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RRHS Physics 11 Unit 3 Slide #1 Unit 4 Energy and Momentum This unit is a continuation of the study of mechanics that trainees studied in Units 2 and 3.

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Presentation on theme: "RRHS Physics 11 Unit 3 Slide #1 Unit 4 Energy and Momentum This unit is a continuation of the study of mechanics that trainees studied in Units 2 and 3."— Presentation transcript:

1 RRHS Physics 11 Unit 3 Slide #1 Unit 4 Energy and Momentum This unit is a continuation of the study of mechanics that trainees studied in Units 2 and 3. It involves looking at mechanics from a different point of view (using work end energy) that can simplify some otherwise complicated situations.

2 RRHS Physics 11 Unit 3 Slide #2 Module 4.1 – Momentum and Impulse Momentum and Impulse is another way of looking at the motion of objects, and is a continuation of the last unit. There are some situations where it is difficult to determine all of the individual forces acting on the objects. Looking at these situations from the point of view of momentum and impulse can allow us to more easily analyze the interaction between objects and to predict their behaviour.

3 RRHS Physics 11 Unit 3 Slide #3 Module 4.1 Objectives 1.To be able to calculate the momentum of an object. 2.To understand the relationship between change in momentum and impulse. 3.To be able to apply conservation of momentum to a variety of one- dimensional collisions and explosions.

4 RRHS Physics 11 Unit 3 Slide #4 Momentum Momentum depends on mass and velocity Momentum is a vector and is defined as: where momentum p is measured in kg·m/s when mass is measured in kg and velocity is measured in m/s

5 RRHS Physics 11 Unit 3 Slide #5 Example Calculate the momentum of a 1200 kg car travelling at 90.0 km/h north. Solution:

6 RRHS Physics 11 Unit 3 Slide #6 Check Your Learning 1.Which has more momentum, a falling raindrop or a transport truck parked along the side of the road? Since the truck is not moving (using the Earth as the frame of reference), it has zero momentum; the raindrop does have momentum, since it is actually moving. Even though the truck has a much larger mass, it does not have any momentum.

7 RRHS Physics 11 Unit 3 Slide #7 Check Your Learning 2.Calculate the momentum of a kg cricket ball that has a velocity of 110 km/h. The positive sign on the momentum indicates the direction of the momentum. The velocity could also be left in km/h, which would result in a momentum of

8 RRHS Physics 11 Unit 3 Slide #8 Change in Momentum Change in velocity causes change in momentum This product is referred to as the impulse

9 RRHS Physics 11 Unit 3 Slide #9 Impulse Change in Momentum and Impulse are equivalent: an impulse of 1 N·s will provide a change in momentum of 1 kg·m/s.

10 RRHS Physics 11 Unit 3 Slide #10 Collisions Collisions usually occur in a short time; force changes quickly The area of the graph represents the impulse

11 RRHS Physics 11 Unit 3 Slide #11 Application Smaller time Larger force Larger timeSmaller force since impulse is usually fixed

12 RRHS Physics 11 Unit 3 Slide #12 Example 1 An egg when thrown against a brick wall will break; however, an egg thrown at the same speed into a sagging sheet will likely survive. Explain why using the concepts of impulse and momentum. Solution: The change in momentum of the egg is the same in each case, since the mass of the egg does not change and the egg goes from the same initial velocity to rest in each case. Since the change in momentum is the same, the impulse must also be the same; however, the brick wall applies the impulse in a very short time, which means the force must be large. The large force causes the egg to break. The sheet brings the egg to a stop more gradually, thereby lengthening the time that the impulse is applied over. This allows a smaller force to be applied to the egg and does not cause the egg to break.

13 RRHS Physics 11 Unit 3 Slide #13 Example 2 A tennis player hits a ball with a racket, giving it a speed of 43 m/s. If the ball has a mass of kg and is in contact with the racket for 4.5×10 -3 s, what is the average force exerted on the ball by the racket?

14 RRHS Physics 11 Unit 3 Slide #14 Check Your Learning A 0.15 kg cricket ball is approaching a bat with a speed of 95 km/h. After it is hit by the bat, it has a speed in the opposite direction of 115 km/h. The interaction between the bat and the ball lasted s. a)What was the change in momentum of the ball? It is important to remember that since the ball changes direction, the final velocity must be made negative (if we take the ball’s original direction to be positive)

15 RRHS Physics 11 Unit 3 Slide #15 Check Your Learning b)What impulse was applied to the ball? Since impulse is equivalent to change in momentum, c)What was the average force exerted on the ball?

16 RRHS Physics 11 Unit 3 Slide #16 Conservation of Momentum

17 RRHS Physics 11 Unit 3 Slide #17 Conservation of Momentum Law of Conservation of Momentum - the total momentum of an isolated system remains constant.

18 RRHS Physics 11 Unit 3 Slide #18 Example 1 Someone throws a heavy ball to you when you are standing on a skateboard. You catch it and roll backward with the skateboard. If you were standing on the ground, however, you would be able to avoid moving. Explain both using momentum conservation. Solution: Consider first the case of the person standing on the skateboard when he catches the ball. In this case there is very little friction, so little that it can be ignored compared to the force exerted on the person by the heavy ball. If we define our system to be the ball and the person on the skateboard, conservation of momentum says that the total momentum must stay the same. Before the collision of the ball with the person, the ball has momentum since it is moving. After the person catches the ball, the ball and the person act as one object. The combined momentum of the ball and the person must still be equal to the momentum of the ball before the collision; therefore, the person and the ball must be moving.

19 RRHS Physics 11 Unit 3 Slide #19 Example 1 Now consider the case of the person standing on the ground. If we again define our system to be the ball and the person, there is now a significant outside force acting on the person since friction is now increased. Since there is a significant outside force, the system is no longer isolated and conservation of momentum does not apply. We can, however, define a different system where conservation of momentum will apply. If we now consider our system to be the ball, the person, and the Earth then the friction between the person and the Earth is now an internal force and our system is again isolated. Now the momentum of the ball, person, and the Earth (since they can be considered to be one object after the collision) will be the same as the momentum of the ball before the collision; however, because this combined mass is so much larger than the mass of the ball alone the velocity will be so small that it will be negligible.

20 RRHS Physics 11 Unit 3 Slide #20 Example 2 In the previous example of the person catching the ball while on the skateboard, assume that the ball has a mass of 6.0 kg and is thrown with a speed of 3.4 m/s. The combined mass of the person and the skateboard is 68.0 kg and the person is initially at rest. Find the velocity of the person on the skateboard after catching the ball. Solution: Before the collision, only the ball is moving. After the collision, the ball, person, and the skateboard are moving as one object. It is the total velocity of this combined mass that we are trying to find

21 RRHS Physics 11 Unit 3 Slide #21 Example 3 The person on the skateboard is now standing on the skateboard at rest and holding the heavy ball. The mass of the person and the skateboard is still 68.0 kg and the mass of the ball is still 6.0 kg. The person throws the ball, giving it a velocity of 4.2 m/s to the right, and is amazed to see that he begins rolling to the left. What is the person’s velocity after throwing the ball? Solution: This is an example of an “explosion”. The initial momentum of the system is zero, and must remain at zero since it is an isolated system. We will take right to be the positive direction.

22 RRHS Physics 11 Unit 3 Slide #22 Example 3

23 RRHS Physics 11 Unit 3 Slide #23 Check Your Learning A 455 g rubber ball is thrown at a 2.10 kg block with a speed of 5.0 m/s. Assume that the block is at rest on a frictionless surface. The rubber ball rebounds in the opposite direction with a speed of 3.2 m/s. What is the speed of the block after the collision?

24 RRHS Physics 11 Unit 3 Slide #24 Module Summary In this module, you have learned that Momentum is calculated using the formula The change in momentum can be calculated from the change in velocity Impulse is equal to the change in momentum and is given by the formula

25 RRHS Physics 11 Unit 3 Slide #25 Module Summary Momentum is conserved in all isolated systems. This is known as the law of conservation of momentum

26 RRHS Physics 11 Unit 3 Slide #26 Module 4.2 Work, Power and Energy Until now, forces and acceleration have been used to describe and understand motion. We will now look at another model that is used to analyze motion – work and energy. This new model allows us to study situations where the forces are changing throughout the problem, and simplifies some otherwise complex situations. You will also learn in this module what machines can and cannot do to help us to complete tasks.

27 RRHS Physics 11 Unit 3 Slide #27 Mechanical Energy Kinetic Energy Potential Energy Work Transfer of Mechanical Energy

28 RRHS Physics 11 Unit 3 Slide #28 Work Requirements Work requires 3 things: There must be a force There must be a displacement The displacement must be a result of (or caused by) the force.

29 RRHS Physics 11 Unit 3 Slide #29 Work General Equation

30 RRHS Physics 11 Unit 3 Slide #30 Special Cases 1.Force and Displacement in the same direction (θ = 0°, cos θ = 1)

31 RRHS Physics 11 Unit 3 Slide #31 Example 1 A person uses their hand to lift a 5.3 kg box 1.2 m into the air (at a constant speed). How much work did the person do? Solution:

32 RRHS Physics 11 Unit 3 Slide #32 Special Cases 2.Force and Displacement are perpendicular to each other (θ = 90°, cos θ = 0)

33 RRHS Physics 11 Unit 3 Slide #33 Example 2 A person walks 3.2 m across the room carrying a 5.3 kg box at a constant speed. How much work was done by the person? Solution:

34 RRHS Physics 11 Unit 3 Slide #34 Special Cases 3.Force and Displacement are in opposite directions (θ = 180°, cos θ = -1)

35 RRHS Physics 11 Unit 3 Slide #35 Example 3 A person holding a 5.3 kg box in the air lowers the box 1.2 m. How much work is done by the person? Solution:

36 RRHS Physics 11 Unit 3 Slide #36 Check Your Learning A 12 kg box is being pulled by Bill across the floor. Bill is exerting a force of 45 N, and the coefficient of friction is He pulls the box a distance of 7.2 m. a)How much work was done by Bill?

37 RRHS Physics 11 Unit 3 Slide #37 Check Your Learning b)How much work was done by friction?

38 RRHS Physics 11 Unit 3 Slide #38 Check Your Learning c)Calculate the net work done on the box. or Although Bill did 320 J of work, only 140 J of it was actually transferred to the box – the rest was lost to friction as heat energy

39 RRHS Physics 11 Unit 3 Slide #39 Example 1 Consider a 75 kg person who walks up a flight of stairs (shown below) compared to a similar person (same mass) who runs up the stairs. The person walking takes 6.3 s, and the person running takes 2.6 s. Assume both are moving at a constant speed. How much work does each person do?

40 RRHS Physics 11 Unit 3 Slide #40 Example 1 - Solution Person 1 Since time was not used, person 2 does the same amount of work!!

41 RRHS Physics 11 Unit 3 Slide #41 Power Since Person 2 did the work more quickly, we say that they used more power. Power is defined as the rate of doing work (or transferring energy): 1 watt (W) = 1 J/s

42 RRHS Physics 11 Unit 3 Slide #42 Example 1 - Continued Person 1Person 2

43 RRHS Physics 11 Unit 3 Slide #43 Example 2 A truck must exert 3750 N of force to pull a trailer at a speed of 90.0 km/h. What power output (in horsepower) is required of the truck? Solution:

44 RRHS Physics 11 Unit 3 Slide #44 Example 2 New Equation

45 RRHS Physics 11 Unit 3 Slide #45 Check Your Learning A person is pulling a sled that has a mass of 93 kg. The sled starts from rest and you can assume that there is no friction. The person is pulling with a horizontal force of 225 N over a distance of 12.0 m. What was the average power developed by the person? We must first find the work done by the person:

46 RRHS Physics 11 Unit 3 Slide #46 Check Your Learning We must now use the rest of the information to find the time so that we can get the power:

47 RRHS Physics 11 Unit 3 Slide #47 Kinetic Energy Energy = ability to do work Kinetic Energy is energy of motion Consider the case where a car with an initial velocity is being pushed horizontally along a table with a net force. Since there is a net force in the direction of motion, there is net work being done.

48 RRHS Physics 11 Unit 3 Slide #48 Kinetic Energy Since If we define kinetic energy to be

49 RRHS Physics 11 Unit 3 Slide #49 Work-Kinetic Energy Theorem or

50 RRHS Physics 11 Unit 3 Slide #50 Example 1 A kg rugby ball is thrown with a horizontal speed of 15.0 m/s. a)What is its kinetic energy? b)What was the work done by the person throwing the ball? Solution: a)

51 RRHS Physics 11 Unit 3 Slide #51 Example 1 b)The only horizontal force acting on the ball was the one exerted by the person, so the work done by the person is the net work.

52 RRHS Physics 11 Unit 3 Slide #52 Example 2 How much work is required to accelerate a 1250 kg car from 50.0 km/h to 95.0 km/h? Solution:

53 RRHS Physics 11 Unit 3 Slide #53 Check Your Learning A kg car is travelling at a speed of 50.0 km/h when the driver applies the brakes. The car comes to a stop after travelling 15 m. a)How much work was done by friction in stopping the car?

54 RRHS Physics 11 Unit 3 Slide #54 Check Your Learning b)What was the magnitude of the average frictional force applied to the car?

55 RRHS Physics 11 Unit 3 Slide #55 Check Your Learning c)If the same car were travelling at 100. km/h when the driver applied the brakes and the car experienced the same frictional force, how far would the car travel before coming to a stop? Notice that by doubling the car’s speed, the stopping distance is multiplied by four.

56 RRHS Physics 11 Unit 3 Slide #56 Gravitational Potential Energy Potential energy is stored energy – elastic, chemical, gravitational An object with gravitational potential energy has the ability to do work if it is dropped Suppose that we are lifting an object vertically at a constant speed; we must exert an upward force equal to the force of gravity

57 RRHS Physics 11 Unit 3 Slide #57 Gravitational Potential Energy If we define Gravitational Potential Energy to be Then the work done by the person is now Height h used in the calculation of gravitational potential energy must be measured from some reference level, since there is no such thing as an absolute zero position or potential energy.

58 RRHS Physics 11 Unit 3 Slide #58 Example 1 A 225 g book is sitting on a table that is 1.2 m above the floor. It must be lifted to a shelf that is 2.1 m above the floor. a)What is the book’s final potential energy relative to the floor? b)What is the book’s final potential energy relative to the table? c)How much work was required to lift the book to the shelf?

59 RRHS Physics 11 Unit 3 Slide #59 Solution a) b)

60 RRHS Physics 11 Unit 3 Slide #60 Solution c)We will choose the floor to be our reference level:

61 RRHS Physics 11 Unit 3 Slide #61 Work-Energy Theorem System – an object or group of objects Any force exerted by an object in the system on another object in the system is an internal force; any force exerted by an object that is not a part of the system is an external force. Consider a person lifting a box with a force greater than the weight of the box – the potential energy will be increasing (because the box is changing height) and the kinetic energy will be changing because the box will be accelerating.

62 RRHS Physics 11 Unit 3 Slide #62 Work-Energy Theorem Define our system to be the box and the Earth and we will ignore friction for now. The force that is being applied by the person is considered an external force. No Friction

63 RRHS Physics 11 Unit 3 Slide #63 Example 2 An 1125 kg elevator is being lifted to a height of 12.5 m from rest. If the cable on the elevator exerts an upward force of 1.2 x10 4 N what is the final speed of the elevator? Solution:

64 RRHS Physics 11 Unit 3 Slide #64 Example 2

65 RRHS Physics 11 Unit 3 Slide #65 Check Your Learning A 34.0 kg child cycles up a hill to a point that is 6.45 m above her starting point. Find a)the change in the child’s gravitational potential energy. We will use her starting point as the reference level,

66 RRHS Physics 11 Unit 3 Slide #66 Check Your Learning b)the amount of work done by the child against gravity.

67 RRHS Physics 11 Unit 3 Slide #67 Mechanical Advantage Simple machines allow people to use less force in completing tasks, not less work Actual Mechanical Advantage (AMA) is defined as: Ideal MachineNo Friction Work input is equal to work output

68 RRHS Physics 11 Unit 3 Slide #68 Mechanical Advantage Ideal Mechanical Advantage (IMA)

69 RRHS Physics 11 Unit 3 Slide #69 Mechanical Advantage Smaller force must be applied over larger distance so that the work is the same! MA of 2 means that we apply half the force over twice the distance – work done is the same

70 RRHS Physics 11 Unit 3 Slide #70 Example 1 An ideal machine allows the user to exert a force of 350 N in order to lift a 110 kg mass to a height of 5.2 m. a)What is the mechanical advantage of this system? b)Over what distance must the input force be applied? Solution: a)

71 RRHS Physics 11 Unit 3 Slide #71 Example 1 b)Since the machine is ideal, the input work must equal the output work

72 RRHS Physics 11 Unit 3 Slide #72 Efficiency Anytime we do work to transfer energy, some of our work is wasted Useful work output (or useful energy) is less than work input (total energy) in the real world Efficiency defined as or

73 RRHS Physics 11 Unit 3 Slide #73 Example 2 An incandescent light bulb transforms 120 J of electric energy into 5.0 J of light energy. A fluorescent bulb requires 65 J of electrical energy to produce the same amount of light. a)Calculate the efficiency of each type of bulb. b)Why is the fluorescent bulb more efficient than the incandescent bulb? Solution: a)In this example, the electrical energy used by the light is the total energy (or input work) and the light energy is the useful energy or work output.

74 RRHS Physics 11 Unit 3 Slide #74 Example 2

75 RRHS Physics 11 Unit 3 Slide #75 Example 2 b)The fluorescent bulb is more efficient because it does not generate as much heat as the incandescent bulb. This heat is energy that is wasted.

76 RRHS Physics 11 Unit 3 Slide #76 Check Your Learning A pulley system is being used to raise a 78 kg mass to a height of 4.2 m. The input force must be exerted over a distance of 9.0 m. The efficiency of the system is 74%. a)How much work (input) is required to lift this mass?

77 RRHS Physics 11 Unit 3 Slide #77 Check Your Learning b)How much force must be applied? c)What is the actual mechanical advantage? The actual mechanical advantage (AMA) can be found by looking at the ratio of the forces:

78 RRHS Physics 11 Unit 3 Slide #78 Module Summary In this module, you have learned that: Work can be calculated using the formula Power is the rate at which work is done and can be calculated using Kinetic Energy is the energy of motion and can calculated using the formula

79 RRHS Physics 11 Unit 3 Slide #79 Module Summary The net work on an object is equal to its change in kinetic energy The gravitational potential energy of an object can be calculated using the formula The work-energy theorem states that the total external work done on a system is equal to the change in energy

80 RRHS Physics 11 Unit 3 Slide #80 Module Summary The actual mechanical advantage of a machine can be found using the formula The Efficiency of a machine (or any other energy conversion) is given by the equation

81 RRHS Physics 11 Unit 3 Slide #81 Module 4.3 Heat When dealing with equipment, it is often important to know how it will react to heat and changes in temperature. Heat can cause changes in the size of materials, and also must be considered when looking at the total work and energy involved in a system.

82 RRHS Physics 11 Unit 3 Slide #82 Module 4.3 Objective 1 Upon completion of this module, the participant will be able to: Understand the difference between heat, temperature and thermal energy. Upon completion of this module, the participant will be able to: Understand the difference between heat, temperature and thermal energy.

83 RRHS Physics 11 Unit 3 Slide #83 Heat as Energy Transfer Heat is – amount of energy that is being transferred from one object to another because of a difference in temperature. kilocalorie (kcal) – the amount of heat necessary to raise the temperature of 1 kilogram of water by 1 Celsius degree Work can also cause an increase in temperature Mechanical Equivalent of Heat:

84 RRHS Physics 11 Unit 3 Slide #84 Expanded Work-Energy Theorem Work can result in a change in kinetic energy, a change in potential energy, or heat Where Q is the heat produced by friction The external work done on a system or body is equal to the total change in energy of the system

85 RRHS Physics 11 Unit 3 Slide #85 Example 1 Suppose that a 2.0 kg box is pushed along a horizontal surface with a force of 75 N over a distance of 3.8 m. Starting from rest, the box reaches a speed of 10.0 m/s at the end of the distance. How much heat (in kilocalories) was produced and shared between the box and the surface? Solution: If we consider the box and surface to be our system, the force being applied by the person can be used to calculate the external work on the system.

86 RRHS Physics 11 Unit 3 Slide #86 Example 1 Since the box is not changing height, there is no change in gravitational potential energy. There is, however, a change in kinetic energy:

87 RRHS Physics 11 Unit 3 Slide #87 Example 1

88 RRHS Physics 11 Unit 3 Slide #88 Definitions Thermal energy refers to the total energy of all the molecules in an object; Temperature is a physical measurement that is related to the average kinetic energy of individual molecules. Heat refers to a transfer of energy from one object to another because of a difference in temperature.

89 RRHS Physics 11 Unit 3 Slide #89 Types of Heat Transfer Conduction – happens through collisions between the molecules in an object. Molecules at the hot end of an object have a lot of kinetic energy and are moving faster. As they collide with nearby molecules, they transfer some of their kinetic energy to these molecules (solids) Convection - heat flows by the movement of molecules from one place to another (liquids and gases) Radiation - heat that is transferred through electromagnetic waves; unlike conduction and convection, no medium is required.

90 RRHS Physics 11 Unit 3 Slide #90 Check Your Learning 1.What happens to the work done when a jar of juice is shaken? The work done in this case would increase the temperature of the juice, since there is no increase in potential energy and no increase in kinetic energy (once the juice stops moving). 2.If two objects of different temperatures are placed in contact, will heat naturally flow from the object with the higher thermal energy to the one with the lower thermal energy? No, heat naturally flows from the object with a higher temperature to the one with a lower temperature. If 50 g of water at 30 o C is mixed with 250 g of water at 24 o C, heat will flow from the warmer water to the cooler water even though the thermal energy of the cooler water is much greater because there is so much more of it.

91 RRHS Physics 11 Unit 3 Slide #91 Check Your Learning 3.Is it possible for heat to flow even if the thermal energies of two objects are the same? Yes, heat flow depends on temperature difference not a difference in thermal energies. If a small piece of iron and a large piece of iron have the same thermal energy, the smaller one will be at a higher temperature and heat will flow into the cooler piece of iron. 4.Identify the mechanism for heat transfer in each of the following situations: a)Heat travelling from the Sun to the Earth. This is an example of radiation – there is no medium (or molecules) to transmit heat using either conduction or convection.

92 RRHS Physics 11 Unit 3 Slide #92 Check Your Learning b)When boiling a pot of water, the handle of the pot gets hot. This is an example of conduction – the molecules in the pot (a solid) are not free to move around over large distances. Heat is transferred as they collide with neighboring molecules. c)When boiling a pot of water, the water at the bottom gets hot first and then the water at the top gets hot. This is an example of mostly convection. As the water at the bottom of the pot heats up, it becomes less dense and rises. This carries heat to the top of the water and forces cooler water down. d)When you hold your hand near a stove element, you feel heat. This is mostly radiation, as hot objects emit electromagnetic radiation. If you are holding your hand above the element, there may also be some convection as the hot air rises.

93 RRHS Physics 11 Unit 3 Slide #93 Module 4.3 Objective 2 Upon completion of this module, the participant will be able to: To understand the mathematical relationship between heat and temperature. Upon completion of this module, the participant will be able to: To understand the mathematical relationship between heat and temperature.

94 RRHS Physics 11 Unit 3 Slide #94 Specific Heat Capacity Heat flows from an object with a high temperature to one with a low temperature. Change in temperature depends on mass and specific heat capacity (refers to the amount of heat Q needed to raise a particular mass of a certain material by 1 degree Celsius.) If heat is in J and mass is in kg then c is the specific heat capacity measured in

95 RRHS Physics 11 Unit 3 Slide #95 Table of Specific Heat Capacities

96 RRHS Physics 11 Unit 3 Slide #96 Example 1 A container of water containing 755 g of water is heated, raising its temperature from 5.0°C to 85.0°C. How much heat was absorbed by the water? Solution:

97 RRHS Physics 11 Unit 3 Slide #97 Heat Transfer in Systems Since heat is a transfer of energy, heat gained by one object can be transferred from another object Several categories of systems: –An open system is one in which mass and energy can enter or leave the system. –A closed system is one in which energy can enter or leave the system, but mass cannot. –A closed system is said to be isolated if no energy can enter or leave the system either. In a closed and isolated system

98 RRHS Physics 11 Unit 3 Slide #98 Example 2 When a 290 g piece of iron is placed at o C is placed in a 95 g aluminium cup containing 250 g of glycerine at 10.0 o C, the final temperature is observed to be 38 o C. Calculate the specific heat of the glycerine. Solution: Since the glycerine is already in the aluminium cup, the glycerine and aluminium can be assumed to be at the same temperature.

99 RRHS Physics 11 Unit 3 Slide #99 Example 2

100 RRHS Physics 11 Unit 3 Slide #100 Example 2

101 RRHS Physics 11 Unit 3 Slide #101 Check Your Learning A piece of lead at o C is placed in 1.3 kg of water that has an initial temperature of 22.0 o C. After a period of time, the water and lead reach a common temperature of 23.5 o C. Assume a closed and isolated system and ignore the effect of the container holding the water. a)How much heat did the water gain?

102 RRHS Physics 11 Unit 3 Slide #102 Check Your Learning b)How much heat did the lead lose? Since the system is closed and isolated, any heat gained by the water must have been lost by the lead; therefore, the heat lost by the lead was

103 RRHS Physics 11 Unit 3 Slide #103 Check Your Learning c)What was the mass of the lead?

104 RRHS Physics 11 Unit 3 Slide #104 Module 4.3 Objective 3 Upon completion of this module, the participant will be able to: Understand the mathematical relationship between heat and change of state. Upon completion of this module, the participant will be able to: Understand the mathematical relationship between heat and change of state.

105 RRHS Physics 11 Unit 3 Slide #105 Latent Heat Latent Heat – the amount of heat needed to change the state of a particular mass of a substance Energy goes into separating molecules rather than increasing the temperature Heat required to change the state of a substance can be found using Where ∆H is the latent heat and is measured in J/kg

106 RRHS Physics 11 Unit 3 Slide #106 Latent Heat If a substance is melting (going from a solid to a liquid) we must use the latent heat of fusion: If a substance is being vaporized (going from a liquid to a gas) we must use the latent heat of vaporization: For the reverse process (liquid to solid or gas to liquid), energy is released

107 RRHS Physics 11 Unit 3 Slide #107 Table of Latent Heats

108 RRHS Physics 11 Unit 3 Slide #108 Phase Change Diagram for Water

109 RRHS Physics 11 Unit 3 Slide #109 Example How much heat does it take to melt 2.0 kg of ice that is at a temperature of -12.0°C? Solution: There are two parts to this problem: 1.Before the ice can melt, the temperature must be raised to the melting point. 2.Once at the melting point, we can calculate how much energy is required for the change of state.

110 RRHS Physics 11 Unit 3 Slide #110 Example Part 1 Part 2

111 RRHS Physics 11 Unit 3 Slide #111 Example The total heat can then be found:

112 RRHS Physics 11 Unit 3 Slide #112 Check Your Learning Water is being heated with 4.0×10 6 J of energy, turning it into steam. The mass of the water is 1.5 kg and the water is initially at a temperature of 20.0 o C. a)How much energy did it take for the water to reach its boiling point? For this part, the water must reach a temperature of o C; there is no change of state involved.

113 RRHS Physics 11 Unit 3 Slide #113 Check Your Learning b)How much energy did it take to convert the water into steam? Once the water has reached its boiling point, there is no change in temperature in converting the water into steam.

114 RRHS Physics 11 Unit 3 Slide #114 Check Your Learning c)What was the final temperature of the steam? Since we know the total energy applied to the water, and we know how much energy it took to raise the temperature of the water and to convert it into steam, the remainder of the energy was used to raise the temperature of the steam.

115 RRHS Physics 11 Unit 3 Slide #115 Module 4.3 Objective 4 Upon completion of this module, the participant will be able to: Calculate the expansion or contraction of a solid or liquid when it undergoes a change in temperature. Upon completion of this module, the participant will be able to: Calculate the expansion or contraction of a solid or liquid when it undergoes a change in temperature.

116 RRHS Physics 11 Unit 3 Slide #116 Linear Expansion When designing bridges and other types of equipment an allowance must be made for expansion and contraction The change in length of a solid can be given by Where α is the coefficient of linear expansion and is measured in °C -1

117 RRHS Physics 11 Unit 3 Slide #117 Table of Coefficients of Expansion

118 RRHS Physics 11 Unit 3 Slide #118 Example 1 The steel bed of a suspension bridge is m long at 20.0 o C. If the extremes of temperature to which it is exposed are o C and 40.0 o C, how much will it contract and expand? Solution: First, we will look at how much it will contract.

119 RRHS Physics 11 Unit 3 Slide #119 Example 1 Now, we will look at how much it will expand.

120 RRHS Physics 11 Unit 3 Slide #120 Example 1 So the steel can decrease in length by as much as 12 cm and increase in length by as much as 4.8 cm.

121 RRHS Physics 11 Unit 3 Slide #121 Volume Expansion Change in volume can be applied to liquids as well as solids (gases will be looked at in Chemisty) The change in volume of a substance that undergoes a temperature change is given by Where β is the coefficient of volume expansion and is measured in °C -1 Hollow objects are found to expand similar to solid objects that have the same volume and are made of the same material. A steel tank will expand the same amount as a block of solid steel having the same dimensions.

122 RRHS Physics 11 Unit 3 Slide #122 Example 2 The 70.0 L steel gas tank of a car is filled to the top with gasoline at 20.0 o C. The car is then left to sit in the sun and the tank reaches a temperature of 40.0 o C. How much gasoline will overflow from the tank? Solution: We must calculate the increase in volume for both the gas tank and the gas.

123 RRHS Physics 11 Unit 3 Slide #123 Example 2 Gas Tank

124 RRHS Physics 11 Unit 3 Slide #124 Example 2 Gasoline

125 RRHS Physics 11 Unit 3 Slide #125 Check Your Learning A concrete highway is built of slabs 14 m long, at 20.0 o C. How wide should the expansion cracks be between the slabs to prevent buckling if the range of temperatures is o C to o C? When the temperature goes down below 20.0 o C, the slabs will contract and the gaps will widen. So we must make sure that the gaps are large enough to allow the slabs to expand linearly without touching each other before they reach a temperature of 50.0 o C. So the amount of expansion should be equal to the size of the gaps.

126 RRHS Physics 11 Unit 3 Slide #126 Check Your Learning

127 RRHS Physics 11 Unit 3 Slide #127 Module Summary In this module you learned: The difference between heat, temperature, and thermal energy. The work-energy theorem can be expanded to include the heat generated by friction That the heat applied to or removed from an object is related to the change in temperature using the equation

128 RRHS Physics 11 Unit 3 Slide #128 Module Summary That the latent heat needed for a change of state is given by the equation That the linear expansion of a solid can be calculated using the equation That the volume change of a solid or a liquid can be calculated using the equation

129 RRHS Physics 11 Unit 3 Slide #129 Module 4.4 Conservation of Energy Conservation of energy allows us to analyze many situations that would be very difficult to analyze using kinematics and dynamics because conservation of energy allows us to simply compare two positions in a system. It is not necessary to consider directions and variations at every step of the motion.

130 RRHS Physics 11 Unit 3 Slide #130 Conservation of Mechanical Energy In the absence of friction, any external work produces a change in mechanical energy If no work is done, then

131 RRHS Physics 11 Unit 3 Slide #131 Conservation of Mechanical Energy Conservation of Mechanical Energy - the total mechanical energy at any initial point in an isolated system is equal to the total mechanical energy at any later point (in the absence of friction) Energy can be transformed from one type to another, but the total amount of energy stays the same Example – a falling book starts with potential energy; as it falls, potential energy gets transformed into kinetic energy

132 RRHS Physics 11 Unit 3 Slide #132 Example 1 Two balls of the same mass are rolled down two different tracks, as shown below. One track is shorter than the other track; the top of each track is at the same height, and the bottom of each track is at the same height. The balls are released from rest. Which ball is going faster when it reaches the bottom, the ball on track A or the ball on track B?

133 RRHS Physics 11 Unit 3 Slide #133 Example 1 Solution: This is an example of a situation that is much easier to analyze using conservation of energy, since a kinematics analysis would involve calculating the accelerations using vectors and using the distances to calculate final velocities. Since both balls are released from rest, they have no initial kinetic energy. Their initial potential energy is the same, since they are released from the same height. When they reach the bottom, neither ball has any potential energy (relative to the bottom of the ramp). Since they had the same initial energy, they must have the same final energy (due to conservation of energy). Since all of the energy is kinetic, having the same kinetic energy (and same mass) means that they are going the same speed. So each ball will reach the bottom with the same speed.

134 RRHS Physics 11 Unit 3 Slide #134 Example 2 A roller coaster cart starts from rest at the top of a hill (point A) that is 15.0 m above the ground, as shown below. How fast will it be going at point C, which is 11.0 m above the ground?

135 RRHS Physics 11 Unit 3 Slide #135 Example 2 Solution: Remember, when dealing with potential energy it is necessary to choose a reference level. We will choose the ground to be the reference level in this solution. We will choose point A as our initial point (since we know information about this point) and point C as our final point (since it is the one we are trying to obtain information about). At point A, the car will have potential energy since it is above the reference level (the ground). It will not have any kinetic energy since it is starting from rest. At point C, the car will have both potential energy (since it is above the reference level) and kinetic energy (since it is moving).

136 RRHS Physics 11 Unit 3 Slide #136 Example 2

137 RRHS Physics 11 Unit 3 Slide #137 Example 3 In a well known lecture demonstration, a bowling ball is hung from a ceiling by a steel wire. The lecturer pulls the ball back and stands against the wall of the room with the ball against his nose. a)If the lecturer simply releases the ball, he will not be injured. Why not? b)If the lecturer pushes the ball, when it returns it will likely hit him and cause injury. Why? Video Demonstration

138 RRHS Physics 11 Unit 3 Slide #138 Example 3 Solution: a)When the ball is released, it is not initially moving so all of its energy is potential. As it swings over and back, its total energy must remain the same (or decrease if it loses any energy) even though the energy gets converted into kinetic energy and back into potential energy. When it returns back to its starting point, all of the energy returns to potential energy. Since it can’t have more energy than it started with, it cannot go any higher than it started. It therefore cannot hit the person. b)If the ball is given a push, its initial energy is the sum of its initial potential and kinetic energies. After it swings across the room and returns to its starting point, it still has kinetic energy left so it can continue above its starting point. This extra energy that the push gave the ball allows it to swing higher than its original height so it keeps going and hits the person.

139 RRHS Physics 11 Unit 3 Slide #139 Conservation of Energy Applet The following simulation can be used to generate real- time graphs of kinetic energy, potential energy, and total energy to demonstrate Conservation of Mechanical Energy Conservation of Energy Applet

140 RRHS Physics 11 Unit 3 Slide #140 Check Your Learning For each of the following two questions, use the same roller coaster that was used in the example problem. 1.Do the Example 2 again, but this time choose your reference level to be point B. Do you get the same answer?

141 RRHS Physics 11 Unit 3 Slide #141 Check Your Learning Since our reference level is point B, all height must be measured from this point. We will still use point A as our initial point and point C as our final point. Yes, we do get the exact same answer. It does not matter where we choose our reference level – we will always get the same answer.

142 RRHS Physics 11 Unit 3 Slide #142 Check Your Learning 2.If the car has a speed of 12.0 m/s at point A, a)What will its speed be at point C? The car now has kinetic energy at point A since it is moving. We will use the ground as our reference level and we will use point A as our initial position and point C as our final position.

143 RRHS Physics 11 Unit 3 Slide #143 Check Your Learning b)What is the highest hill (above the ground) that the car could reach on this roller coaster? We will use point A as our initial position again, but for our maximum height we must use some unknown position that occurs later on the roller coaster. At the maximum height, all of the energy will be potential since the kinetic energy will go to zero as the car stops.

144 RRHS Physics 11 Unit 3 Slide #144 Conservation of Total Energy Consider the situation shown below, where a ball is released from the top of the semi-circular track: Ball would continue to go back and forth forever as the energy is converted back and forth between potential and kinetic energy In reality the ball will eventually stop, meaning that it has lost all of its mechanical energy. Where did this energy go?

145 RRHS Physics 11 Unit 3 Slide #145 Conservation of Total Energy In the presence of friction, any external work produces a change in mechanical energy If no work is done, then

146 RRHS Physics 11 Unit 3 Slide #146 Conservation of Total Energy Still allows for friction within the system, but no work outside the system Principle of Conservation of Total Energy – even in the presence of friction, the total energy in a closed and isolated system remains constant

147 RRHS Physics 11 Unit 3 Slide #147 Example 1 A 250 g car rolls down a ramp. The car starts from a height of 32 cm, and reaches a speed of 1.7 m/s at the bottom of the ramp. Was mechanical energy conserved? Solution: To determine if mechanical energy was conserved, we must calculate the total initial energy and the total final energy. We will use the bottom of the ramp as the reference level.

148 RRHS Physics 11 Unit 3 Slide #148 Example 1 Since E i ≠ E f, mechanical energy was not conserved. There were 0.42 J of energy lost. This energy was generated as heat due to the friction due to the moving parts in the car and between the car and the ramp.

149 RRHS Physics 11 Unit 3 Slide #149 Example 2 Consider the same roller coaster that we looked at in the previous section: If a 215 kg car starts from rest at the top of the first peak (point A) but only reaches a speed of 6.0m/s at point C, how much energy was lost?

150 RRHS Physics 11 Unit 3 Slide #150 Example 2 Solution: The energy lost here was lost due to friction between the car and the track, and will appear as heat. We are therefore trying to find Q. We will use the ground as our reference level, and use point A as our initial point and point C as our final point.

151 RRHS Physics 11 Unit 3 Slide #151 Problem-Solving Hints In this unit, we have discussed using both the work-energy theorem and the principle of conservation of energy to solve problems. In many situations, either approach might be taken, depending on what you choose as your system. If you choose your system to be an object or objects on which there is an external force doing work, then you must use the work-energy theorem since there will be a change in energy of the system. If you expand your system to include the objects applying the forces (making the forces internal) then there is no external work being done on the system by external forces and conservation of energy can be used.

152 RRHS Physics 11 Unit 3 Slide #152 Conservation of Energy Hints 1.Draw or visualize a picture of the situation to help indicate where the object is at different times. 2.Determine the system for which energy will be conserved; in other words, determine the system for which there will be no outside forces acting on any of the objects. 3.Choose initial and final positions – these will be the two positions for which you are comparing the total energy. 4.Choose a reference level – this is the point where the gravitational potential energy is zero. Although any level can be chosen, the lowest point in the problem or the ground is often a convenient choice. 5.Ask yourself what kinds of energy (potential and/or kinetic) are present at each of the positions chosen in step 3. These will be included in the conservation of energy equation in the next step. 6.Apply the appropriate conservation of energy equation.

153 RRHS Physics 11 Unit 3 Slide #153 Check Your Learning In the roller coaster used in the example, suppose that 3250 J of heat is generated between points A and B. If the 215 kg car starts from rest at point A, how fast will it be going at point B?

154 RRHS Physics 11 Unit 3 Slide #154 Check Your Learning We will use point A as the initial point and point B as the final point. We will use the ground as the reference level.

155 RRHS Physics 11 Unit 3 Slide #155 Energy and Collisions Momentum is conserved in all collisions when applied to closed and isolated systems. Is energy? Consider two balls going toward one another as shown in the diagram below only type of mechanical energy present before the collision is kinetic energy.

156 RRHS Physics 11 Unit 3 Slide #156 Energy and Collisions Consider what happens in slow motion and look at the balls on a small scale, we see that they momentarily stop when they come in contact: edges of the ball compress in this process - the kinetic energy of the moving ball is transformed into a type of potential energy known as elastic potential energy.

157 RRHS Physics 11 Unit 3 Slide #157 Energy and Collisions When the balls bounce away from one another, then the potential energy that was temporarily stored in this collision is transformed back into kinetic energy If this energy conversion is 100% efficient (meaning that no kinetic energy was lost in this process of transforming kinetic energy into potential energy and back to kinetic energy) than the collision is said to be elastic

158 RRHS Physics 11 Unit 3 Slide #158 Energy and Collisions An elastic collision can be defined as one in which the total amount of kinetic energy is conserved A completely inelastic collision is one in which the objects stick together. Although some kinetic energy must be lost in this type of collision, all of the kinetic energy does not have to be lost since the objects can continue to move together. Even though kinetic energy is lost, it is changed into other forms of energy (often heat) so the total energy is still conserved as always

159 RRHS Physics 11 Unit 3 Slide #159 Energy and Collisions In a closed and isolated system, –Momentum is conserved in all collisions –Mechanical energy is only conserved if the collision is elastic

160 RRHS Physics 11 Unit 3 Slide #160 Example 1 Ball A, mass 3.0 kg, is moving to the left at 2.5 m/s when it collides with ball B, mass 4.0 kg, moving to the right at 4.0 m/s. Ball A rebounds in the opposite direction with a speed of 2.0 m/s. a)Find the velocity of ball B after the collision. b)Was the collision elastic? Show your work.

161 RRHS Physics 11 Unit 3 Slide #161 Example 1 Solution: a)Since we have a collision, the first thing we should do is apply conservation of momentum. Remember that momentum involves vectors, so direction is important. We will use the right as positive.

162 RRHS Physics 11 Unit 3 Slide #162 Example 1 In order to determine if the collision is elastic, we must determine whether or not the total initial kinetic energy is equal to the total final kinetic energy. Since we are calculating the energy we do not have to worry about direction. Since there is less total kinetic energy after the collision than before the collision, the collision was not elastic.

163 RRHS Physics 11 Unit 3 Slide #163 Example 2 Consider a collision between two billiard balls, which we will assume to be elastic. Suppose that the cue ball (ball 1), moving at a speed of 3.0 m/s, collides with another ball (ball 2) of the same mass that is at rest. Find the velocity of each ball after the collision. Solution: Since momentum is conserved in all collisions, we will apply this principle first. For consistency with energy, we will use subscripts instead of primes for all quantities.

164 RRHS Physics 11 Unit 3 Slide #164 Example 2 There are obviously two unknowns in this equation. In order to solve for two unknowns, it is necessary to have two equations. Since the collision can be considered to be elastic, we can apply conservation of kinetic energy:

165 RRHS Physics 11 Unit 3 Slide #165 Example 2 Rearranging the momentum equation we get

166 RRHS Physics 11 Unit 3 Slide #166 Example 2 Substituting this into the energy equation gives

167 RRHS Physics 11 Unit 3 Slide #167 Example 2 We see that mathematically, there are two solutions to the final velocity of ball 2, 0 and 3.0 m/s. It is necessary to look at the situation, however. Ball 2 was initially at rest. It does not make sense that it would remain at rest after the collision; therefore, the velocity of ball 2 must be Substituting this back into the momentum equation gives

168 RRHS Physics 11 Unit 3 Slide #168 Check Your Learning A 53.0 g marble moving with a velocity of 2.00 m/s collides with a stationary marble of mass 87.0 g. If the first marble rolls backward with a velocity of m/s, was the collision elastic? Since we have a collision, the first thing we should do is apply conservation of momentum to find the speed of the second marble. Remember that momentum involves vectors, so direction is important.

169 RRHS Physics 11 Unit 3 Slide #169 Check Your Learning In order to determine if the collision is elastic, we must determine whether or not the total initial kinetic energy is equal to the total final kinetic energy.

170 RRHS Physics 11 Unit 3 Slide #170 Check Your Learning Since the initial and final energy are within J (less than 1%) we can assume that the difference is due to rounding (especially since the final energy is slightly larger than the initial energy); since the energies are the same, the collision was elastic.

171 RRHS Physics 11 Unit 3 Slide #171 Module Summary In this module, you learned The total mechanical energy in a closed and isolated system (in other words, where there is no external work done) remains constant in the absence of friction: If there is friction, then the total energy in the system still remains constant if the heat that is generated is included:

172 RRHS Physics 11 Unit 3 Slide #172 Module Summary An elastic collision is one in which the total mechanical (kinetic) energy remains constant. A completely inelastic collision is one where the objects remain stuck together after the collision. Some kinetic energy is always lost in inelastic collisions.

173 RRHS Physics 11 Unit 3 Slide #173 Unit Summary Module 4.1 – Momentum and Impulse Momentum is calculated using the formula The change in momentum can be calculated from the change in velocity Impulse is equal to the change in momentum and is given by the formula Momentum is conserved in all isolated systems. This is known as the law of conservation of momentum

174 RRHS Physics 11 Unit 3 Slide #174 Unit Summary Module 4.2 – Work, Power and Energy Work can be calculated using the formula Power is the rate at which work is done and can be calculated using Kinetic Energy is the energy of motion and can calculated using the formula The net work on an object is equal to its change in kinetic energy

175 RRHS Physics 11 Unit 3 Slide #175 Unit Summary The gravitational potential energy of an object can be calculated using the formula The work-energy theorem states that the total external work done on a system is equal to the change in energy The actual mechanical advantage of a machine can be found using the formula The Efficiency of a machine (or any other energy conversion) is given by the equation

176 RRHS Physics 11 Unit 3 Slide #176 Unit Summary Module 4.3 – Heat The difference between heat, temperature, and thermal energy. The work-energy theorem can be expanded to include the heat generated by friction That the heat applied to or removed from an object is related to the change in temperature using the equation That the latent heat needed for a change of state is given by the equation That the linear and volume expansion of a substance can be calculated using the equations

177 RRHS Physics 11 Unit 3 Slide #177 Unit Summary Module 4.4 – Conservation of Energy The total mechanical energy in a closed and isolated system (in other words, where there is no external work done) remains constant in the absence of friction: If there is friction, then the total energy in the system still remains constant if the heat that is generated is included: An elastic collision is one in which the total mechanical (kinetic) energy remains constant. A completely inelastic collision is one where the objects remain stuck together after the collision. Some kinetic energy is always lost in inelastic collisions.


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