1. What is the relevance of Physics in day to day activities?
2. How importance is understanding Physics?
3. If you are traveling in an airplane and dropped a ball where will it hit? why?
4. A large rock is dropped from a bridge into the river below.
a) if the time required to fall is 1.7 seconds, with what velocity, in m/sec, does it hit the water? b) what is the height, in meters, of the bridge above the water?
5. A car possesses 20 000 units of momentum. What would be the car's new momentum if its velocity were doubled.
6. Find the height (in m) from which you would have to drop your textbook so that it would have a speed of 14.2 m/s just before it hits the floor?
7. Differentiate work from power
8. What can make any work be more efficient?
9. If the population of the U.S. was 200,000,000, what was the average rate of energy consumption per person?
Tuesday, March 11, 2008
Tuesday, February 26, 2008
announcements
MARCH 2 and 9 deadlines i will be at the school for any clarifications
MARCH 16 - Finals first part
MARCH 23- Part II finals
MARCH 16 - Finals first part
MARCH 23- Part II finals
march 2
| If you push vigorously against a brick wall, how much work do you do on the wall? |
A person pushes a 10 kg cart a distance of 20 meters by exerting a 60 Newton horizontal force. The frictional resistance force is 50 Newtons. How much work is done by each force acting ont he cart? How much kinetic energy does the cart have at the end of the 20 meters if it started from rest:
An automobile of mass 1200 kg has a speed of 30 m/s on a horizontal road when the engine is developing 37300 Watts (50.0 horsepower). What is the speed, with the same power output, if the automobile now climbs a hill inclined at 30o?
In the early 1980's, the total consumption of electrical energy in the U.S. was on the order of 1 X 1019 joules per year.
a) What was the average rate of energy consumption in watts? kilowatts?
b) If the population of the U.S. was 200,000,000, what was the average rate of energy consumption per person?
c) If the sun transfers energy to the earth by radiation at a rate of 1.4 kW per square meter of surface, how great an area would be required to collect the energy cited above?
The human heart is a powerful and reliable pump. Each 24 hour day, it takes in and discharges over 7500 liters of blood. If the work done by the heart is equal to the work required to lift this amount of blood a height equal to the average American female (1.63 m), and if the density of blood is the same as that of water,
a) how much work does the heart do in a day?
b) what is the power output in watts? horsepower?
example of power
Problem 3.1:
An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip?
Solution for Problem 3.1
The work done by the elevator over the 100 meters is easily calculable: W = mgh = (1000)(9.8)(100) = 9.8×105 Joules. The total time of the trip can be calculated from the velocity of the elevator: t = 25 s. Thus the average power is given by: P == 3.9×104 Watts, or 39 kW.
An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip?
Solution for Problem 3.1
The work done by the elevator over the 100 meters is easily calculable: W = mgh = (1000)(9.8)(100) = 9.8×105 Joules. The total time of the trip can be calculated from the velocity of the elevator: t = 25 s. Thus the average power is given by: P == 3.9×104 Watts, or 39 kW.
power
Power
Mechanical systems, an engine for example, are not limited by the amount of work they can do, but rather by the rate at which they can perform the work. This quantity, the rate at which work is done, is defined as power.
Equations for Power
From this very simple definition, we can come up with a simple equation for the average power of a system. If the system does an amount of work, W, over a period of time, T, then the average power is simply given by:
=
It is important to remember that this equation gives the average power over a given time, not the instantaneous power. Remember, because in the equation w increases with x, even if a constant force is exerted, the work done by the force increases with displacement, meaning the power is not constant. To find the instantaneous power, we must use calculus:
P =
In the sense of this second equation for power, power is the rate of change of the work done by the system.
From this equation, we can derive another equation for instantaneous power that does not rely on calculus. Given a force that acts at an angle θ to the displacement of the particle,
P = = = F cosθ
Since = v ,
P = Fv cosθ
Though the calculus is not necessarily important to remember, the final equation is quite valuable. We now have two simple, numerical equations for both the average and instantaneous power of a system. Note, in analyzing this equation, we can see that if the force is parallel to the velocity of the particle, then the power delivered is simply P = Fv.
Units of Power
The unit of power is the joule per second, which is more commonly called a watt. Another unit commonly used to measure power, especially in everyday situations, is the horsepower, which is equivalent to about 746 Watts. The rate at which our automobiles do work is measured in horsepower.
Power, unlike work or energy, is not really a "building block" for further studies in physics. We do not derive other concepts from our understanding of power. It is far more applicable for practical use with machinery that delivers force. That said, power remains an important and useful concept in classical mechanics, and often comes up in physics courses.
Mechanical systems, an engine for example, are not limited by the amount of work they can do, but rather by the rate at which they can perform the work. This quantity, the rate at which work is done, is defined as power.
Equations for Power
From this very simple definition, we can come up with a simple equation for the average power of a system. If the system does an amount of work, W, over a period of time, T, then the average power is simply given by:
=
It is important to remember that this equation gives the average power over a given time, not the instantaneous power. Remember, because in the equation w increases with x, even if a constant force is exerted, the work done by the force increases with displacement, meaning the power is not constant. To find the instantaneous power, we must use calculus:
P =
In the sense of this second equation for power, power is the rate of change of the work done by the system.
From this equation, we can derive another equation for instantaneous power that does not rely on calculus. Given a force that acts at an angle θ to the displacement of the particle,
P = = = F cosθ
Since = v ,
P = Fv cosθ
Though the calculus is not necessarily important to remember, the final equation is quite valuable. We now have two simple, numerical equations for both the average and instantaneous power of a system. Note, in analyzing this equation, we can see that if the force is parallel to the velocity of the particle, then the power delivered is simply P = Fv.
Units of Power
The unit of power is the joule per second, which is more commonly called a watt. Another unit commonly used to measure power, especially in everyday situations, is the horsepower, which is equivalent to about 746 Watts. The rate at which our automobiles do work is measured in horsepower.
Power, unlike work or energy, is not really a "building block" for further studies in physics. We do not derive other concepts from our understanding of power. It is far more applicable for practical use with machinery that delivers force. That said, power remains an important and useful concept in classical mechanics, and often comes up in physics courses.
examples
Problem 1.1:
A 10 kg object experiences a horizontal force which causes it to accelerate at 5 m/s2, moving it a distance of 20 m, horizontally. How much work is done by the force? \
The magnitude of the force is given by F = ma = (10)(5) = 50 N. It acts over a distance of 20 m, in the same direction as the displacement of the object, implying that the total work done by the force is given by W = Fx = (50)(20) = 1000 Joules.
A 10 kg object experiences a horizontal force which causes it to accelerate at 5 m/s2, moving it a distance of 20 m, horizontally. How much work is done by the force? \
The magnitude of the force is given by F = ma = (10)(5) = 50 N. It acts over a distance of 20 m, in the same direction as the displacement of the object, implying that the total work done by the force is given by W = Fx = (50)(20) = 1000 Joules.
work
Definition of Work
Work, though easily defined mathematically, takes some explanation to grasp conceptually. In order to build an understanding of the concept, we begin with the most simple situation, then generalize to come up with the common formula.
The Simple Case
Consider a particle moving in a straight line that is acted on by a constant force in the same direction as the motion of the particle. In this very simple case, the work is defined as the product of the force and the displacement of the particle. Unlike a situation in which you hold something in place, exerting a normal force, the crucial aspect to the concept of work is that it defines a constant force applied over a distance. If a force F acts on a particle over a distance x, then the work done is simply:
W = Fx
Since w increases as x increases, given a constant force, the greater the distance during which that force acts on the particle, the more work is done. We can also see from this equation that work is a scalar quantity, rather than a vector one. Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account.
What are the units of work? The work done by moving a 1 kg body a distance of 1 m is defined as a Joule. A joule, in terms of fundamental units, is easily calculated:
W = Fx = (m) =
The joule is a multipurpose unit. It serves not only as a unit of work, but also of energy. Also, the joule is used beyond the realm of physics, in chemistry, or any other subject dealing with energy.
In dynamics we were able to define a force conceptually as a push or a pull. Such a concise definition is difficult to generate when dealing with work. To give a vague idea, we can describe work as a force applied over a distance. If a force is to do work, it must act on a particle while it moves; it cannot just cause it to move. For instance, when you kick a soccer ball, you do no work on the ball. Though you produce a great deal of motion, you have only instantaneous contact with the ball, and can do no work. On the other hand, if I pick the ball up and run with it, I do work on the ball: I am exerting a force over a certain distance. In technical jargon, the "point of application" of the force must move in order to do work.
Work, though easily defined mathematically, takes some explanation to grasp conceptually. In order to build an understanding of the concept, we begin with the most simple situation, then generalize to come up with the common formula.
The Simple Case
Consider a particle moving in a straight line that is acted on by a constant force in the same direction as the motion of the particle. In this very simple case, the work is defined as the product of the force and the displacement of the particle. Unlike a situation in which you hold something in place, exerting a normal force, the crucial aspect to the concept of work is that it defines a constant force applied over a distance. If a force F acts on a particle over a distance x, then the work done is simply:
W = Fx
Since w increases as x increases, given a constant force, the greater the distance during which that force acts on the particle, the more work is done. We can also see from this equation that work is a scalar quantity, rather than a vector one. Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account.
What are the units of work? The work done by moving a 1 kg body a distance of 1 m is defined as a Joule. A joule, in terms of fundamental units, is easily calculated:
W = Fx = (m) =
The joule is a multipurpose unit. It serves not only as a unit of work, but also of energy. Also, the joule is used beyond the realm of physics, in chemistry, or any other subject dealing with energy.
In dynamics we were able to define a force conceptually as a push or a pull. Such a concise definition is difficult to generate when dealing with work. To give a vague idea, we can describe work as a force applied over a distance. If a force is to do work, it must act on a particle while it moves; it cannot just cause it to move. For instance, when you kick a soccer ball, you do no work on the ball. Though you produce a great deal of motion, you have only instantaneous contact with the ball, and can do no work. On the other hand, if I pick the ball up and run with it, I do work on the ball: I am exerting a force over a certain distance. In technical jargon, the "point of application" of the force must move in order to do work.
Tuesday, January 29, 2008
feb 3 problems
1. A baseball player decides to show his skills by catching a 0.142-kg baseball dropped from the top of a 96-m tall building. What is the speed (in m/s) of the baseball just before it strikes his glove?
2. A child and sled with a combined mass of 35.9 kg slide down a frictionless hill. The sled starts from rest and has acquired a speed of 3.7 m/s by the time it reaches the bottom of the hill. What is the height (in m) of the hill? Enter your answer, accurate to the fourth decimal place.
3. A 39.3-kg high jumper leaves the ground with a speed of 5.7 m/s. How high (in m) can he leap? Enter your answer, accurate to the third decimal place.
4. Find the height (in m) from which you would have to drop your textbook so that it would have a speed of 14.2 m/s just before it hits the floor?
2. A child and sled with a combined mass of 35.9 kg slide down a frictionless hill. The sled starts from rest and has acquired a speed of 3.7 m/s by the time it reaches the bottom of the hill. What is the height (in m) of the hill? Enter your answer, accurate to the fourth decimal place.
3. A 39.3-kg high jumper leaves the ground with a speed of 5.7 m/s. How high (in m) can he leap? Enter your answer, accurate to the third decimal place.
4. Find the height (in m) from which you would have to drop your textbook so that it would have a speed of 14.2 m/s just before it hits the floor?
Potential and kinetic energy
Potential Energy:
Potential energy exists whenever an object which has mass has a position within a force field. The most everyday example of this is the position of objects in the earth's gravitational field.
The potential energy of an object in this case is given by the relation:
PE = mgh
where
PE = Energy (in Joules)
m = mass (in kilograms)
g = gravitational acceleration of the earth (9.8 m/sec2)
h = height above earth's surface (in meters)
Kinetic Energy:
Kinetic Energy exists whenever an object which has mass is in motion with some velocity. Everything you see moving about has kinetic energy.
The kinetic energy of an object in this case is given by the relation:
KE = (1/2)mv2
where
KE = Energy (in Joules)
m = mass (in kilograms)
v = velocity (in meters/sec)
Conservation of Energy
This principle asserts that in a closed system energy is conserved. This principle will be tested by you, using the experimental apparatus below. In the case of an object in free fall. When the object is at rest at some height, h, then all of its energy is PE.
As the object falls and accelerates due to the earth's gravity, PE is converted into KE. When the object strikes the ground, h=0 so that PE=0, the all of the energy has to be in the form of KE and the object is moving it at its maximum velocity. (In this case we are ignoring air resistance).
Potential energy exists whenever an object which has mass has a position within a force field. The most everyday example of this is the position of objects in the earth's gravitational field.
The potential energy of an object in this case is given by the relation:
PE = mgh
where
PE = Energy (in Joules)
m = mass (in kilograms)
g = gravitational acceleration of the earth (9.8 m/sec2)
h = height above earth's surface (in meters)
Kinetic Energy:
Kinetic Energy exists whenever an object which has mass is in motion with some velocity. Everything you see moving about has kinetic energy.
The kinetic energy of an object in this case is given by the relation:
KE = (1/2)mv2
where
KE = Energy (in Joules)
m = mass (in kilograms)
v = velocity (in meters/sec)
Conservation of Energy
This principle asserts that in a closed system energy is conserved. This principle will be tested by you, using the experimental apparatus below. In the case of an object in free fall. When the object is at rest at some height, h, then all of its energy is PE.
As the object falls and accelerates due to the earth's gravity, PE is converted into KE. When the object strikes the ground, h=0 so that PE=0, the all of the energy has to be in the form of KE and the object is moving it at its maximum velocity. (In this case we are ignoring air resistance).
Practice questions pict only one number pass it on feb 3
1. Determine the momentum of a ...
a. 60-kg halfback moving eastward at 9 m/s.
b. 1000-kg car moving northward at 20 m/s.
c. 40-kg freshman moving southward at 2 m/s.
2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...
a. its velocity were doubled.
b. its velocity were tripled.
c. its mass were doubled (by adding more passengers and a greater load)
d. both its velocity were doubled and its mass were doubled.
a. 60-kg halfback moving eastward at 9 m/s.
b. 1000-kg car moving northward at 20 m/s.
c. 40-kg freshman moving southward at 2 m/s.
2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...
a. its velocity were doubled.
b. its velocity were tripled.
c. its mass were doubled (by adding more passengers and a greater load)
d. both its velocity were doubled and its mass were doubled.
impulse and momentum
The Impulse-Momentum Change Theorem
Momentum
The sports announcer says "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter."
Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team which is on the move has the momentum. If an object is in motion (on the move) then it has momentum.
Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.
Momentum = mass • velocity
In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as
p = m • v
where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.
The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units which are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum.
Momentum is a vector quantity. As discussed in an earlier unit, a vector quantity is a quantity which is fully described by both magnitude and direction. To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg•m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction which an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg•m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction.
From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object which is at rest is 0. Objects at rest do not have momentum - they do not have any "mass in motion." Both variables - mass and velocity - are important in comparing the momentum of two objects.
The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.
Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p = m•v serves as a "guide to thinking" and not merely a "plug-and-chug recipe for algebraic problem-solving."
Momentum
The sports announcer says "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter."
Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team which is on the move has the momentum. If an object is in motion (on the move) then it has momentum.
Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.
Momentum = mass • velocity
In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as
p = m • v
where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.
The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units which are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum.
Momentum is a vector quantity. As discussed in an earlier unit, a vector quantity is a quantity which is fully described by both magnitude and direction. To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg•m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction which an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg•m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction.
From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object which is at rest is 0. Objects at rest do not have momentum - they do not have any "mass in motion." Both variables - mass and velocity - are important in comparing the momentum of two objects.
The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.
Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p = m•v serves as a "guide to thinking" and not merely a "plug-and-chug recipe for algebraic problem-solving."
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