2??Has Work Been Done??1) A teacher applies a force to a wall and become exhausted2) A book falls off a table and free falls to the ground3) A waiter carries a tray full of meals above his head by one arm across the room4) A rocket accelerates through space
3force, displacement, and cause. Work is defined as a force acting upon an object to cause a displacement.There are three key words in this definitionforce, displacement, and cause.In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement.
4If an object does not move then no work has been done. Force and displacement must both be in the same direction for work to have been done.Example: A Flag in a parade goes horizontally but the force is vertical. Therefore no work is done.
5??Has Work Been Done??1) A teacher applies a force to a wall and become exhaustedNO2) A book falls off a table and free falls to the groundYES3) A waiter carries a tray full of meals above his head by one arm across the roomNO4) A rocket accelerates through spaceYES
6Work Facts The formula for work is simply force times distance. The unit for Work is Nm or J(Joule)1 Joule is the work done by a Force of 1 Newton in moving an object a distance of 1 meter.
7Example: What Work is done by pushing a physics text book with a Force of 20 N a Distance of 3m in an attempt to avoid doing homework.
8Example:Mr. Harper lifts up a 65kg student to a height of 0.50 m.a) What Force does Mr. Harper use to lift the student?F = Weight = mg =(65kg)(9.8 m/s2) = 637 N =640 Nb) What work does he do?W=Fd=(637N)(0.5m) = J = 320 J
9Example: If two physics students are rearranging a room and they decide to move a desk across the room, a total distance of 3.0 m. If they move the desk at a constant velocity by each exerting horizontal force of 200 N. Calculate the amount of work that was done to move the desk across the room.
10Example: A child ties a ball to the end of a 1 m long piece of string and swings the ball in a full circle. If the string exerts a continues force on the ball of 10 N, how much work does the string due on the ball during one full revolution?No distance. No work
17Positive and Negative Work Consider a weightlifter bench pressing a barbell weighing 650 N to a height of 0.55 m. There are two distinct motions, the first is when the barbell goes up and second is when the barbell is lower back down. Calculate the work done by the weightlifter during the two separate motions.Work done going upWork done going downTotal work done ( 0 J )
19EnergyThere are many different forms of energy however all of which are measured in units of Joules.In this chapter we will look at two different forms of energy. Kinetic energy and potential energy. As well as how they are related to the concept of work.
20Kinetic EnergyKinetic energy: which is the energy of an object due to its motion.
21Example: A kg hockey puck, initially at rest, is accelerated to 27.0 m/s. Calculate the kinetic energy of the puck both at rest, and in motion.At RestIn Motion
23The Work and Kinetic Energy Theorem In order to do work on object there must be a force applied to the object.When a force is applied to an object it will accelerate.When it accelerates there is an increase in velocity.An increase in velocity will cause an increase in kinetic Energy.Therefore, the work done on object is equal to the change in the kinetic Energy of the object.
24Example: A shot putter heaves a 7. 26 kg shot with a final speed of 7 Example: A shot putter heaves a 7.26 kg shot with a final speed of 7.51 m/s.a) What was the kinetic energy of the shot?b) If the shot was initially at rest how much work was done on it to give its it this kinetic energy?
25Example: A physics student does work on a 2 Example: A physics student does work on a 2.5 kg curling stone by exerting a 40 N force to it horizontally over a distance of 1.5 m.a) Calculate the work done by the student on the stone.60 Jb) Assuming that the stone started from rest, calculate the velocity of the stone at the point of release. Consider the ice surface to be frictionless.6.9 m/s
26Example:A 75 kg skateboarder initially moving at 8.0 m/s, exerts an average force of 200 N by pushing on the ground, over a distance of 5.0 m. Find the new kinetic energy of the skateboarder.3.4 kJ
28What do all of these things have in common? What do all of these things have in common?
29Potential EnergyPotential energy: is stored energy, or when an object has the potential to do work.There are many different types of potential energy such as a battery, a waterfall, a compressed spring, gasoline or anything that has the potential to do work.In this chapter will concentrate on what is called gravitational potential, or energy due to an objects position on earth.Often we refer to what is called the total mechanical energy of the system. Which is simply the total combined kinetic and gravitational potential energies.
30Gravitational potential energy is directly proportional to an object’s mass and height. The higher an object is lifted the more gravitational potential energy it will have.Also a more massive object will have a larger gravitational potential energy that a less massive object at the same height.Gravitational potential energy can be found using the following formula.
31Example: While setting up a tent you use a 3 Example: While setting up a tent you use a 3.0 kg rock to drive the tent pegs into the ground. If you lift the rock to a height of 0.68 m, what gravitational potential energy will the rock have?
32*** Caution ***When talking about gravitational potential energy you have to specify what the height is relative to.ie: the ground, the table, the top of the hill, the bottom of the hill, ect ....
33Example:A 2.0 kg textbook is lifted from the floor to a shelf 2.1 m above the floor.a) What is the gravitational potential energy relative to the floor?41 Jb) What is the gravitational potential energy relative to the head of a 1.65 m tall person?8.8 J
35Gravitational Potential Energy and Work When you do work on an object by lifting it to a new relative height. The object will as a result have an increase in gravitational potential energy thus the work done on an object is equal to the change in the gravitational potential energy of the object.
36Example:A 65 kg rock climber did 16 kJ of work against gravity to reach a ledge. How high did the rock climber accend?25 m
37Question:You carry a heavy box up a flight of stairs. Your friend carries an identical box on an elevator to reach the same floor as you. Which one, you or your friend, did the greatest amount of work on the box against gravity?Because the change in gravitational potential energy of the two different boxes is the same, the work done on the two boxes are equal.
38So far we have discussed both the work/kinetic energy theorem and the work/potential energy theorem. In both cases the amount of work done on the system was equal to the change in energy of the system.As it turns out both theorems are a part of a single all encompassing theorem called the work /energy theorem.Where the work done on a system is equal the the change in the total mechanical energy of the system.
41Work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement.Power is the rate a which work is done. How fast is the work being done.It can be found using the following equation.
42The metric unit of power is the Watt ( W ). As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time, thus, a Watt is equivalent to a Joule/secondFor historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to approximately 750 Watts.
43Example:A crane is capable of doing 1.5 x 105 J of work in 10 s. What is the power of the crane in watts?1.5 x 104 W
44Example: A cyclist and her mountain bike have a combined mass of 60 kg Example: A cyclist and her mountain bike have a combined mass of 60 kg. She’s able to cycle up a hill that changes her altitude by 400 m in 1min.a) How much work did she do against gravity in climbing the hill?2.4 x 105 Jb) How much power is she able to generate?3.9 x 103 W
45Example: Two physics students, Jacob and Ryan are in the weight lifting room. It takes Jacob 3 sec to lift the 100 kg barbell over his head a distance of m. It takes Ryan 2 sec to lift the same barbell over his head a distance of 0.55 m.a) Which student does the most work?b) Which student delivers the most power?
47EfficiencyEfficiency is the ratio of useful energy or work output to the total energy or work input.
48Example: A model rocket engine contained explosives stores 3 Example: A model rocket engine contained explosives stores 3.50x103 J of chemical potential energy. When launched the stored chemical energy is transformed into gravitational potential energy. Calculate how efficiently the rocket transforms the stored chemical energy into gravitational potential energy if the 0.50 kg rocket is propelled to a height of 100 m.14%
49Do Practice Problems Page 270 (pdf 37) #’s 44 - 50 ** EXTRA QUESTIONS **(only if students want to)End of Chapter Review Pg 274 (pdf 38)#’s 1, 2, 4, 5, 6, 15, 16, 17,18, 20, 21, 23, 24, 25, 26, 27, 28
50Conservation of Energy It’s not just a good idea, it’s the law!!
51“Energy cannot be created or destroyed simply transformed.” The Law of conservation of energy states that;“Energy cannot be created or destroyed simply transformed.”Example: A ball sitting on top of a table has gravitational potential energy.The ball then falls off the table and hits the ground.Was all the gravitational potential energy lost?Was all of the energy lost?
52Explanation: As the ball falls from its initial position it started to gain velocity. This gain in velocity then results in a gain in kinetic energy.So as the ball falls further, it goes faster, thus the gravitational potential energy decreases and the kinetic energy increases.There has been a transfer of energy from gravitational potential energy to kinetic energy. With no net energy loss.
53Question: What caused the transformation of energy (of the pig)? Example: A square pig of mass 50 kg is released from rest at a height of 10.5 m above the ground. With what velocity would pig hit the ground?14 m/sQuestion: What caused the transformation of energy (of the pig)?Answer: The force of gravity
54In the previous example what would happen if the pig was to be pushed down a ramp so that his final velocity would be zero when he reaches the bottom of the ramp?Is energy conserved now?At the bottom of the ramp the pig would have zero gravitational potential energy and zero kinetic energy. Therefor energy is not conserved.Question: So where did the energy go?Answer: All of the energy was lost due to friction.
55For the above examples we have two different types of forces For the above examples we have two different types of forces. The first force, force of gravity, is what we call a conservative force. The second force, the force friction is a non-conservative force.A conservative force (such as gravity) is a force that does not remove energy from the system. It is also path independent. This means that the path the object travels does not affect the final energy of the system.A non-conservative force (such as friction) is a force that does remove energy from a system, and is path dependent. The longer the path the more energy that will be removed.
56The Law of Conservation of (Mechanical) Energy When all the work done through a process is done by conservative forces, the total mechanical energy of the system after the process is equal to the total mechanical energy of the system before the process.**This however can not be said for non conservative forces.**EK + Eg = EK‘ + Eg‘The total energy before is equal to the total energy after.**The net energy is constant.**
57As the rock falls its potential energy is transformed by gravity, a conservative force, into kinetic energy.
58Question? Imagine an amusement park that has rides like those illustrated below. The masses of the car and occupants are the same for each case. Assume the wheels and track are effectively frictionless. Each car starts from rest at level A. What are the respective speeds of the cars when they reach level B?Answer: At level B all cars will have the same speed
59Question. While playing catch by yourself, you throw ball straight up Question? While playing catch by yourself, you throw ball straight up. Neglecting friction how does the speed of the ball, what returns to your hand, compared to the speed with which it left your hand?Answer: The ball will have the same speed except going in the opposite direction.
60Example: A skier is gliding along with a speed of 2 Example: A skier is gliding along with a speed of 2.0 m/s at the top of the ski hill, 40.0 m high. The skier then begins to slide down the icy, frictionless, hill.a) what will be the skiers speed at a height of 25 m?17 m/sb) what will be the height of the skier at a speed of 10.0 m/s?35 m
61Example: A person riding a bike approaches a hill with a speed of 8 Example: A person riding a bike approaches a hill with a speed of 8.5 m/s. The total mass of the bike and rider is 85 kg.a) Find the kinetic energy of the bike and rider.6.1 kJb) The rider coasts up the Hill. Assume there is no friction, at what height will the bike come to a stop?7.4 mc) Does your answer depend on the mass of the bike and rider? explain.No, b/c mass cancels out
62Example: Tarzan, mass of 85 kg, swings down from a tree limb on the end of the 20 m vine. his feet touch the ground 4 m below the limb.a) How fast is Tarzan moving when he reaches the ground?b) Does your answer depend on Tarzan’s mass? Explain.c) Does your answer depend on the length vine? Explain.
63Example: A skier starts from rest at the top of the 45 m hill, skis down a 30̊ incline into a valley continues up a 40 m Hill. Both heights are measured from the Valley floor, assume you can neglect to friction.a) How fast is the skier moving at the bottom of the Valley?b) What is the skiers speed at the top of the next hill?
64Do Practice Problems Page 287 (pdf 40) #’s 1, 2, 3, 4, 6, 8 Section Review Page 300 (pdf 40) #’s 1 - 6Conservation of Energy Work Sheet64
65Conservation of Energy “with non-conservative forces” Key termsInternal force: any force exerted on an object in the system due to another object in the system.External force: any force exerted by a object that is not part of the system on an object within the system.Open system: a system that can exchange both matter and energy with its surroundings.
66Closed system: a system that can exchange energy with its surroundings, but not with matter. Isolated system: a system that does not exchange either matter or energy with its surroundings.
67So far we have only looked at conservation of energy for conservative forces. Now we will look at what happens the total energy of a system when a non-conservative force acts on the system.The work done by non-conservative forces is equal to the difference of the final mechanical energy and the initial mechanical energy of a system.For example, when you do work on a system you increase its final energy.
68We have already seen this in the form of the “Work Energy Theorem” We have already seen this in the form of the “Work Energy Theorem”. Where the work done is equal to the change in the energy, if no work is done then there is no change in energy.
69Example: A 65.0 kg skydiver stepped of a hot air balloon that is 500 m above the ground. After freefalling a short distance, she deploys a parachute, finally reaching the ground of the velocity of 8.00 m/sa) Find the gravitational potential energy of the skydiver, relative to the ground, before she jumps.3.19 x 105 Jb) Find the kinetic energy of the skydiver just before she lands on the ground.2.08 kJc) How much work was done by air friction.3.17 x 105 J
70Example: A roller coaster car of mass of 200 kg is moving at a speed of 4.00 m/s at point A in the diagram. This point is 15 m above the ground. The car then heads down the slope toward point B, which is 6 m above the ground. If 3.40 x 103 J of work are done by friction between points A and B, determine the speed of the car at point B.15 m/s
71Do Practice Problems Page 308 (pdf 41) #’s 18 -23 End of Chapter Review Pg 327 (pdf 44)#’s 1, 2, 3, 5, 8, 12, 22, 23, 24, 2571