstellar medium is about one hydrogen atom per cubic
centimeter. Imagine a 1000 tonne (1 tonne is 1000 kg), 4 by 6 meter, classroom-sized interstellar spacecraft traveling at 60,000 km/s on its way to Proxima Centauri (the nearest solar system to our own (4.0 E13 km away)).
A. How long would it take our hypothetical spacecraft to complete its hypothetical journey?
impulse–momentumB. Determine the momentum of our spacecraft.
C. What mass of interstellar medium is swept up during the journey?
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D. What impulse does the interstellar medium deliver to the spacecraft?
E. How does this impulse compare to the momentum of the spacecraft?
work–energyF. Determine the kinetic energy of our spacecraft.
G. What is the effective drag force of the interstellar medium during the journey?
H. How much work does the interstellar medium do on the spacecraft?
I. How does this work compare to the kinetic energy of the spacecraft?
%%%
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conceptual
1.In older passenger cars, body panels were attached to a single frame around the perimeter, making them very rigid. This is known as body-over-frame construction. In newer cars, different body parts have stress-bearing elements within them and these parts are then welded to each other. This is known as unitized body construction. Repairing "unibody" cars after collision is comparatively difficult as stress (and thus damage) are distributed throughout the different parts. Why then are cars now built this way
2.To escape from a horrible fire, two people are forced to jump from the third story of a burning building on to solid concrete. Which person is more likely to sustain serious injuries: the jumper who comes to an abrupt halt when he lands or the jumper who bounces after impact?
numerical
1.When hit, the velocity of a 0.145 kg baseball changes from +20 m/s to −20 m/s. What is the magnitude of the impulse delivered by the bat to the ball?
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2. A rubber ball of mass 0.025 kg traveling at 4.0 m/s down strikes the floor and bounces straight up at 2.0 m/s. Find the magnitude of the impulse that the floor gave to the ball.
3. A model rocket has mass of 1.5 kg. The engine exerts an effective upward thrust of 120 N for 3.2 seconds. (Assume a negligible amount of air resistance while the rocket is ascending.) Determine …
A. the net force on the rocket
B. the net impulse on the rocket
C. the speed of the rocket when the engine stopped
D. the height of the rocket above the ground when the engine stopped
E. After the engine shuts down, the rocket is still moving upward. What maximum height above the ground did the rocket reach?
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Momentum in Two Dimensions
Working with collisions that happen in two dimensions is similar to what we have done in the past. The major difference is that we must keep track of both horizontal and vertical aspects of momentum.
1. Imagine that two masses A (10 kg) and B (4 kg) are moving horizontally with a speed of 2 meters per second. Then, while they are moving a small explosion pushes then apart. Object A ends up going upwards with a speed of 2 m/sec. Find the downward speed of B. (assume positive is upwards and forwards
Initially the vertical momentum of the combined blob is? ___________In the vertical direction conservation of momentum looks like
Conservation statement Before = After
General equation 0 = ma va + mb vb
Substitutions 0 = _____ _____ + _____ v b
vb in the vertical direction is __________________
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2. In the previous problem the final speeds of the two blobs are actually greater than 2 m/sec and 5 m/sec. This is because the true speeds are combinations of horizontal and vertical components.
Magnitude(Vel A)2 = 22 + 22
Vel A = _________
Direction Tan = opp / adj
Tan = 2/2
= _________
(Vel B)2 = 22 + 52
Vel A = _________
Direction Tan = opp / adj
Tan = 5/2
= _________
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3. Two masses collide and stick as shown below. Initial conditionsmass A (20 kg) is moving to the right with a speed of 12 m/sec mass B (8 kg) is moving upward with a speed of 20 m/sec
Find the final magnitude and direction of the combined mass.
Horizontal Vertical Before = After Before = After
MaVa + MbVb = ( Ma+ Mb)Vbf MaVa + MbVb = ( Ma+ Mb)Vbf
____ ____ + ____ ____ = ( _____ + _____) Vbf _____ _____ + _____ _____ = ( _____ + _____) Vbf
Vbf giruz = __________ Vbf vert = __________
Recombine to create the resultant vector
V2vert + V2
hor = V2resultant
_________ + _________ = V2resultan
Vresultant = ____________
Tan = Vver / Vhor = _____ / _____
= ______
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3. Two masses collide and stick as shown below.
mass A (20 kg) is moving with a speed of 12 m/sec mass B (8 kg) is moving with a speed of 20 m/sec
They hit and stick, find the magnitude and direction of the velocity of the combined mass.
Va horiz = __________
Va vert. = ___________
Vb horiz = __________
Vb vert. = ___________
Momentum in the Vertical Momentum in the Horizontal
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4. Imagine you are playing billiards. The white ball starts in motion and hits a stationary black ball. Each ball has a mass of 0.4 kg Find the magnitude and direction of the velocity of the white ball.
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Center of MassOf you look at a the generic conservation of momentum law it looks something like this
Before After
ma va + mb vb + mc vc + . . . = ma va + mb vb + mc vc + .
Momentum is (m v) has units of kg m /sec as expressed below
kga m/sa + kgb m/sb + kgc m/sc + . . . = kga m/sa + kgb m/sb + kgc m/sc
If both sides of the equation are multiplied by time, the equation reduces to
kga ma + kgb mb + kgc mc + . . = kga ma + kgb mb + kgc mc + . . .
Let us return to the generic form of the above equation where distance is given by the letter r
ma ra + mb rb + mc rc + . . . = ma ra + mb rb + mc rc + . If you combined all of the term on the before side the equation reduces to
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Mtotal rCofM = ma ra + mb rb + mc rc + . .
The above equation is the statement of center of mass for a system.
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1. Let us find the location of the center of mass of a barbell composed of two equal masses (3 kg each) separated by a distance of 2 meters.
A
B
C
2. The center of mass of a system is the point at which the system will balance. Find the balance point for the following:
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3. A mallet is composed of a cylindrical wooden handle and a wooden head. The
handle is 22 cm long and has a mass of 0.5 kg. The wooden head has a diameter of 8 cm and a mass of 2 kg. Find the center of mass of the mallet.
4. Objects orbit around their center of mass. This explains why there is a high tide on the side of the earth that is facing away from the moon. Find the center of mass for the Earth and Moon. Mass moon is 7.26 E22 kg Mass of Earth is 5.97 E 24 kg
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Find the center of mass in the following situations:5.
6.
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A girl stands on the end of a barge. The bottom of the barge is on a frictionless surface and the barge is free to move. She starts running toward the dock at the other end of the barge. It takes her 2 seconds to reach the other end of the barge. Answer the following:
8. What is the location of the original center of mass of the system?
9. How fast is she moving relative to the barge?
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10.How fast is she moving relative to the dock that was at the end of the barge?
11.How far away from the dock will she be when she reaches the end of the barge?
12.When she stops at the end of the barge, what happens to the barge?
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Chapter 3 Springs Giancoli Ch 6
In this section we will be investigating the nature of springs. Pick up a sample spring from the front of the classroom and hang it from a fixed horizontal rod. Place a hanger at the end of the spring and measure the distance to the ground (or reference point)
Fg Do Df Disp (X)
1 Newton
2 Newtons
3 Newtons
4 Newtons
5 Newtons
6 Newtons
Graph the line and determine its slope.
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The relationship between spring restorative force and its stretch is given by Hook’s Law.
Fsp = - k x
“k” is called the spring constant and it is a measure of the stiffness of the spring. Notice that Hook’s law is stated as a negative. This is because the direction a spring acts in always in opposition to the displacement of the spring. Pull on a spring and it pulls back. Push on a spring and it springs back. Lets look at several places where springs are used.
1. Explain how you would find the spring constant of a short length of surgical tubing.
2. Find the spring constant of a piece of surgical tubing.Data: Eq: Ans:
3. Find the spring constant of twice the length of the same tubing.Data: Eq: Ans:
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4. A retractable ball point pen has a spring inside of it. Find the spring constant of this spring.
Data: Eq: Ans:
5. A car has four springs, one for each wheel. They are used to cushion the car’s ride. When a 100 kg person sits down on inside a car the car settles down 1 cm.
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Energy stored in a spring
Energy can be stored in springs. The stored energy in a spring is given by the following equation.
PEsp = ½ k x2
Imagine you have a spring that has a spring constant of 500 n/meter is being stretched one cm at a time. Graph the energy stored in the spring.
Notice that most of the energy stored in a spring is stored in the last part of its stretch.
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Springs and energy problems.
1. A 10 gram pellet is fired from a pellet gun with a speed of 50 m/sec. The gun’s energy comes from a spring that is stretched 10 cm. What is the spring constant of the spring?
2. A 2 kg block is pushed back against a spring. The spring compresses 20 cm. If the spring constant is 300 n/m, how fast will the block be going when it is released?
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Given the picture below, find the following for a 4 kg block at the top of a 10 meter high hill:
3. The speed of the block at the bottom of the hill.Equation Substitutions Answer
4. The magnitude of the spring constant needed to stop the block in a distance of 1 meter.
Equation Substitutions Answer
5. The acceleration of the block when the spring has compressed 0.25 meters.Equation Substitutions Answer
6. The acceleration of the block when the spring has compressed 0.5 meters.Equation Substitutions Answer
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Continued from previous page
7. The acceleration of the block when the spring has compressed 0.75 meters.Equation Substitutions Answer
8. The acceleration of the block at the moment the block has stopped moving into the spring.
Equation Substitutions Answer
9. How fast is the block going when the spring is compressed by 0.5 meters?Equation Substitutions Answer
10. How much work has been done on the block by the time it has compressed the spring by 0.5 meters?
Equation Substitutions Answer
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11. A spring with a spring coefficient of 100 is compressed by a distance of 0.5 meters and placed between two blocks. Each block has a mass of 1 kg. When the blocks are released the spring decompresses. How fast are the blocks going afterwards?
Conservation of Energy Conservation of Momentum
12. A spring with a spring coefficient of 100 is compressed by a distance of 0.5 meters and placed between two blocks. One of the blocks (A) has a mass of 120 grams, and the other block (B) has a mass of 200 grams. When the blocks are released the spring decompresses. How fast are the blocks going afterwards?
Conservation of Energy Conservation of Momentum
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13.A 10 kg block is pushed with a force of 50 Newtons into a spring that has a spring constant of 62.5 and then released. The block slides away and runs into another block which has a mass of 8 kg. The two blocks stick together. Find the following:
A. The distance the spring compresses.
B. The initial acceleration of the spring when the block is released
C. The total energy stored in the compressed spring system
D. The velocity of the block upon leaving the spring
E. The velocity of the combined blocks after the collision
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Problem 13 continued
F. The impulse experienced by the 10 kg block during the collision
G. If the collision lasts 0.2 seconds what average force did the 10 kg block experience during the collision?
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14.A 10 kg block is moving around a loop the loop. At the top it has a speed of 8 m/sec. The loop has a radius of 5 meters.
Find the following:A. The centripetal acceleration at
the top of the loop.
B. The apparent weight of the block at the top (Froad)
C. The velocity at the bottom of the loop.
D. The coefficient of friction needed to stop the block in 5 meters distance.
E. The maximum acceleration of the block while in contact with the spring if the spring has a “k” of 250.
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15.Two masses (mass A is 12 kg, mass B is 3 kg) are stationary on a frictionless flat surface. Between them is a spring that has been compressed 10 centimeters. The spring has a spring constant of 200. The spring expands sending the two blocks in opposite directions. Find the velocity of each.
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Chapter 4 Density
Terms we will be using
Terms Sub-units M-K-S units
Mass m = kg
Surface area A = m2
Volume V = m3
Surface density (mass/area) = kg/m2
Volumetric density (mass/volume) kg/m3
Pressure (Force/Area) Pascal = N/m2
Pressure (Energy/volume) Joules/meter3
VolumeOne of the most important concepts to understand in if we are going to be living in and studying in a three dimensional world is volume. You have been introduced to this concept numerous times in math classes and introductory science classes. We will quickly review the basics. In the M-K-S (Meter, Kilogram, Second) version of the metric system the unit of distance is the meter. Volume starts with this basic unit cubed. This is fairly large so we often break it down into smaller units. The cubic decimeter – often called the liter – is a common one.
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Conversions
Cubic meters Cubic Decimeters Cubic Centimeters Cubic Millimeters ( m3 ) ( liters ) (CCs or milliliters)
1 meter cubed ____________ ____________ ____________
____________ 1 liter ____________ ____________
____________ ____________ 1 cc ____________ ____________ ____________ ____________ 1 cubic millimeter
Lets do some quick measurements of volume. How would you find the volume of each of the following and what equipment would you need?
This classroomExplain the method
Its approximate volume is _________________ meters3 ± __________m3
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Finding the volume of your lungsExplain the method
Its approximate volume is _________________ meters3 ± __________m3
An irregular granite rockExplain the method
Its approximate volume is _________________ meters3 ± __________m3
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Real Life Measurements
Length is a measure of distance. We measure length in meters. Your ability to approximate length is important.
Guess Actual length
The length of your pencil ______________ ________________
Your height ______________ ________________
The dimensions of this room
East West _______________ ________________
North-South _______________ ________________
Ground-Ceiling _______________ ________________
Volume is the amount of space that is occupied or enclosed. We measure volume in cubic meters. The liter is 1/1000 of a cubic meter. It is equal to a cube that is 10 cm on a side.
1. How many gallons are in one cubic meter?
Rough guess ____________________
We will use ratios
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2. What is the volume of a basketball?
Rough guess ____________________
A standard basketball has a radius of 12 cmThe volume of a sphere is 4/3 r3
= ____________________ meters 3
3. What is the volume of the cubic zirconium sphere that your teacher is holding?
Rough guess ____________________
A standard basketball has a radius of 12 cmThe volume of a sphere is 4/3 r3
= ____________________ meters 3
4. What is the approximate volume of this room?
Rough Guess ____________________
(East West) x ( North South) x (height)
__________ x _____________ x ____________ = __________________ m3
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Density
Density is a measure of how much material is found in a given space. We will deal with two forms of density; surface density and volume density.
Surface density = kg/m2
Below are pictures of 1 kg disks placed evenly on table tops. The first table is 16 meters by 16 meters. Find the density of the material in each figure.
Figure 1 Figure 2 Figure 3
Density ______________ Kg/m2 ______________ Kg/m2 ______________ Kg/m2
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Volumetric density kg/m3
Below are pictures of 64 one kg spheres placed evenly inside of cubes. The first cube is 16 meters on a side. The second is 8 meters on a side, The third is 4 meters on a side. Find the density of the material in each figure.
Figure 1 Figure 2 Figure 3
Density ____________ Kg/m3 ____________ Kg/m3 ___________ Kg/m3
Density ____________ Kg/liter ____________ Kg/liter ____________ Kg/liter
Density ____________ Kg/cc _ ___________ Kg/cc __________ Kg/cc
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Below are the densities of several common materials. If you were to have a cube of each. How big would it have to be to have a mass of 1 kg? (remember that on the Earth one kilogram will have a weight of approximately 2.2 lbs)
Density Volume of 1 kg Volume of 1 kg Volume of 1 kg Meters cubed Liters CCs
Air 1.2 kg/m3 ______________ _______________ _________________
Water 1000 kg/m3 ______________ _______________ _________________
Aluminum 2.7 gr/cm3 ______________ _______________ _________________
Granite 2.75 gr/cm3 ______________ _______________ _________________
Steel 8 gr/cm3 ______________ _______________ _________________
Mercury 13.5 gr/cm3 ______________ _______________ _________________
Uranium 19.1 gr/cm3 ______________ _______________ _________________
Gold 19.3 gr/cm3 ______________ _______________ _________________
Granite Mercury Gold
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Volumetric Density
= Mass / Volume Units are Kg/m3
1. What is the approximate mass of the air in this room?
Rough guess ______________________
Equation Substitutions Answer _________________
2. What is the mass of a block of goldthat is the size of a Xerox paper box?
Rough guess ______________________
Equation Substitutions Answer _________________
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3. What is the density of the block of foam that your teacher is holding?
Rough guess ______________________
Equation Substitutions Answer __________________
4. What is the density of the cubic zirconium ball that your teacher is holding?
Rough guess ______________________
Equation Substitutions Answer __________________
5. Find the following information about the rock sitting on Mr. Stimson’s desk
Volume _______________________
Mass _______________________
Density _______________________
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Chapter 5: Pressure
Pressure is a measure of how much force is acting at a representative surface.
Pressure = Force / Area
Pascal = 1 Newton / meter2
PSI = 1 lb/in2 1 pascal = 0.000145037738 PSI
We are going to do a lot of interesting stuff with pressure. However, before we do, we need to get some practice dealing with pressure. Lets look at some situations.
1. At sea level the pressure of the atmosphere acting downward on a one square meter section of pavement is 101325 Pascals ( Newtons / square meter )
a. How much total force is this?
_______________________b. How many kilograms of air are there above the 1 square meter
section of pavement? ______________________
2. Your sister is of average size, 65 kg. Normally she wears tennis shoes. Each shoe has a bottom surface area of 116 cm2. What average pressure does she exert on the ground when she stands on one foot?
Your sister’s weight _______________ Newtons
The pressure is she exerting on the ground while standing on one foot
________________ Pascals
________________ PSI
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Your sister buys a cool pair of Stiletto high heels. The stiletto heels get very narrow where they hit the ground. The bottom surface is only 0.3 cm2. Your sister is of average size, 65 kg. When she walks she places all of her weight on the heel. Answer the following:
Your sister’s weight _______________ Newtons
The pressure is she exerting on the ground while standing on one foot
________________ Pascals
________________ PSI
BTW, when Disneyland first opened to the public in 1955. Stiletto heels were in fashion. The weather was hot. And, the asphalt was fresh. Women were leaving deep marks in the asphalt. Some had their shoes get stuck in the asphalt.
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3. The neat thing about liquids is that if there is not a substantial elevation difference. The pressure at one location is the same as the pressure at all other locations. This is how a hydraulic lift works. Below is a picture of a system with a 10 kg mass on top of a small diameter piston. Pressure caused by this mass translates into a substantial lifting force acting on the larger piston.
a. What is the weight of the 10 kg mass?
___________________
b. What is the pressure created by the mass
___________________
c. What is the upwards force acting on the large piston?
___________________
d. How many kg of material must be on the larger piston to keep it from moving upwards?
___________________
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4. Remember when we determined the mass of my car by pushing it and measuring the car’s acceleration? Well, it turns out that you can determine the car’s mass by measuring the contact area between the wheels and the ground. This can be done if you know the air pressure in the tires. Lets try this out. What we will do is to lift one side of the car and place carbon paper underneath the wheels. Then we will lower the car down.
a. What is the contact area of the front wheel? ________________
b. What is the contact area of the back wheel? ________________
c. What is the pressure in the tires? ________________
d. What is the mass of the car? ___________________
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Energy and Fluids
In the past we studied energy as it related to particles. There was kinetic energy (KE) which is a measure of the energy in a particle’s motion and gravitational potential energy PEg which is a measure of the energy stored in a particle’s elevation. The equations for these two forms of energy are
PEg = m g h
KE = 1/2 m v2
With fluids it is often impossible to measure the total energy in a fluids system because one does not know exactly how many particles there are in the system. For example, how do we measure the total energy in the Mississippi River system? Fortunately, we can do a lot of energy stuff , even if we don’t know about every drop in the stream. What we do need to know is the energy density of the fluid.
Energy of Energy densityparticles of fluids
PEg = m g h PEg/vol = g h
KE = 1/2 m v2 KE/vol = 1/2 v2
Let us start by looking at PEg/vol (Pressure) Below are three glass tubes. Each has a certain amount of water in it. The pressure of the water at the bottom is due only to the height of the water above. PEg/vol = g h
Pressure due to water = ____________, _____________, _____________
Absolute pressure = ____________, _____________, _____________
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5. Lets combine what we have just learned with some older material. Imagine that your neighbor has a tall water tank (shown below) She is bored and decides to shoot it up with her assault rifle. She fires five holes into the tank at different heights. Find the following for each hole: the pressure (Energy/vol), the velocity at which the water leaves (KE/vol), and the distance away from the tank at which each stream of water lands.
We do not need to find absolute pressure – can you explain why?
Hole height H20Pressure Velocity Distance
6 m _________________ __________________ _________________
5 m _________________ __________________ _________________
4 m _________________ __________________ _________________
3 m _________________ __________________ _________________
2 m _________________ __________________ _________________
1 m _________________ __________________ _________________
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6. At the surface of the ocean you are experiencing a pressure of 1 atmosphere. As you go deeper into the ocean, the weight of the water above you increases. Consequently you feel greater pressure. Water has a density of 1000 kg/m3. How far down do you have to go for the total pressure to be equal to 2 atmospheres?
Hint: one atmosphere is 101325 Pascals
_________________ meters
_________________ feet
7. The Marianna Trench is the deepest point in the pacific ocean. It is 10,924 meters deep (35,840 feet). What is the pressure at the bottom of the trench?
a. __________________ Pascals
b. __________________ PSI.
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8. The street water pressure to the average house is 280,000 pascals (40 psi.) above air pressure The pressure gets less as you go from the first to the second floor. If it is 3 meters from the ground to the shower on the second floor. Find the following
a. The pressure of the water at the ground level in Pascals ___________________
b. The pressure of the water at the shower head in Pascals ___________________
c. The maximum height that water can be pumped to
___________________
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Above is a sphere with particles bouncing around inside. Pressure on the sphere is related to the outwards force of the collisions of the myriads of molecules that are inside of the sphere. But, moving molecules have KE. This implies that there must be a relationship between pressure and energy. Below we will be answering these questions in a not so scientific but very effective way.
Unit matching – An effective tool in science
In college it was common practice among students, when you were up against a wall and could not solve a problem, to reverse engineer through units. The procedure went like this. Look at the units of the answer that you are trying to find. Next, look at the units of what you are given. Then manipulate the units so that they end up in the form of those found in the answer. This approach bothered me. For, while it consistently worked, it demonstrated no real knowledge of the underlying physics. This being said, I have found unit matching a valuable problem solving method. The following is a good example. We have been studying pressure. So, is pressure more like energy or more like force? Lets look at the units and how they came about.
Pressure Constituent Units
Pascals Kg / ( m sec2 )
We will derive them on the next page using Newton’s second law and then see how they are related to energy.
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Pressure in terms of Force
We know that Force = m a Newtons = kg (m/sec2)
We can view pressure in terms of force
Pressure = Force / Area
Pascals = (Newtons) / ( m2 )
Pascals = (Kg m/sec2) / m2
Pascals = Kg / ( m sec2 )
From this point of view Pressure acts as a force if we are looking at its effects on a surface..
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Pressure in terms of Energy
We know that PEg = m g h
So, Joules are kg m/sec2 m
Or Joules are Kg m2 /sec2
Below we shall see how Force/Area is the unit equivalent of Energy/Vol
Force / Area Energy / Volume
Newtons / meter2 (Joules) / ( m3 )
Kg (m/ sec2) / ( m2 ) = (Kg m2/sec2) / ( m3 )
Kg / ( m sec2 ) = Kg / ( m sec2 )
Basically,
Pressure is Force if we are looking at its effect on surfaces.
But,
Pressure is Energy if we are looking at its effect on volumes.
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Lets see how we can use the pressure – energy relationship in problem solving. To the left is a picture of a high pressure gas cylinder. We decide to fill it up with air. How much energy is stored in the cylinder?
Volume Max Pressure
English 2640 ci 3600 psi
Metric 0.0433 m3 2.48 E 7 J/m3
Remember that the pressure is a measure of Energy per unit volume.
Pressure = Energy / volume or
Energy = (Pressure) x ( volume)
Energy = ____________ J/m3 _______________ m3
Energy = ______________ Joules
This sounds like a lot of energy but it compares to only 1/8th of a cup of gasoline.
Inside closed systems, Energy is conserved. If we slowly empty the cylinder into a large mylar balloon how much volume would it fill when the pressure is finally reduced to one atmosphere.
Before (in cylinder) = After (in balloon)
___________ = ( Pressure ) ( Volume )
___________ = ___________ J/m3 ( Volume )
Volume = ________________ m3
this would fill a spherical balloon with a radius of ______________ meters.
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Lets talk about waterAn Incompressible fluid
In 1738 Daniel Bernoulli published a conservation of energy law for fluids. It came to be known as Bernoulli’s principle. Basically he figured that fluids held energy in three different ways: Gravitational, Kinetic, and Pressure.
PEg = m g h for a static material
g h per unit volume is measured in kg/m3
KE = 1/2 m v2 static material
= 1/2 v2 per unit volume is measured in kg/m3
Pressure= for all fluids
Below is a simplified version of Bernoulli’s principle. Essentially it is the law of conservation of energy applied to liquids. It works any incompressible fluid, as long as there is little or no friction or drag (Wout).
Before = After
PEgo + KEo + Po = PEgf + KEf + Pf
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1. Lets start with a very simple problem. We will be looking at a one-liter chunk of water. We will see what happens to it as it flows through pipes of varying sizes.
Initially the water is flowing through pipes at a rate of 1 liter per second (1 kg per second). The nominal diameter of the pipes is 4 cm (0.04 meters) at a pressure of 400,000 N/m2. Answer the following:
What is the cross-sectional area of the pipe?
____________________
Remember: 1 liter is one thousandth of a cubic meter.
What is the linear velocity of the water? Hint: Vol = Vol
(Area) (distance) (Area) (x) = (1E-3 m3)
____________________
What is the energy per unit volumeFor the water in the system?
____________________ Hint: P + KE = Energy
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The one liter slug of fluid encounters a constriction in the pipe. The pipe narrows to a diameter of 0.02 meters.
What is the cross-sectional area of the constriction?
____________________
What is the linear velocity of Hint: Vol in big pipe = Vol in small pipethe water?
____________________
What is the new pressure at theconstriction point?
____________________
Hint: Po + KEo = Pf + KEf
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2. On the previous page the water pressure in the pipe determined the height of the water in the stand pipes. The higher the water level, the greater the pressure. How high would the water be in each pipe
Po = PEgo
Water height above _________________ = g h the larger pipe
_________________ = _________________ ______ h
ho = _____________________________
Pf = PEgf
Water height above _________________ = g h the narrower pipe
_________________ = _________________ ______ h
ho = ______________________________
If a hole were to be drilled into the top portion of the pipe, water would shoot out and attain a maximum height calculated in this problem.
For an applet showing what tube constriction does to the flow of fluids go to http://library.thinkquest.org/27948/bernoulli.html
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The following problem was taken from the free response section of the 2008 AP Exam. Show your work on the back of the previous page.
An underground pipe carries water of density 1000 kg m3 to a fountain at ground level, as shown above.
At point A, 0.50 m below ground level, the pipe has a cross-sectional area of 1.0 -4 m2.
At ground level, the pipe has a cross-sectional area of 0.50 E-4 m2.
The water leaves the pipe at point B at a speed of 8.2 m/s.
(a) Calculate the speed of the water in the pipe at point A.
_______________________
(b) Calculate the absolute water pressure in the pipe at point A.
_______________________
(c) Calculate the maximum height above the ground that the water reaches. Assume air resistance is negligible and that the water goes straight up.
_______________________
(d) Calculate the horizontal distance from the pipe that is reached by water exiting the pipe at 60 degree from the horizontal. Assume air resistance is negligible.
________________________
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3. (From 2008 AP Exam) A fountain with an opening of radius 0.015 m shoots a stream of water vertically from ground level at 6.0 m/s. The density of water is 1000 kg m3 Show your work on the back side of the previous page.
(a) Calculate the volume rate of flow of water.
________________________
(b) The fountain is fed by a pipe that at one point has a radius of 0.025 m and is 2.5 m below the fountain’s opening. Calculate the absolute pressure in the pipe at this point.
________________________
(c) The fountain owner wants to launch the water 4.0 m into the air with the same volume flow rate. A nozzle can be attached to change the size of the opening. Calculate the radius needed on this new nozzle.
________________________
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4. (From 2007 AP Exam) The large container shown in the cross section above is filled with a liquid of density 1.1E3 kg/m3. A small hole of area 2.5E-6 m2 is opened in the side of the container a distance h above the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. The amount of liquid collected in the beaker in 2.0 minutes is 7.2E-4 m3.
(a) Calculate the volume rate of flow of liquid from the hole in m3/sec.
____________________
(b) Calculate the speed of the liquid as it exits the hole.
____________________
(c) Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b).
____________________
(d) Suppose that there is now less liquid in the container so that the height h is reduced to h/2. In relation to the beaker, where will the liquid hit the tabletop?
____ Left of the beaker ___ Right of the beaker ___ Into the beaker
Justify you answer.
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Physics of flight
Bernoulli’’s principle is directly involved in aviation. It is central to the physics of wings and propellers. The exact calculations are very difficult and require very sophisticated mathematics. However, a first approximation is relatively easy to make and surprisingly accurate.
We start by assuming that air that makes up our atmosphere acts like an incompressible liquid when something slices through it. Imagine our atmosphere sort of like a giant jello mold. If a knife blade were to be passed horizontally through the jello atmosphere, jello will be temporarily displaced by the blade. It cannot go downward because the ground is below and we are assuming that air will not compress. All of the air above gets pushed upward only to fall back down on top of the same lower block of air after the blade goes by.
Notice how each vertical section of air moves up as the blade passes by.
The picture above shows what is happening from the point of view of a person who is stationary with the atmosphere. From the blade’s frame of reference, the blade is stationary and the blocks of air are moving from the left side of the page to the right side.
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Below is a picture of a typical airfoil. Notice that the distance from front to back over the top of the airfoil is greater than the distance below (y > x). . Because both blocks of air will come back together after the foil has moved past, the top block of air must have a greater apparent surface speed (Vtop > Vbottom) than the block of air on the bottom.
The fact that Vabove is greater than Vbelow has great consequences on the foil. Remember that the conservation laws of physics must always be obeyed. This includes Bernoulli’s principle. Let us look at a representative block of air that will pass underneath the airfoil. It has both KE and Pressure. Because the speed of the block remains the same below the foil the air pressure remains the same as well.
Below the airfoil
KEo + Po = KEbottom + Pbottom
If we look at a block of air passing over the top of the foil we see that the KE increases over the airfoil. There is no free lunch in physics. For energy to be conserved the gain in KE must be balanced by an equal loss in pressure.
Above the airfoil
KEo + Po = KEabove + Pabove
Notice that there is less downward pressure on the wing from above than there is upward pressure from below.
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Air pressure acts on all surfaces of the foil. Because there is less pressure acting down on the top surface of the foil than acting upward on the bottom of the foil, there is a net upward push.
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Sample problem: Imagine you own a private airplane. The physical specifications of the plane are given below.
%Wingspan = 30.0 feet (9.2 m)
Wing area: 160 sq ft (15.14 m2)
Distance X = 1.65 m
Distance Y = 2.145 m
Empty weight: 1201 lb (544 kg)
Loaded weight: 2150 lb (975 kg)
There is a minimum speed that you must have for the plane to lift off from the ground. On the next pages we are going to do the physics necessary in order to calculate the minimum takeoff speed of your airplane. Here we go.
We start with that air particles cross the top surface of the wing in exactly the same amount of time as it takes for air molecules to pass underneath the bottom surface of the wing.
(A) Remember that the geometry of the wing determines how fast the air must move on the top and bottom surfaces of the wing. Where the air moves more quickly the kinetic energy of the air is greater and the potential energy (air pressure) is less. Let us start by determining the relative air speeds. Remember that air traveling over the top of the wing must reach the trailing (back) edge of the wing at the same time as air that moves under the wing.
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Time above = Time belowSince D = v t
Dabove / Vabove= Dbelow / Vbelow
Solve for Vabove in Terms of Vbelow
V above = ( ) Vbelow
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Write Bernoulli’s equation for air flowing over the top of the wing vs. air flowing underneath the wing.
Energy Below Wing = Energy Above Wing
Pbelow + KEx = Pabove + KEy
(B) Wings provide lift because the bottom surfaces have a larger air pressure pushing them upward than the top surfaces feel pushing downwards. We can find this difference in pressure by bringing all the pressures to the left side of the equation and the KE’s to the right.
Pbelow - Pabove = KEy - KEx
P (N/m2) = 1/2 vabove2 - 1/2 vbelow
2
(C) The density of dry air at sea level is = 1.2 kg/m3
P (N/m2) = 1/2 (1.2) vabove2 - 1/2 (1.2) vbelow
2
P (N/m2) = ________ ( vabove2 - vbelow
2 )
(D) We can now find how much lift you get per square meter of wing in terms of the plane’s speed. Substituting for Vabove we get
P (N/m2) = _____ ((____ vx) 2 - vx2 ))
P (N/m2) = _____ (vx) 2 (___ - 1 )
P (N/m2) = ________ ( vx2 )
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(E) P is the pressure difference between the top and bottom of the wing. Since pressure is measured in force per unit area, since we know the total area (bottom surface) of the wing we can find the amount of lift at any speed.
P = Force / surface area
________ = Forcelift / __________
Forcelift = ___________
(F) When fully loaded the plane has a mass of 975 kg. What is its weight in Newtons?
Fg = m g
Fg = _________ _________
Fg = _________
(G) Calculate the minimum take off velocity for a fully loaded plane.(should be somewhere near 40 m/sec)
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Sample car problem:
Below is the Stimson mobile. It is an airfoil of sorts. Because of this it appears to get lighter at higher speeds (not necessarily a good thing) When the car is moving at 40 m/sec, how much lighter will it be? And, should we worry?
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1. Another plane problem. The MIT human powered plane the Daedalus weighs only 69 lbs. Fully loaded it is about 117 kg. In 1988 it flew the 199 km from the Iraklion Air Force Base on Crete in the Mediterranean Sea, to the island of Santorini. What minimum speed does it need to be able to fly?
The wing has a span of 34 meters and a surface area of 31 m2
The average underside wing width is 1 meter
The average topside width is 1.35 meters
Answer ________________m/sec _________________ mph
Show you work on the back side of this page
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2. Below is a picture of a spin ball which is moving toward the right side of the page. Using you knowledge of the Bernoulli principle, draw the direction of the net force acting on the ball. Explain your answer.
.
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3. You have probably been into stores during the summer where there is a beach ball floating up above a fan. This is usually done as an advertisement for fans. Not only is it an effective advertisement for fans but it is also a great demonstration of physics. Explain.
Notice that the ball still floats even when
the fan is at an angle
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We need to pause for a moment. While it is true that airfoils use Bernoulli’s principle for a substantial portion of their lift, it is also true that the attack angle of a wing has an enormous impact on lift. Angle of attack works on the concept of Impulse (remember conservation of momentum?)
.
Diverting the mass of the air flow downward places an upward force on the wing. In the next section three we shall review the one dimensional studies of momentum and extend them to two and three dimensions.
For a more complete explanation of the physics of flight go to
http://www.amasci.com/wing/airfoil.html
http://www.av8n.com/how/htm/airfoils.html
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Buoyant Forces
Have you ever wondered why some objects float and other objects sink? For that matter, have you ever wondered big ships float when they are made out of steel, a material that sinks in water? The math for determining is very simple understanding.
Objects sink into water until the mass of the water they displace is equal to the object’s own mass. It is all a matter of density (sometimes known as specific gravity. In the metric system the specific gravity of water is measured in one of three ways
1000 kg/m3 or 1 kg /decimeter3 or 1 gram per cm3
1 kg / liter 1 gr/ml
Below are objects that will be placed into water.
1. Balsa wood has a density of 110 kg/m3. A block of balsa (one meter on a side) is placed in water. How deep will it sink in the water?
2. Oak has a density of 770 kg/m3. A block of oak (one meter on a side) is placed in water. How deep will it sink in the water?
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3. Steel has a density of 7800 kg/m3. A block of steel (one meter on a side) is placed in water. How deep will it sink in the water?
4. You have a sheet of steel that is 0.002 meters thick. The steel is folded into the shape of a cube with the top side missing. The dimensions of the cube are one meter by one meter by one meter.
a. What will the mass of the hollow cube be?
b. How deep will the cube sink into the water?
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Buoyant force = the weight of the displaced fluid
5. Let us revisit the one cubic meter block of balsa wood from problem 1. a. Calculate the buoyant force on the block.
Picture Equations and work
b. If the block is pulled completely under water by a rope force the rope must exert.
Picture Equations and work
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6. Let us revisit the one cubic meter block of oak from problem 2a. Calculate the buoyant force on the block.
Picture Equations and work
b. If the block is pulled completely under water by a rope calculate the force the rope must exert.
Picture Equations and work
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7. Let us revisit the one cubic meter block of steel from problem 3a. Calculate the buoyant force on the block.
Picture Equations and work
b. Calculate the net force on the block. Picture Equations and work
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