1. Preview Section 1 Work Section 2 Energy
Section 3 Conservation of Energy
Section 4 Power

2. What do you think?
List five examples of things you have done in the last year that you would consider work.
Based on these examples, how do you define work?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
Likely answers are: homework, babysitting, jobs, studying physics, and so on. After listening and discussing, let the students know that in physics, the definition of work is much more precise and many things that they consider work will not fit that definition. In physics, work produces a change in energy. Work is defined in terms of force and displacement on the next slide.

3. Work
In physics, work is the magnitude of the force (F) times the magnitude of the displacement (d) in the same direction as the force.
W = Fd
What are the SI units for work?
Force units (N)  distance units (m)
N•m are also called joules (J).
How much work is 1 joule?
Lift an apple weighing about 1 N from the floor to the desk, a distance of about 1 m.
Units are sometimes confusing. It would be a good idea to show students that 1 J = 1 N•m = 1 kg•m2/s2 at this time. Give them a chance to figure it out for themselves from the definition of a newton (F=ma). This is important because they will later learn that kinetic energy and potential energy are measured in joules as well, and the equations lead to kg•m2/s2. Students need to understand the fundamental SI units behind all of the derived units such as newtons, joules, watts and so on.

4. Work
Pushing this car is work because F and d are in the same direction.
Why aren’t the following tasks considered work?
A student holds a heavy chair at arm’s length for several minutes.
A student carries a bucket of water along a horizontal path while walking at a constant velocity.
In the first case, no work is done because the object does not move (d = 0).
In the second case, no work is done because the distance moved is not in the direction of the force (the force is vertically upward while the distance is horizontal). There is no component of the force in the horizontal direction.

5. Work How would you calculate the work in this case?
What is the component of F in the direction of d?
F cos 
If the angle is 90°, what is the component of F in the direction of d?
F cos 90° = 0
If the angle is 0°, what is the component of F in the direction of d?
F cos 0° = F
Discussion of the component of F along the direction of d should lead to the equation on the next slide.

6. Work
Students should already have deduced this equation from the last slide.

7. Work is a Scalar
Work can be positive or negative but does not have a direction.
What is the angle between F and d in each case?
Show students that the two diagrams on the left show force and distance in opposite directions, while those on the right show force and distance in the same direction.
Ask the angle between the force and distance in the top left diagram. It looks like it is roughly 135°. Point out to them that the cos(135°) is a negative number.
The angle on the top right is about 45° (cos is +).
The angle on the bottom left is about 225° (cos is -).
The angle on the bottom right is about 315° (cos is +).
For the bottom pictures, it will be harder for students to determine the angle unless they draw the force and distance starting at a common point.

8. Classroom Practice Problem
How much work is done on a vacuum cleaner pulled 3.0m by a force of 50.0N at an angle of 30.0º above the horizontal?
Answer: 130J
Students may use the mass instead of the weight (20.0 kg x 9.81 m/s2). This is a good time to remind them that mass and weight are different although related quantities.

9. Classroom Practice Problem
A tugboat pulls a ship with a constant net horizontal force of 5,000N, causing the ship to move through a harbor. How much work does the tugboat do on the ship if each moves a distance of 3.00 km?
Answer: 1.5 X 10 7 J or 15,000,000 J
Students may use the mass instead of the weight (20.0 kg x 9.81 m/s2). This is a good time to remind them that mass and weight are different although related quantities.

10. Classroom Practice Problem
A weightlifter lifts a set of weights a vertical distance of 2.00 m. If a constant net force of 350 N is exerted on the weights, how much work does the weight lifter do on the weights?
Answer: 7.0 X 10 2 J or 700 J
Students may use the mass instead of the weight (20.0 kg x 9.81 m/s2). This is a good time to remind them that mass and weight are different although related quantities.

11. Classroom Practice Problem
A shopper in a supermarket pushes a cart with a force of 35N directed at an angle of 25º downward from the horizontal. Find the work done by the shopper on the cart as the shopper moves along a 50.0m length of aisle?
Answer: 1.6 X 10 3 J or 1,600 J
Students may use the mass instead of the weight (20.0 kg x 9.81 m/s2). This is a good time to remind them that mass and weight are different although related quantities.

12. What do you think? You have no doubt heard the term kinetic energy.
What is it?
What factors affect the kinetic energy of an object and in what way?
You have no doubt heard the term potential energy.
What factors affect the potential energy of an object and in what way?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
Kinetic energy comments will likely center around the velocity and probably not mention the mass of the object. Students may mention the different types of potential energy (gravitational, electrostatic, elastic, etc). Listen to ideas about these topics and then begin the lecture/discussion.

13. Kinetic Energy Since then or
Show students the steps and substitutions needed to derive the final equation for work. Make sure they see the use of F = ma in the first equation and the substitution for ax from the 2nd equation into the third equation. Help students see the transformation of the 3rd equation into the 4th equation.
Have them note that calculating the work no longer requires knowledge of the force but, instead, can be determined by the effect of the force or the change in velocity.
Mention that a name has been given to the quantity 1/2 mv2 . It is called kinetic energy. So, work is the change in KE. Then show them the next slide, which introduces the kinetic energy equation.

14. Kinetic Energy What are the SI units for KE? kg•m2/s2 or N•m or J
Ask students to determine the units from the equation before showing this on the slide. Have them see that, since N are kg•m/s2, the units of N•m are equivalent to kg•m2/s2.

15. Work and Kinetic Energy
KE is the work an object can do if the speed changes.
Wnet is positive if the speed increases.
Discuss the many examples of moving objects doing work on other objects. For example, a moving baseball bat does work on a ball as it exerts a force on the ball, and the ball moves a distance in the direction of the force. Conversely, the ball does work on the bat as it exerts a force opposite to the direction the bat is moving. Work has a negative value in this case. A change in speed for an object allows it to do work on its environment.

16. Classroom Practice Problems
A 6.00 kg cat runs after a mouse at 10.0 m/s. What is the cat’s kinetic energy?
Answer: 3.00 x 102 J or 300 J
Suppose the above cat accelerated to a speed of 12.0 m/s while chasing the mouse. How much work was done on the cat to produce this change in speed?
Answer: 1.32 x 102 J or 132 J
For the second problem, students should just use the change in KE (432 J – 300 J = 132 J).
Sometimes they make the mistake of thinking that they can use the change in speed (2 m/s) in the equation for KE and end up with an answer of 12 J for the work done. This does not work because ( )2 is not equal to ( ).

17. Classroom Practice Problems
Calculate the kinetic energy of a 8.0 X kg airliner flying at km/h.
Answer: 1.1 x 109 J or 1,100,000,000 J
For the second problem, students should just use the change in KE (432 J – 300 J = 132 J).
Sometimes they make the mistake of thinking that they can use the change in speed (2 m/s) in the equation for KE and end up with an answer of 12 J for the work done. This does not work because ( )2 is not equal to ( ).

18. Classroom Practice Problems
Two bullets have masses of 3.0g and 6.0g, respectively. Which bullet has more kinetic energy? What is the ratio of their kinetic energy.
Answer: the bullet with the greater mass has more kinetic energy….2 to 1
For the second problem, students should just use the change in KE (432 J – 300 J = 132 J).
Sometimes they make the mistake of thinking that they can use the change in speed (2 m/s) in the equation for KE and end up with an answer of 12 J for the work done. This does not work because ( )2 is not equal to ( ).

19. Classroom Practice Problems
Two 3.0 g bullets are fired with velocities of 40.0 m/s and 80.0 m/s, respectively. What are their kinetic energies?Which bullet has more kinetic energy? What is the ratio of their kinetic energy.
Answer: 2.4 J and 9.6 J. The bullet with the greater velocity has more kinetic energy….4 to 1
For the second problem, students should just use the change in KE (432 J – 300 J = 132 J).
Sometimes they make the mistake of thinking that they can use the change in speed (2 m/s) in the equation for KE and end up with an answer of 12 J for the work done. This does not work because ( )2 is not equal to ( ).

20. Classroom Practice Problems
A running student has half the kinetic energy that his younger brother has. The student speeds up by 1.3 m/s, at which point he has the same kinetic energy as his brother. If the student’s mass is twice as large as his brother’s mass, what were the original speeds of both the student and his brother?
Answer: 3.2 m/s; 6.4 m/s
For the second problem, students should just use the change in KE (432 J – 300 J = 132 J).
Sometimes they make the mistake of thinking that they can use the change in speed (2 m/s) in the equation for KE and end up with an answer of 12 J for the work done. This does not work because ( )2 is not equal to ( ).

21. Potential Energy
Energy associated with an object’s potential to move due to an interaction with its environment
A book held above the desk
An arrow ready to be released from the bow
Some types of PE are listed below.
Gravitational
Elastic
Electromagnetic
Hold a book above the desk. The book has the potential to move due to an interaction with Earth (gravity).
Stretch a rubber band with a wad of paper held in it like a sling shot. The paper has the potential to move due to an interaction with its environment (the rubber band).

22. Gravitational Potential Energy
What are the SI units?
kg•m2/s2 or N•m or J
The height (h) depends on the “zero level” chosen where PEg = 0.
This equation comes from W = Fd = (ma)d = mgh, so PEg is simply the work done in lifting an object.
To help students understand the fact that the zero level is arbitrary, hold a book over the desk and ask them what they would use for h in order to calculate the PE. Then, maintaining the book at the same height, move it over the floor and ask the students once again what value they would use for h.
Point out that, in general, our primary concern in physics involves changes in PE, not the actual amount of PE. The change in PE is always the same regardless of what zero level is assigned. Generally, the zero level is assigned to the lowest point the object will reach. For example, the desk if the book is held over the desk, and the floor if the book is held over the floor.

23. Elastic Potential Energy
The energy available for use in deformed elastic objects
Rubber bands, springs in trampolines, pole-vault poles, muscles
For springs, the distance compressed or stretched = x
Point out that x in the diagram is the “Distance compressed.” This will be used in the equation for elastic potential energy (slide 10).
Discuss the transfer of the elastic potential energy to the block when the deformed spring returns to its original configuration.

24. Spring Constant(k) Click below to watch the Visual Concept.

25. Elastic Potential Energy
The spring constant (k) depends on the stiffness of the spring.
Stiffer springs have higher k values.
Measured in N/m
Force in newtons needed to stretch a spring 1.0 meters
What are the SI Units for PEelastic?
Help students find the SI units as N•m or Kg•m2/s2 or J.
Now would be a good time to remind them that work, KE, and PE are all measured in joules (kg•m2/s2).

26. Classroom Practice Problems
A 70 kg stuntman is attached to a bungee cord with an unstretched length of 15m. He jumps off a bridge spanning a river from a height of 50.0m. When he finally stops, the cord has stretched the length of 44.0m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?
3.43 X 10 4 J or 34,300 J
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

27. Classroom Practice Problems
If the stuntman above changes bungee cords, what will the spring constant of the new cord be if the total potential energy is the same as before and the man stops 12.0m above the water level? Assume the unstretched cord has a length of 15.0m.
98.7 N/m
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

28. Classroom Practice Problems
The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150m?
3.1 X J or .031 J
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

29. Classroom Practice Problems
A spring with a force constant of 5.3 N/m has relaxed length of 2.45m/ When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57m Calculate the elastic potential energy stored in the spring?
3.3J
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

30. Classroom Practice Problems
A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions?
When the rope is horizontal
785 J
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

31. Classroom Practice Problems
A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions?
When the rope is at a 30.0º with the vertical
105 J
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

32. Classroom Practice Problems
A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions?
When the rope is at the bottom of the circular arc
0 J
Point out to students that the zero level is at the table for gravitational PE. Also, they must use meters, not centimeters, in order to have joules as units in the final answers.

33. Now what do you think? What is kinetic energy?
What factors affect the kinetic energy of an object and in what way?
How are work and kinetic energy related?
What is potential energy?
What factors affect the gravitational potential energy of an object and in what way?
What factors affect the elastic potential energy of an object and in what way?
Mass and KE are directly proportional. KE is directly proportional to the velocity squared.
The net work done on an object equals the change in kinetic energy.
Factors affecting gravitational PE are mass, acceleration due to gravity, and height above the zero level. All are directly related.
Factors affecting elastic PE are the spring constant (directly related) and the displacement from equilibrium position (directly related to the displacement squared).

34. What do you think?
Imagine two students standing side by side at the top of a water slide. One steps off of the platform, falling directly into the water below. The other student goes down the slide. Assuming the slide is frictionless, which student strikes the water with a greater speed?
Explain your reasoning.
Would your answer change if the slide were not frictionless? If so, how?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
A variety of answers are possible. Encourage them to explain their ideas so you can better understand the answer they chose. Remind the students that friction is not a consideration. Water slides are not truly frictionless, so they might want to imagine an air hockey table as a slide. Students will find out later in the lesson that, in the absence of friction, the speeds would be the same. The only difference would be the direction each is traveling at impact.

35. What do you think?
What is meant when scientists say a quantity is conserved?
Describe examples of quantities that are conserved.
Are they always conserved? If not, why?
Students may be familiar with conservation of mass as a conservation law. Try to get them to explain what it means for something to be “conserved.” There are many ways to describe conservation but students often struggle. They may say that they know what it means but they can’t figure out how to say it or write it. These questions should be a good exercise for them to express their ideas in writing.
Conservation may be expressed as “no change in the quantity” or “before = after” or “the total always stays the same.”

36. Mechanical Energy (ME)
ME = KE + PEg + PEelastic
Does not include the many other types of energy, such as thermal energy, chemical potential energy, and others
ME is not a new form of energy.
Just a combination of KE and PE
Discuss ME as a useful tool for studying motion. Do not tell students yet that ME is conserved. They will determine this from the coming slides and calculations.
As an example. toss a ball in the air and talk about the potential energy and kinetic energy as it rises and falls. As another example. show students a pendulum and talk about the PE and the KE changing as it swings.

37. Classroom Practice Problems
Starting from rest, a child zooms down a frictionless slide with an initial height of 3.00m What is her speed at the bottom of the slide? Assume she has a mass of 25 kg.
7.67 m/s.
Students should use PE = mgh to get the PE at each point.
To calculate the KE they need to first find the velocity. The easiest way to get v2 is using vf2 = 2gy. (Note that the initial velocity is zero, so it was eliminated from the equation). After getting the velocity, use the equation KE = 1/2 mv2.
After making these calculations, show students the chart on the next slide.

38. Classroom Practice Problems
A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00kg fish. Assuming the altitude of the bird is 5.40m, and disregarding friction, what is the speed of the fish when it hits the water?
20.7 m/s.
Students should use PE = mgh to get the PE at each point.
To calculate the KE they need to first find the velocity. The easiest way to get v2 is using vf2 = 2gy. (Note that the initial velocity is zero, so it was eliminated from the equation). After getting the velocity, use the equation KE = 1/2 mv2.
After making these calculations, show students the chart on the next slide.

39. Classroom Practice Problems
A 755N diver drops from a board 10.0m above the water’s surface. Find the divers speed 5.00m above the water’s surface? Find the diver’s speed just before striking the water
9.9m/s and 14.0 m/s
Students should use PE = mgh to get the PE at each point.
To calculate the KE they need to first find the velocity. The easiest way to get v2 is using vf2 = 2gy. (Note that the initial velocity is zero, so it was eliminated from the equation). After getting the velocity, use the equation KE = 1/2 mv2.
After making these calculations, show students the chart on the next slide.

40. Classroom Practice Problems
An olympic high jumper leaps over a horizontal bar. The jumper’s center of mass is raised 0.25m during the jump. Calculate the minim speed with which the athlete must leave the ground to perform this feat.
2.2m/s
Students should use PE = mgh to get the PE at each point.
To calculate the KE they need to first find the velocity. The easiest way to get v2 is using vf2 = 2gy. (Note that the initial velocity is zero, so it was eliminated from the equation). After getting the velocity, use the equation KE = 1/2 mv2.
After making these calculations, show students the chart on the next slide.

41. Classroom Practice Problems
A pendulum 2.0m long is released from rest when the support string is at an angle of 25.0 º with the vertical. What is the speed of the bob at the bottom of the swing?
1.9m/s
Students should use PE = mgh to get the PE at each point.
To calculate the KE they need to first find the velocity. The easiest way to get v2 is using vf2 = 2gy. (Note that the initial velocity is zero, so it was eliminated from the equation). After getting the velocity, use the equation KE = 1/2 mv2.
After making these calculations, show students the chart on the next slide.

42. Conservation of Mechanical Energy
The sum of KE and PE remains constant.
One type of energy changes into another type.
For the falling book, the PE of the book changed into KE as it fell.
As a ball rolls up a hill, KE is changed into PE.
Conservation of energy provides an easier method of solving some problems. For example, go back to the table created for the falling book, and point out the fact that they never needed to calculate the velocity. If they know the PE, then the KE is simply (19.6 J - PE).

43. Conservation of Mechanical Energy
Click below to watch the Visual Concept.
Visual Concept

44. Conservation of Energy
Acceleration does not have to be constant.
ME is not conserved if friction is present.
If friction is negligible, conservation of ME is reasonably accurate.
A pendulum as it swings back and forth a few times
Consider a child going down a slide with friction.
What happens to the ME as he slides down?
Answer: It is not conserved but, instead, becomes less and less.
What happens to the “lost” energy?
Answer: It is converted into nonmechanical energy (thermal energy).
Previously, in Chapter 2, the equations developed required that acceleration be constant. That is not the case for conservation of ME.
The example with friction provides an opportunity to point out that even though ME may not be conserved, energy in general is conserved. With friction, the material becomes warmer due to the contact between the surfaces, which causes the speed of the vibrating molecules to increase.

45. Classroom Practice Problems
On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10?
2.4m
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

46. Classroom Practice Problems
A 2.0 X 103 kg car moves down a level highway under the actions of two forces. One is a 1140N forward force exerted on the wheels by the road. The other is a 950 N resistive force exerted on the car by the air. Use the work-kinetic energy theorem to find the speed of the car after it has moved a distance of 20.0m , assuming the car starts from rest?
1.9 m/s
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

47. Classroom Practice Problems
A 2.10 X 103 kg car starts from rest at the top of a driveway 5.0 m long that is sloped at 2.0.0º with the horizontal. If an average friction force of 4.0 X 103 N impedes the motion, what is the speed of the car at the bottom of the driveway?
3.8 m/s
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

48. Classroom Practice Problems
A 10.0 kg crate is pulled up a rough incline with a initial speed of 1.5 m/s. The pulling force is 100.0N parallel to the incline, which makes an angle of 15.0 º with the horizontal . Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following…the work done by the Earth’s gravity on the crate
-1.9 X 102 J
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

49. Classroom Practice Problems
A 10.0 kg crate is pulled up a rough incline with a initial speed of 1.5 m/s. The pulling force is 100.0N parallel to the incline, which makes an angle of 15.0 º with the horizontal . Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following…the work done by the force of friction on the crate
-2.8 X 102 J
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

50. Classroom Practice Problems
A 10.0 kg crate is pulled up a rough incline with a initial speed of 1.5 m/s. The pulling force is 100.0N parallel to the incline, which makes an angle of 15.0 º with the horizontal . Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following…the work done by the puller on the crate
7.5X 102 J
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

51. Classroom Practice Problems
A 10.0 kg crate is pulled up a rough incline with a initial speed of 1.5 m/s. The pulling force is 100.0N parallel to the incline, which makes an angle of 15.0 º with the horizontal . Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following…the change in kinetic energy of the crate
2.8 X 102 J
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

52. Classroom Practice Problems
A 10.0 kg crate is pulled up a rough incline with a initial speed of 1.5 m/s. The pulling force is 100.0N parallel to the incline, which makes an angle of 15.0 º with the horizontal . Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following…the speed of the crate after it is pulled 7.5 m
7.6 m/s
Remind students that J require the use of kg and m (not g and cm).
Answer 1: 0.36 J. Students should use PE = 1/2 kx2 to determine the energy stored in the rubber band.
Answer 2: 0.36 J. Students should use conservation of mechanical energy. PE lost = KE gained
Answer 3: 8.5 m/s. Students should use the KE from the last question and KE = 1/2 mv2 to determine v
Answer 4: 3.7 m. Students may revert to equations from Chapter 2. but this is most easily solved using conservation of ME. They only need to find the height required for KE lost = Peg gained = 0.36 J = mgh.

53. Now what do you think?
Imagine two students standing side by side at the top of a water slide. One steps off of the platform, falling directly into the water below. The other student goes down the slide. Assuming the slide is frictionless, which student strikes the water with a greater speed?
Explain your reasoning.
Would your answer change if the slide were not frictionless? If so, how?
Both strike at the same speed. One hits the water vertically, while the other slides in nearly horizontally. The PE lost is the same for both and, therefore, the KE gained is the same as well.
With friction, the student falling straight down would be moving faster, because energy is not conserved for the student on the slide. Some of the lost PE is converted into thermal energy

54. Now what do you think?
What is meant when scientists say a quantity is “conserved”?
Describe examples of quantities that are conserved.
Are they always conserved? If not, why?
Conserved means that the quantity does not change. The quantity neither increases nor decreases. Common examples are the conservation of mass and the conservation of energy. Mechanical energy is considered to be conserved in cases where friction is negligible. If friction is significant, mechanical energy is not conserved.

55. What do you think?
Two cars are identical with one exception. One of the cars has a more powerful engine. How does having more power make the car behave differently?
What does power mean?
What units are used to measure power?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
For this question, students may discuss the top speed of the cars or the time required for the cars to accelerate from 0 to 60 miles per hour. See if you can get them to talk about energy consumption. How quickly would each consume a gallon of gas? They likely will mention horsepower, but it is unlikely that any will consider watts as a unit of power.

56. Power The rate of energy transfer Energy used or work done per second
Remind students that W = Fd, and ask them to substitute this for W in the power formula. Then ask what d/t represents. At this point, move on to the next slide, which shows the alternative form of the power equation (P = Fv).

57. Power SI units for power are J/s.
Called watts (W)
Equivalent to kg•m2/s3
Horsepower (hp) is a unit used in the Avoirdupois system.
1.00 hp = 746 W
Be sure students understand that P = Fv is not a new definition. It is simply a different but equivalent formula that makes calculations easier in some cases.
Do not just show the units. Ask students to figure them out. It should be easy for them to get J/s but the basic units of kg•m2/s3 will be more difficult. This provides a good opportunity to review the units for joules and newtons.
The horsepower was based on the work a good horse could do lifting coal out of a mine. A good horse could lift 275 pounds of coal at 2.0 ft/s, so it could do 550 ft•lb/s. This is equivalent to 746 J/s or 746 W.

58. Watts These bulbs all consume different amounts of power.
A 100 watt bulb consumes 100 joules of energy every second.
Remind students that a joule is the energy required to lift an apple a distance of a meter. Light bulbs are also rated for the amount of light they produce. Fluorescent bulbs using 22 watts can produce as much light as incandescent bulbs using 100 watts.

59. Classroom Practice Problems
A 193 kg curtain needs to be raised 7.5 m in as closed to 5.0s as possible. Three mototes are available. The power rating for the threemotors are listed as 1.0 kW, 3.5 kW, and 5.5 kW. Which motor is best for the job
Answers: 3.5 kW
These two problems can be done in either order. Most students will use P = Fv, and then use P = W/t to get the work done. However, it is worth showing them that they could calculate the work done by using W = Fd, where d = 2.0 m/s  s = 1.2 x 103 m. After getting the work, then they could use the fundamental definition of power (P=W/t) to get the power.

60. Classroom Practice Problems
A 1.0 X 10 kg elevator carries a maximum load of kg. A constant frictional force of 4.0 X 10 N slows the elevator’s motion upward. What minimum power, in kW, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?
Answers: 66 kW
These two problems can be done in either order. Most students will use P = Fv, and then use P = W/t to get the work done. However, it is worth showing them that they could calculate the work done by using W = Fd, where d = 2.0 m/s  s = 1.2 x 103 m. After getting the work, then they could use the fundamental definition of power (P=W/t) to get the power.

61. Classroom Practice Problems
A 1.5 X 10 3 kg car accelerates uniformly from rest to 10.0 m/s in 3.00 s. What is the work done on the car in this time interval? What is the power delivered by the engine in this time interval
Answers: 75,00 J and 25,000 w
These two problems can be done in either order. Most students will use P = Fv, and then use P = W/t to get the work done. However, it is worth showing them that they could calculate the work done by using W = Fd, where d = 2.0 m/s  s = 1.2 x 103 m. After getting the work, then they could use the fundamental definition of power (P=W/t) to get the power.

62. Classroom Practice Problems
A car with a mass of 1,500 kg starts from rest and accelerates to 18.0 m/s in 12.0 s. Assume that air resistance remains constant at N during this time interval. What is the average power developed by the engine?
Answers: 23,800 W or 23.8 kW
These two problems can be done in either order. Most students will use P = Fv, and then use P = W/t to get the work done. However, it is worth showing them that they could calculate the work done by using W = Fd, where d = 2.0 m/s  s = 1.2 x 103 m. After getting the work, then they could use the fundamental definition of power (P=W/t) to get the power.

63. Classroom Practice Problems
A rain cloud contains 2.66 X kg of water vapor. How long would it take for a 2.00 kW pump to raise the same amount of wter to the cloud’s altitude, 2.00 km?
Answers: 2.61 X 10 8 s or 8.27 years
These two problems can be done in either order. Most students will use P = Fv, and then use P = W/t to get the work done. However, it is worth showing them that they could calculate the work done by using W = Fd, where d = 2.0 m/s  s = 1.2 x 103 m. After getting the work, then they could use the fundamental definition of power (P=W/t) to get the power.