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Work and Energy

Slide 3 / 127 How to Use this File

· Each topic is composed of brief direct instruction · There are formative assessment questions after every topic denoted by black text and a number in the upper left. >Students work in groups to solve these problems but use student responders to enter their own answers. >Designed for SMART Response PE student response systems. >Use only as many questions as necessary for a sufficient number of students to learn a topic. · Full information on how to teach with NJCTL courses can be found at njctl.org/courses/teaching methods

Slide 4 / 127 Table of Contents

· Energy and Work · Conservative and Non-Conservative Forces · Two Dimensional Forces and Work · Position Dependent Potential Energy Graphical Analysis

Click on the topic to go to that section

· Work done by a Position Dependent Force · Power · Placeholder · Conservation of Total Mechanical Energy

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Energy and Work

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• f Contents

Slide 6 / 127 Energy

The concept of energy is so fundamental, like space and time, that there is no real good definition of what it "is." However, just like space and time, that doesn't stop us from doing very useful calculations with energy. There are some things we can say about it: · It is the ability to do work. · It can be stored. · It can be changed from one form to another (light to thermal energy, mechanical to thermal energy, gravitational potential energy to kinetic energy). · It can be measured and compared. Did you notice a term in the bullet list above that hasn't been defined yet?

Slide 7 / 127 Work

Work has the ability to increase or decrease the amount of Energy at a certain position and time in space. Work has the same units as energy - Joules. What is Work? It is not what is talked about in common language. It is unfortunate that sometimes Physics uses words that are used everyday - in a quite different fashion. For example - if you're holding up a heavy box, do you think you're doing work?

Slide 8 / 127 Work

You're not! Work, in physics terms, is defined as the exertion of a force

• ver a displacement where only the component of the force in

the direction of the motion is relevant. For now, we'll assume a constant force. If you're just holding a box, you are certainly exerting an upward force on the box (to keep gravity from pulling it to the ground), but it's not moving, so there is no displacement. Therefore, there is no work. You learned the following equation: W = Fdparallel = Fdcosθ where θ is the angle between the Force and the displacement.

Slide 9 / 127 Work

A more elegant way to represent this is by using vector notation and a specific type of vector multiplication called the vector dot product, or just dot product. The result of the dot product of two vectors is a scalar - and is the length of one vector projected on the other - so this achieves the goal of just using the component of force along the direction of the displacement. But where is the work being done? Where is the energy being increased or decreased? The entire universe? We need two more definitions to bound these questions.

Slide 10 / 127 System and Environment

The system and the environment. A system is a small segment of the universe that will be considered in solving a specific problem, and we will erect a boundary around it. Any force or object outside this boundary will not be considered. The environment is everything outside the system boundary. The system can be a particle, a group of particles, an object, an area of space, and its size and shape is totally determined by how you want to solve the problem. Why are we defining a system and its environment?

Slide 11 / 127 System and Environment

So we can make the problem solvable. By defining an appropriate system, we can isolate the forces that are within the system from the forces that act on the system from the environment. If the forces are internal to the system, then there is no change in the energy of the system (as long as we don't consider thermal energy - which we won't for now). If the forces are external, then there will be a change in the energy of the system.

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1 Which of the following are characteristics of energy? A Thermal energy can be changed to mechanical energy. B Mechanical energy can be changed to thermal energy. C Energy can be stored. D Energy has the ability to do work. E All of the above.

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1 Which of the following are characteristics of energy? A Thermal energy can be changed to mechanical energy. B Mechanical energy can be changed to thermal energy. C Energy can be stored. D Energy has the ability to do work. E All of the above.

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Answer E

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2 A system is defined as: A All the forces that are external to the boundary between it and the rest of the universe. B A small segment of the universe that has no internal forces. C A small segment of the universe that is chosen to solve a

• problem. Forces internal to the system can change its total

mechanical energy. D A small segment of the universe that is chosen to solve a

• problem. Forces internal to the system cannot change its

total mechanical energy. E Uniquely for a problem. Only one specific system can be used to solve a problem.

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2 A system is defined as: A All the forces that are external to the boundary between it and the rest of the universe. B A small segment of the universe that has no internal forces. C A small segment of the universe that is chosen to solve a

• problem. Forces internal to the system can change its total

mechanical energy. D A small segment of the universe that is chosen to solve a

• problem. Forces internal to the system cannot change its

total mechanical energy. E Uniquely for a problem. Only one specific system can be used to solve a problem.

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Answer D

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3 In solving an energy problem, the environment is defined as: A An area that contains no forces. B An area that is partially in, and partially outside the system. C The source of the external forces on the system. D The source of the internal forces on the system. E A small area within the system.

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3 In solving an energy problem, the environment is defined as: A An area that contains no forces. B An area that is partially in, and partially outside the system. C The source of the external forces on the system. D The source of the internal forces on the system. E A small area within the system.

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Answer C

Slide 15 / 127 Work

· If the force acts in the same direction as the object's motion, then the work done is positive, and the energy of the object increases. · If the force acts in the opposite direction as the object's motion, then the work done is negative and the energy of the

• bject decreases.

· If the object does not move, then zero work is done. · Work is a scalar - it has magnitude, but not direction. · The unit of work is the Joule - just like energy.

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This equation gives us the units of work. Since force is measured in Newtons (N) and distance is measured in meters (m) the unit of work is the Newton-meter (N-m). And since N = kg-m/s2; a N-m also equals a kg-m2/s2. In honor of James Joule, who made critical contributions in developing the idea of energy, the unit of work and energy is also known as the Joule (J). 1 Joule = 1 Newton-meter = 1 kilogram-meter2/second2 1 J = 1 N-m = 1 kg-m2/s2

Units of Work and Energy Slide 17 / 127

Joule was instrumental in showing that different forms of energy can be converted into other forms - most notably mechanical to thermal energy. Before Joule, it was commonly accepted that thermal energy is conserved. This was disproved by Joule's extremely accurate and precise measurements showing how thermal energy is just another form of energy. This was made possible by his experience as a brewer which relied on very accurate measurements of temperature, time and volume!

James Prescott Joule Slide 18 / 127

4 Which is a valid unit for work? A N/m B N-s C W D J/s E J

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4 Which is a valid unit for work? A N/m B N-s C W D J/s E J

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Answer E

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F #x v F #x v F #x v

Force and Work Slide 20 / 127

5 A 36.0 N force is applied to an object that moves 11.0 m in the

same direction as the applied force on a frictionless surface. How much work is done on the object?

F

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5 A 36.0 N force is applied to an object that moves 11.0 m in the

same direction as the applied force on a frictionless surface. How much work is done on the object?

F

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Answer

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6 A 36.0 N force is applied to an object that moves 11.0 m in the

• pposite direction of the applied force on a frictionless surface. How

much work is done on the object?

F v

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6 A 36.0 N force is applied to an object that moves 11.0 m in the

• pposite direction of the applied force on a frictionless surface. How

much work is done on the object?

F v

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Answer

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F

7 A 36 N force is applied to an object that remains stationary. How

much work is done on the object by the applied force?

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F

7 A 36 N force is applied to an object that remains stationary. How

much work is done on the object by the applied force?

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Answer

Slide 23 / 127 Work

You have to very specific about using work. The system or environment that the work is acting on needs to be

• specified. For example:

"An applied force does 12 J of work on a box." "Gravity does -5 J of work on a box that is being raised up." This next sentence is not a complete statement. What's missing? "An external force does 6 J of work." The system or the environment that the work is acting on must be described.

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8 In which of the following cases is positive work done by an external force? A A softball player catches a ball in her glove. B A home owner is pushing a lawnmower from rest. C A drive applies the break to his car. D A student holds her textbook and is not moving. E A ball falls from a height. The ground applies a force to stop the ball.

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8 In which of the following cases is positive work done by an external force? A A softball player catches a ball in her glove. B A home owner is pushing a lawnmower from rest. C A drive applies the break to his car. D A student holds her textbook and is not moving. E A ball falls from a height. The ground applies a force to stop the ball.

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Answer B

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9 A 2 kg block slides 4.5 m to the right on a frictionless table with a

constant velocity of 5 m/s. What is the net work on the block?

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9 A 2 kg block slides 4.5 m to the right on a frictionless table with a

constant velocity of 5 m/s. What is the net work on the block?

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Answer Since the block is moving with a constant velocity, there is zero net force on the block.

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10 A book is held at a height of 2.0 m for 20 s. How much work is

done on the book? A 400 J B 200 J C 40 J D 20 J E 0 J

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10 A book is held at a height of 2.0 m for 20 s. How much work is

done on the book? A 400 J B 200 J C 40 J D 20 J E 0 J

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Answer E

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11 An athlete is holding a football. He then throws it to a teammate

who catches it. Describe the work done on the football by both players starting from when the football is at rest before it is thrown and after it is caught.

Students type their answers here

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11 An athlete is holding a football. He then throws it to a teammate

who catches it. Describe the work done on the football by both players starting from when the football is at rest before it is thrown and after it is caught.

Students type their answers here

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Answer When the ball is at rest before it is thrown and after it is caught, there is no work done as there is no

• displacement. Positive work is

done as it is thrown since the force is in the same direction as the displacement of the ball. Negative work is done by the receiver in catching the ball as the force is

• pposite the displacement.

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Conservative and Non-Conservative Forces

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Slide 29 / 127 Work - Energy Equation

When a net external force acts on an object in the same direction of its displacement, positive work is done - just think

• f pushing a ball on the floor by applying a constant force.

Δx parallel to a constant force enables us to use kinematics equation 3 solving for aΔx substituting into W equation

Recognize anything?

Slide 30 / 127 Work - Energy Equation

By applying a constant force over a displacement, we've derived both the Work - Energy Equation and found the expression for kinetic energy. Net positive work applied to a system increases its kinetic energy. What if an object is moving in the positive x direction with a velocity, v0, and a force is applied opposite its displacement?

Slide 31 / 127 Work - Energy Equation

Negative work is done on the object, so its kinetic energy, and velocity decreases; vf < v0

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12 How much net external force must be applied to an object such that

it gains 100.0 J of kinetic energy over a displacement of 20.0 m, parallel to the direction of the external force?

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12 How much net external force must be applied to an object such that

it gains 100.0 J of kinetic energy over a displacement of 20.0 m, parallel to the direction of the external force?

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Answer

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13 A net external force of 5.00 N is applied to a 5.10 kg object and it

moves parallel to the force application a displacement of 20.0 m. What is the object's final velocity if it started at rest?

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13 A net external force of 5.00 N is applied to a 5.10 kg object and it

moves parallel to the force application a displacement of 20.0 m. What is the object's final velocity if it started at rest?

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Answer

(F parallel to Δx)

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14 Over what displacement must a 400.0 N net external force be

applied, in parallel with the displacement to an object such that it gains 1600 J of kinetic energy?

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14 Over what displacement must a 400.0 N net external force be

applied, in parallel with the displacement to an object such that it gains 1600 J of kinetic energy?

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Answer

(F parallel to Δx)

Slide 35 / 127 Gravitational Potential Energy

Kinetic energy is the energy of motion. Gravitational potential energy is the energy of an object due to its position. This is derived by examining the work done by the gravitational force on an object that is changing its distance from the center of the earth. Caution - this will only apply for objects near the surface of the earth (or other celestial bodies). This restriction will be lifted in the Universal Gravitation unit. Fapp Fg = mg a = 0 An object of mass m, is being lifted at a constant velocity near the surface of the earth by an external force, Fapp = F. It starts at height h0, with v0 = 0 and finishes at hf, with vf = 0.

Slide 36 / 127 Gravitational Potential Energy

Fapp = F Fg = mg a = 0

h0 hf Negative sign because Fg points down and Δy is in the up direction

The Work done by gravity is negative (hf > h0). Define the gravitational potential energy as the negative of the work done by the gravitational force on the object. This will give us a positive potential energy for an object at a height above the earth - which you've seen in your previous physics course.

Slide 37 / 127 Gravitational Potential Energy

Gravitational Potential Energy depends on the change in height, not the absolute value of the height - an object will have a different value of GPE depending on where h0 is chosen. If h0 = 0, and an object is at height, h, then the familiar expression for GPE shows up: Can potential energies be defined for other forces? Yes - they will be covered soon......

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15 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. How much work does the external force do on the book? A mg B -mg C 0 D mgh E -mgh

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15 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. How much work does the external force do on the book? A mg B -mg C 0 D mgh E -mgh

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Answer D

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16 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. How much work does the gravitational force do on the book? A mg B -mg C 0 D mgh E -mgh

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16 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. How much work does the gravitational force do on the book? A mg B -mg C 0 D mgh E -mgh

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Answer E

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17 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. How much net work is done

• n the book by the external force and the gravitational force?

A mg B -mg C 0 D mgh E -mgh

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17 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. How much net work is done

• n the book by the external force and the gravitational force?

A mg B -mg C 0 D mgh E -mgh

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Answer C

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18 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. What is the gravitational potential energy of the mass when it is lifted to the height, h? A mg B -mg C 0 D mgh E -mgh

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18 A book of mass, m, is lifted upwards at a constant velocity, a

displacement, h, by an external force. What is the gravitational potential energy of the mass when it is lifted to the height, h? A mg B -mg C 0 D mgh E -mgh

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Answer D

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19 What is the change of GPE for a 5.0 kg object which is raised

from an initial height of 1.0 m above the floor to a final height

• f 8.0 m above the floor?

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19 What is the change of GPE for a 5.0 kg object which is raised

from an initial height of 1.0 m above the floor to a final height

• f 8.0 m above the floor?

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Answer

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20 What is the change of GPE for an 8.0 kg object which is lowered

from an initial height of 2.1 m above the floor to a final height of 1.5 m above the floor?

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20 What is the change of GPE for an 8.0 kg object which is lowered

from an initial height of 2.1 m above the floor to a final height of 1.5 m above the floor?

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Answer

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21 What is the change in height of a 2.0 kg object which gained 16 J

• f GPE?

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21 What is the change in height of a 2.0 kg object which gained 16 J

• f GPE?

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Answer

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22 A librarian takes a book off a high shelf and refiles it on a lower

• shelf. Which of the following are correct about the work done on

the book by the librarian and the earth's gravitational field as the book is lowered? Librarian Gravitational Field A Positive Positive B Negative Negative C Positive Negative D Negative Positive E Zero Zero

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22 A librarian takes a book off a high shelf and refiles it on a lower

• shelf. Which of the following are correct about the work done on

the book by the librarian and the earth's gravitational field as the book is lowered? Librarian Gravitational Field A Positive Positive B Negative Negative C Positive Negative D Negative Positive E Zero Zero

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Answer

D

Slide 46 / 127 Conservative and Non Conservative Forces

The gravitational force is a conservative force. The path that an

• bject takes has no bearing on its potential energy - GPE depends
• nly on the initial and final heights.

Work has a non zero value only for the force components in the same direction as the motion. The gravitational force always points down, so lateral motion requires zero work by the gravitational force. The only impact on potential energy is the vertical component of the motion.

h0=0 h

Each path results in the same GPE for an object moving from h0 to h.

Slide 47 / 127 Slide 48 / 127 Conservative and Non Conservative Forces

h0=0 h

As the object goes up, the gravitational force does negative work, decreasing the object's kinetic energy (and speed). On the way down, positive work is done by the gravitational force, increasing its kinetic energy (and speed). The sum of the work done over the closed path is zero.

Slide 49 / 127 Conservative Forces

A conservative force has two properties: · The work done by a conservative force on an object depends only

• n its initial and final position - it is path independent.

· The work done by a conservative force on an object is zero on a closed path (initial and final position are the same).

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23 A vacuum cleaner is moved from the ground floor to the second floor of an apartment building. In which of the following cases is the most work done by the person moving the vacuum? A The vacuum cleaner is pushed up an inclined plane set over the stairs. B The person carries the cleaner up the stairs. C The person brings the cleaner to the third floor, by mistake, then back to the second floor. D The work is the same in each case. E A pulley is set up outside the building and the vacuum is hoisted up to the second floor by a rope.

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23 A vacuum cleaner is moved from the ground floor to the second floor of an apartment building. In which of the following cases is the most work done by the person moving the vacuum? A The vacuum cleaner is pushed up an inclined plane set over the stairs. B The person carries the cleaner up the stairs. C The person brings the cleaner to the third floor, by mistake, then back to the second floor. D The work is the same in each case. E A pulley is set up outside the building and the vacuum is hoisted up to the second floor by a rope.

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Answer D

Slide 51 / 127 Non - Conservative Forces

A non - conservative force does not follow the two properties

• f a conservative force.

The path taken does impact the work done, and the work done

• n a closed path is not equal to zero.

Think about this and please do two things: 1.Propose a force that is non-conservative and show how it doesn't follow the two properties. 2.Discuss what impact this has on deriving a potential energy from the force.

Slide 52 / 127 Non - Conservative Forces

Friction is the most common example of a non-conservative

• force. Friction force always opposes motion. The longer the

path taken by an object, the greater the work done. An easy demonstration is to move your hand over a smooth surface - like your desk. Movement in a straight line may warm your hand a little. But if you move your hand back and forth and take a longer path to get to the final position - it heats up more - your hand experieces more frictional force - more work. You cannot derive a potential energy function for this force

• as the energy difference between two points will not always be

the same - it depends on the path taken.

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24 Which of the following is an example of a conservative force?

A Kinetic friction B Gravitational force C Static friction D Air resistance E Water resistance

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24 Which of the following is an example of a conservative force?

A Kinetic friction B Gravitational force C Static friction D Air resistance E Water resistance

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Answer B

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25 Which of the following is a property of a non conservative force?

A The net work done by this force over a closed path is zero. B The net work done by this force over a closed path is a non zero value. C The work done by this force on an object moving from point A to B is independent of the path taken. D The work done by this force on an object moving from point A to B is always positive. E A potential energy can be associated with a non- conservative force.

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25 Which of the following is a property of a non conservative force?

A The net work done by this force over a closed path is zero. B The net work done by this force over a closed path is a non zero value. C The work done by this force on an object moving from point A to B is independent of the path taken. D The work done by this force on an object moving from point A to B is always positive. E A potential energy can be associated with a non- conservative force.

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Answer B

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Conservation of Total Mechanical Energy

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Slide 56 / 127 Conservation of Total Mechanical Energy

We're now going to derive a conservation law - the Conservation

• f Total Mechanical Energy which is defined as the total potential

energy plus the kinetic energy of an object. In earlier physics courses, you learned the conservation of energy, the conservation of mass, and maybe the conservation of mass- energy. Here, we're just going to look at the energy of objects moving around the surface of the earth, and not worry about their thermal or nuclear or sound or light energy. Start with the Work-Energy equation, and assume an object is being elevated from h0 to hf.

Slide 57 / 127 Conservation of Total Mechanical Energy

Work-Energy Equation Split the net work on the system into the work done by external non conservative forces (such as friction) and conservative forces. This is done because the work done by a conservative force can be replaced by its potential energy. The only conservative force is the gravitational force. Negative sign as gravitational force is opposite the displacement as the object is raised from h0 to hf.

Slide 58 / 127 Slide 59 / 127 Conservation of Total Mechanical Energy

In the absence of a net external non conservative force, we have WNC = 0 and Ef = E0. The initial total mechanical energy equals the final total mechanical energy - it is conserved. If there are other forms of potential energy in the system - like elastic potential energy (spring), then those terms would be added to the total mechanical energy (we will do this a little later).

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Net non-zero work can only be done to a system by an external force; a force from the environment outside the system. So if our system is a box sitting on a table and I come along and push the box, I can increase the kinetic energy of the box - I am doing net non-zero work on the box. Why are none of the internal forces (forces within the box, such as the box molecules moving about and colliding with each

• ther) involved in increasing the energy of the system? The

molecules are certainly exerting forces on each other, and they are causing each other to move.

Internal forces Slide 61 / 127

Newton's Third Law! Every time a molecule in the box strikes another molecule, it exerts a force on it, and moves it. However, the second molecule exerts an equal and opposite force on the first one. Thus, assuming equal masses for the molecules, the work done internal to the system equals zero - it all cancels out. Thermal Energy increases as the molecules vibrate and move faster, but this type of energy is not included in TME by definition.

Internal forces Slide 62 / 127

26 Which law explains why internal forces of a system do not change

its total mechanical energy? A Newtons First Law B Newton's Second Law C Newton's Third Law D Newton's Law of Universal Gravitation E Conservation of Angular Momentum

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26 Which law explains why internal forces of a system do not change

its total mechanical energy? A Newtons First Law B Newton's Second Law C Newton's Third Law D Newton's Law of Universal Gravitation E Conservation of Angular Momentum

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Answer C

Slide 63 / 127 Elastic Potential Energy

We have analyzed two types of energy, kinetic energy (KE) and gravitational potential energy (GPE). In your previous physics classes, you studied elastic potential energy (EPE). Hooke's Law tells us that . The equation for GPE was calculated using the definition of work and the fact that the potential energy resulting from a conservative force is the negative of the work done by the force. The spring force is a conservative force. This allows us to calculate a potential energy. But, what makes this a little trickier than when GPE was derived?

Slide 64 / 127 Elastic Potential Energy

The gravitational force (near the surface of the earth) is assumed to be constant (it doesn't vary much). The spring force is a function of position - it is not constant. At each point of the spring's motion, the force is different. In

• rder to calculate work, the motion must be analyzed at

infinitesimal displacements which are multiplied by the force at each infinitesimal point, and then summed up. What does that sound like?

Slide 65 / 127 Elastic Potential Energy

Calculus - specifically, integration. Rewrite the Work equation to take into account the position varying force, where F(x) is the force in the x direction: For motion in three dimensions, work is expressed as follows:

• r in Cartesian coordinates:

Slide 66 / 127 Elastic Potential Energy

Start at the equilibrium point, x0 = 0, and stretch the spring to xf. EPE has been used in this course, but U is generally the symbol for potential energy. We'll use both. Let's go back to the Work - Energy equation and see how to fit this new potential energy in.

Slide 67 / 127 Total Mechanical Energy (adding a spring)

Work-Energy Equation Split the net work on the system into the work done by external non conservative forces (such as friction) and conservative forces. This is done because the work done by a conservative force can be replaced by its potential energy. Two conservative forces: elastic and gravitational. Taking the general case where the spring is stretched from x0 to xf.

Slide 68 / 127 Slide 69 / 127 Advantages of using Conservation of Mechanical Energy

When trying to solve motion problems, you were first taught to use Newton's Laws and the Kinematics equations. That can get problematic if the forces are not constant, are acting in multiple dimensions, very complex or numerous. Conservation laws enable you to just work with the initial and final conditions - you don't care how or why the object gets to where it is, you just need a snapshot of where it was and where it is now. In addition, energy is a scalar, so you don't have to worry about vectors, and the solutions are typically easier.

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27 When using the Conservation of Total Mechanical Energy to solve

a system problem, what needs to be considered? A The initial and final energy of the system. B The initial and final forces on the system. C Only the initial energy of the system. D Only the final energy of the system. E The magnitude and direction of the internal forces on the system.

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27 When using the Conservation of Total Mechanical Energy to solve

a system problem, what needs to be considered? A The initial and final energy of the system. B The initial and final forces on the system. C Only the initial energy of the system. D Only the final energy of the system. E The magnitude and direction of the internal forces on the system.

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Answer A

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28 A ball is swung around on a string, in a circle, traveling a

displacement of 2.0 m in 5.0 s. What is the work done on the ball by the string? A 0 J B 2.5 J C 5.0 J D 10 J E 25 J

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28 A ball is swung around on a string, in a circle, traveling a

displacement of 2.0 m in 5.0 s. What is the work done on the ball by the string? A 0 J B 2.5 J C 5.0 J D 10 J E 25 J

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Answer A

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29 Assume the earth moves around the sun in a perfect circular orbit

(a good approximation). Use the direction of the gravitational force between the two celestial objects and describe the work done by the sun on the earth and how that impacts the earth's orbital speed. How does your answer change if you don't make the circular orbit assumption?

Students type their answers here

Slide 72 (Answer) / 127

29 Assume the earth moves around the sun in a perfect circular orbit

(a good approximation). Use the direction of the gravitational force between the two celestial objects and describe the work done by the sun on the earth and how that impacts the earth's orbital speed. How does your answer change if you don't make the circular orbit assumption?

Students type their answers here

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Answer

The gravitational force is in a line connecting the earth and the sun, and is perpendicular to the earth's displacement at all times. The sun does no work on the earth. The earth's total mechanical energy is conserved. The earth is the same distance from the sun in its circular orbit, its potential energy is

• constant. Its kinetic energy is constant

and its speed is constant. Since the earth's orbit is not a perfect circle, Fg has a component in the direction of the earth's motion, so work is done by the sun on the earth and the earth's speed changes.

Slide 73 / 127

Two Dimensional Forces and Work

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Slide 74 / 127

In the previous section, we learned that the amount of work done to a system, and therefore the amount of energy increase that the system experiences, is given by We have only dealt with one dimensional movement, and will now generalize to two dimensions. We will actually go backwards a little - we won't use the scalar dot product, but will show how it arises from trigonometry.

Two Dimensional Forces and Work Slide 75 / 127

v

F

A P P

#x

How would we interpret: W = Fdparallel for this case?

Two Dimensional Forces and Work

Instead of pulling the object horizontally, what if it is pulled at an angle to the horizontal?

Slide 76 / 127

F

p a r a l l e l

v F

p e r p e n d i c u l a r

#x

# After breaking FAPP into components that are parallel and perpendicular to the direction of motion, we can see that no work is done by the perpendicular component; work is only done by the parallel component.

Two Dimensional Forces and Work

Using trigonometry, we find that Fparallel = FAPPcosθ

Slide 77 / 127

F

p a r a l l e l

v F

p e r p e n d i c u l a r

#x

# In words, the work done on an object by a force is the product of the magnitude of the force and the magnitude of the displacement times the cosine of the angle between them. Which is exactly what the scalar dot product shows:

Two Dimensional Forces and Work

W = Fparalleld becomes: W = (FAPPcosθ)Δx = FAPPΔxcosθ

Slide 78 / 127

This is really no more difficult a

• case. We just have to find the

component of force that is parallel to the object's displacement.

Two Dimensional Forces and Work

Instead of pulling the object at an angle to the horizontal, what if it is pushed?

Slide 79 / 127

The interpretation is the same, just determine the angle between the force and displacement and use: W = FAPPΔxcosθ

F

p a r a l l e l

F

p e r p e n d i c u l a r

#

FAPP #x

Two Dimensional Forces and Work

Even though Fperpendicular is in the negative direction (it was positive when the object was pulled), it does not affect the work - as only the parallel component contributes to the work.

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30 A 40.0 N force pulls an object at an angle of θ = 37.00 to its

direction of motion. Its displacement is Δx = 8.00 m. How much work is done by the force on the object?

v #x

#

Slide 80 (Answer) / 127

30 A 40.0 N force pulls an object at an angle of θ = 37.00 to its

direction of motion. Its displacement is Δx = 8.00 m. How much work is done by the force on the object?

v #x

# [This object is a pull tab]

Answer

Slide 81 / 127

31 An object is pushed with an applied force of 36.0 N at an angle of

θ = 60.00 to the horizontal and it moves Δx = 3.40 m. What work does the force do on the object? #

FAPP #x

Slide 81 (Answer) / 127

31 An object is pushed with an applied force of 36.0 N at an angle of

θ = 60.00 to the horizontal and it moves Δx = 3.40 m. What work does the force do on the object? #

FAPP #x

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Answer

Slide 82 / 127

Work done by a Position Dependent Force

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Slide 83 / 127 Position Dependent Force

We've already seen one example of a position dependent force - the spring. Let's generalize.

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x(m) F(x) N F(x) = Cx1/2

To find Work on a Force/position diagram, you take the area under the force function. When the force is constant or increases linearly, it is a simple geometry problem.

Slide 84 / 127 Position Dependent Force

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x(m) F(x) N F(x) = Cx1/2

The Force plotted to the left is proportional to the square root of the position; hence integration techniques are used to solve for the work performed by the force

• n this system.

Where have you done this before? In the Kinematics unit of this course with position-time, velocity-time and acceleration- time graphs.

Slide 85 / 127 Position Dependent Force

For the spring problem (where F = -kx), we assumed that the spring was stretched from x = 0 to a given x to find We and then Ue.

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x(m) F(x) N F(x) = Cx1/2

Using the same technique, but changing the limits of integration, enables you to find the work done between any two points in the object's motion. The yellow boxes show an approximate area below the force

• curve. By decreasing the width
• f the boxes (green), you get

closer to the real area. How is a more exact answer achieved?

Slide 86 / 127 Position Dependent Force

By decreasing the width of the boxes to be infinitesimally small - the integration branch of calculus.

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x(m) F(x) N F(x) = Cx1/2 x0 xf

Given where C is a constant, find the work done by the force when the object is moved from x0 to xf. Try this problem:

Slide 87 / 127 Position Dependent Force

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x(m) F(x) N F(x) = Cx1/2 x0 xf

The potential energy due to this force (if it is conservative) would be the negative of the work.

Slide 88 / 127 Position Dependent Force Graphical Analysis

For conservative forces moving in one dimension, the potential energy is found: Use a bit of calculus and take the derivative of both sides with respect to x:

Slide 89 / 127

32 The potential energy (in Joules) of a block as it moves in the

x direction is U(x) = 3x3 + 4x2 + 5. Find the general expression for the force exerted on the block. What is the force, in Newtons, on the block at x = 2.0 m?

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Slide 89 (Answer) / 127

32 The potential energy (in Joules) of a block as it moves in the

x direction is U(x) = 3x3 + 4x2 + 5. Find the general expression for the force exerted on the block. What is the force, in Newtons, on the block at x = 2.0 m?

Students type their answers here

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Answer

Slide 90 / 127 Slide 90 (Answer) / 127 Slide 91 / 127

34 The force exerted by a non-linear spring on a mass is F = -kx2/2.

If k = 200 N/m, find the work done by the spring on a mass from x = 0.08 m to x = 0.10 m. What is the change in potential energy of the spring?

Slide 91 (Answer) / 127

34 The force exerted by a non-linear spring on a mass is F = -kx2/2.

If k = 200 N/m, find the work done by the spring on a mass from x = 0.08 m to x = 0.10 m. What is the change in potential energy of the spring?

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Answer

because Fspring(x) is in the

• pposite direction of dx

Slide 92 / 127

Position Dependent Potential Energy Graphical Analysis

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Slide 93 / 127 Position Dependent Potential Energy Graphical Analysis

A particle's potential energy, U, is plotted with respect to its displacement, as shown to the

• left. The equation below will be

used to determine the motion of a particle with this potential energy:

Slide 94 / 127 Position Dependent Potential Energy Graphical Analysis

First, let's find where this particle is at an equilibrium position. A particle is at equilibrium when the net force on it is zero. Where can you see equilibrium points on the graph using the equation below?

Slide 95 / 127 Position Dependent Potential Energy Graphical Analysis

is the slope of the curve. The slope is zero at the four points indicated by the arrows. If the particle is placed at any

• f those points at rest, it will

not move, since there is no force acting on it. But what if the particle is momentarily pushed by an external force when it is at those points? How will it move?

Slide 96 / 127 Position Dependent Potential Energy Graphical Analysis

At the two arrowed points - where the curve is concave down, points on the curve to the left of the peak have a positive slope; points to the right have a negative slope. What does that tell you about the direction of the force that the particle after it is pushed?

Slide 97 / 127 Unstable Equilibrium

A particle pushed to the left experiences a negative force, moving it to the left - sliding down the potential energy- position curve. A particle pushed to the right experiences a positive force, moving it to the right - sliding down the potential energy - position curve. This is called unstable equilibrium - the particle is stationary, but if an external force is applied, it moves and doesn't come back to the same equilibrium point.

Slide 98 / 127 Position Dependent Potential Energy Graphical Analysis

At the two arrowed points - where the curve is concave up, points on the curve to the left of the peak have a negative slope; points to the right have a positive slope. What does that tell you about the direction of the force that the particle feels from the potential energy after it is pushed?

Slide 99 / 127 Stable Equilibrium

A particle pushed to the left experiences a positive force, moving it to the right - back in the direction of the equilibrium point. A particle pushed to the right experiences a negative force, moving it to the left - back in the direction of the equilibrium point. This is stable equilibrium - the particle is stationary in the potential energy well, and if an external force is applied, it moves, but comes back to the equilibrium point. Will it stay there?

Slide 100 / 127 Simple Harmonic Motion

• No. If it is moving on a

frictionless surface, it will reach the equilibrium point with a velocity and continue past it - until the restoring force stops it and turns it around. It will continue this oscillatory

• motion. Just like a spring.

If friction were present, it would eventually come to rest at the bottom of the potential energy well after the oscillations. Simple Harmonic Motion!

Slide 101 / 127 Position Dependent Potential Energy Graphical Analysis

What's nice about the graph is that you can visualize a particle

• n the curve and get the same

result as the math. If the graph was a drawing of a roller coaster track, a ball put at the top of the curves would move away from the peaks and not come back. A ball pushed from the wells would go up the track, and then return, and oscillate about the bottom of the well.

Slide 102 / 127

35 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which points is the particle not in equilibrium? A A, B, D B A, B, C C B, C, D D C, D, F E A, B, E

Slide 102 (Answer) / 127

35 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which points is the particle not in equilibrium? A A, B, D B A, B, C C B, C, D D C, D, F E A, B, E

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Answer A

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36 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which points is the particle in equilibrium? A A, B, D B A, B, C C B, C, D D C, E, F E A, B, E

Slide 103 (Answer) / 127

36 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which points is the particle in equilibrium? A A, B, D B A, B, C C B, C, D D C, E, F E A, B, E

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Answer D

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37 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the particle in stable equilibrium? A B, D B C, F C C, E D A E E

Slide 104 (Answer) / 127

37 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the particle in stable equilibrium? A B, D B C, F C C, E D A E E

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Answer B

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38 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the particle in unstable equilibrium? A A, E B C, F C C, E D E E A

Slide 105 (Answer) / 127

38 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the particle in unstable equilibrium? A A, E B C, F C C, E D E E A

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Answer D

Slide 106 / 127 Position Dependent Potential Energy Graphical Analysis

Four more curve segments to analyze. The segments at points A and C will be looked at first. What kind of segments are they? What is the sign of their slopes? A B C D

Slide 107 / 127 Position Dependent Potential Energy Graphical Analysis

A C Points A and C are on straight lines with constant positive (but different) slopes. The force on those points is negative - so if a particle is placed at points A or C with zero velocity, it will move down the curve and to the left. If the particle has an initial velocity to the right, the force will accelerate it to the left and slow it down.

Slide 108 / 127 Position Dependent Potential Energy Graphical Analysis

B Point B is on a straight line with a negative slope. What is the direction of the force

• n the particle?

What would be the motion of a particle placed at point B with a negative velocity, zero velocity and positive velocity?

Slide 109 / 127 Position Dependent Potential Energy Graphical Analysis

B D A particle with zero velocity would increase its speed in the positive x direction. A particle with negative velocity would slow down and a particle with positive velocity would speed up. The acceleration would be the same in each case since the slope of a straight line is constant, thus the Force is constant. Last one - what about point D, which is on a parabola?

Slide 110 / 127 Position Dependent Potential Energy Graphical Analysis

E D The potential energy at point D is proportional to x

2, hence the

slope, or force, is proportional to x. Sound familiar? That's a

• spring. The slope at point D is

negative, so the force is to the right. Let's add one more point - point E which is symmetric to point D with respect to the bottom of the well. Describe the motion of an object that is released at point D with zero velocity.

Slide 111 / 127 Position Dependent Potential Energy Graphical Analysis

E D The particle would oscillate between points D and E. The force at point E (and that side of the parabola) is to the left - a restoring force.

Slide 112 / 127 Position Dependent Potential Energy Graphical Analysis

A B C D E As the particle moves along the potential energy curve, what can be said about the kinetic and total mechanical energy?

Slide 113 / 127 Position Dependent Potential Energy Graphical Analysis

A B C D E The total mechanical energy stays constant, and as the potential energy decreases, the kinetic energy will increase. As the potential energy increases, the kinetic energy will decrease.

Slide 114 / 127 Position Dependent Potential Energy Graphical Analysis

A B C D E Be careful with the sign of the potential energy - it can be negative or positive. Kinetic energy can only be positive. A particle moving from point 1 to point 2 is speeding up as the potential energy decreases. A particle moving from point 2 to point 3 is slowing down as the potential energy increases. The roller coaster analogy also works for determining the speed behavior! 1 2 3

Slide 115 / 127

39 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. What is the force on the particle when at point C? A -2.0 N B -1.0 N C 0 N D 1.0 N E 2.0 N

Slide 115 (Answer) / 127

39 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. What is the force on the particle when at point C? A -2.0 N B -1.0 N C 0 N D 1.0 N E 2.0 N

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Answer C

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40 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point B. What is the largest value of x reached by the particle during this motion? A 2.9 m B 4.0 m C 5.5 m D 6.7 m E 8.0 m

Slide 116 (Answer) / 127

40 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point B. What is the largest value of x reached by the particle during this motion? A 2.9 m B 4.0 m C 5.5 m D 6.7 m E 8.0 m

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Answer D

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41 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the force on the particle positive? A E, F B E C C, E D D E A, B

Slide 117 (Answer) / 127

41 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the force on the particle positive? A E, F B E C C, E D D E A, B

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Answer E

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42 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the force on the particle negative? A E, F B E C C, E D D E A, B

Slide 118 (Answer) / 127

42 A conservative force parallel to the x-axis moves a particle along

the x-axis. The potential energy as a function of position is presented by the graph. The particle is released at rest at point A. At which point or points is the force on the particle negative? A E, F B E C C, E D D E A, B

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Answer D

Slide 119 / 127

Power

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Slide 120 / 127 Power

It is often important to know not only if there is enough energy available to perform a task but also how much time will the energy be used. Average Power is defined as the rate that work is done: Since work is measured in Joules (J) and time is measured in seconds (s) the unit of power is Joules per second (J/s). In honor of James Watt, who made critical contributions in developing efficient steam engines, the unit of power is known as a Watt (W).

Slide 121 / 127 Power

Instantaneous power is found by using the same method we found instantaneous velocity and acceleration: Using a little non rigorous calculus in one dimension (don't show this to your calculus teacher): Substitute this into the instantaneous power equation above.

Slide 122 / 127 Power

For a constant force acting in the x direction, the instantaneous power generated is equal to the magnitude of the force times the

• velocity. This is quite a handy equation.

In vector notation, allowing for a non parallel force:

Slide 123 / 127

A third useful expression for power can be derived from the work energy equation when only the KE changes: The power absorbed by a system can be thought of as the rate at which the kinetic energy in the system is changing.

Power Slide 124 / 127

43 A steam engine does 52 J of work in 12 s. What is the power

supplied by the engine? A 3.4 W B 3.9 W C 4.3 W D 4.9 W E 5.7 W

Slide 124 (Answer) / 127

43 A steam engine does 52 J of work in 12 s. What is the power

supplied by the engine? A 3.4 W B 3.9 W C 4.3 W D 4.9 W E 5.7 W

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Answer C

Slide 125 / 127

44 How long must a 350.0 W engine run in order to produce 720.0 kJ

• f work?

A 1987 s B 2057 s C 2146 s D 2305 s E 2861 s

Slide 125 (Answer) / 127

44 How long must a 350.0 W engine run in order to produce 720.0 kJ

• f work?

A 1987 s B 2057 s C 2146 s D 2305 s E 2861 s

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Answer B

Slide 126 / 127

45 A 12.0 kW motor runs a vehicle at a speed of 8.00 m/s. What is

the force supplied by the motor (assume all of the power is delivered in the direction of the vehicle's motion). A 1000 N B 1250N C 1400 N D 1450 N E 1500 N

Slide 126 (Answer) / 127

45 A 12.0 kW motor runs a vehicle at a speed of 8.00 m/s. What is

the force supplied by the motor (assume all of the power is delivered in the direction of the vehicle's motion). A 1000 N B 1250N C 1400 N D 1450 N E 1500 N

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Answer E

Slide 127 / 127