To. Hello, miss. Hello, Jackson. How are you? Great. Good. So we're going to look at work energy efficiency and try a few last questions on this topic. And then what would you like to move on to next week? Okay. You can let me know before the lesson next week what topic youlike to do next per revision. Maybe we should do maybe we should do a bit more work on uncertainties and the practical stuff. Okay, but if you want to do something else, let me know. Okay, so work is force times distance. If we have a force working at an angle, the line of motion is of course, horizontal. So we have to take into account the horizontal component. So the work done is f cotheta times change in distance. So. So we can look at examples of that. A boy is pulling a sledger on a rope. The force of tension in the rope is zation nutrions acting at 45 degrees to the horizontal. So atimes cost 45, is 57 nutions. And he pulls the sludge for half a kilometer. Calculate the work he's done. So it's 57 times 500. So eight, 2008, 28500 joules. And so. Okay, if we have a crate of MaaS 100 kilograms being pushed along a horizontal floor at a steady velocity of 5m per second with a force of 400 musions, what is the magnitude of the frictional force? Rectional forces 400 newtons. Yes, indicate the direction of the frictional force on the diagram, direction opposite to the to the direction of pushing. So to the left, good. Calculate the work done in pushing the crate 5m along the floor. Pushing the great ates. So work is equal to force times change in distance. So what force has been exerted by the man? Yeah. What force? 400 the newtons times 5m equal to 2000 jles joules. Yeah because 2000 jewels have worked, done good. What quantity of heat energy is being transferred to the surrounding as a result of the work done above? So if there's friction, most of the frictional force will result in heat energy being generated. So we worked out that for three Marks. So this is only one mark. So it suggests there's nothing else to calculate. So the work done will equal the frictional force in the opposite direction. So the heat energy will be given to the surroundings, which will be equal to the work done. So 2000 joules of heat energy will be transferred to the surroundings due to friction. He decides to pull the crate at an angle of 30 degrees to the horizontal to make it easier. Why is a larger force now needed to drag the crate along the same floor at the same speed? So instead of pushing at 400 newtons, he's having to pull it now at 460 newtons. 460 newtons. But the woforce is not horizontal. So larger force ces will need to jothe creates because Yeah because that is not in the horizontal. So we should so the so the pulling on pulling force is science cosine 30 degrees times 460 newtons. Which which is less so the pulling force is more because. Costs 30 times 460 is 398. So the greater. Force is needed as it is acting. An angle, how much work is done dragging the crate 5m along the floor? So 460 times costs 30 times five. So it's three, nine, eight times five. So thatbe. One magnet two joules. So compare and account for our answers to three and for c and F. C and f. So it was 2000 nutons and it was 1992 or joules. So what can we say here? Do you think they're similar enough? So with so the question of is is 1992 jeso 2000Yeah is Yeah is similar. If that is just the two significant figure that is the same answer. Work is done to overcome friction. That's the main force that is needed here, okay? Then we have kinetic energy and gravitational potential energy. So gravitational potential energy is the energy of an object due to its position in a gravitational field. And kinetic energy is something an object has because of its movement. So. We can derive the kinetic energy equation. Ek is a half mv squared. If we say v squared is equal to U squared plus two as so, v squared is two as so v squared is two f times distance. F over m acceleration is force per unit MaaS. Rearranging this equation, force times distance is a half mv squared. So work done is equal to the gain in kinetic energy. A bullet of this MaaS is farred from a rifle with a barrel of 80 cm long with a velocity of 500m per second. What is the kinetic energy of the bullet? So a half times the MaaS in kilograms, times the velocity squared. So we get this. What is the average force on the bullish whilst it is accelerating along the barrel? So we know that the barrel is 0.8 of a meter long. So 2500 joules is equal to the work done, which is equal to the force times the distance. So we know the distance of the front of the gun. We can work out the accelerating force. So it's the work divided by the length of the barrel gives us the force done. Very often you're given the work done. You have to find the frictional force or the resistive forces, etcetera. Gravitational potential energy is due to m times, g times, H. So if we have a student of this MaaS climbing 25 steps up a tall ladder, and the rung on the ladder, the rungs are 30 cm apart, what is the increase in the students gravitational potential energy when they're at the top of the ladder? So 25 steps times 30 cm each gives us the distance. M times, g times, vertical height gives us gravitational potential energy. Okay. So a fiwork, let's try this one. A firework is launched vertically with a velocity of 4m per second. Calculate how high it will travel, stating the assumptions you have made. Okay, fireworks vertically were not still 4040. 40m. And how I hide travel. Vertically vilast year 40 how high travel assumption you make? So let's think of the assumption. First of all, assumption, firstly, assumption that there's no air resistance. Perfect. Good. How do you think we'll work out how high it will travel? So what we can do, this is the question we're doing. We can say that the kinetic energy. Equals the gravitational potential energy. So we can say a half mv squared. And we know velocity is. M times g times H. So we can simplify this. We don't need to know the MaaS of the firework. Okay? We know g, we know v, we can work out H. So it's a half and it's 40 chmeters per second times four sheet squared. Divided by 9.81. Okay. So we would expect to get. Four, she squared. Divided by two times nine pointation one. So. So I get ac 1.5 measures. That's about sensible for a firework. Okay. So kinetic and gravitational potential energy. So let's read the first question here. A car of MaaS 18 hundred kilograms, initially traveling at 20m per second, slows down uniformly to rest over a distance of 40m in an emergency stop. Determine the kinetic energy of the car before it starts to decelerate. How will we do that? The Waton by the bricks is equal to the initial energy of core result for the mean breaking force. Okay, about this, about breaking force. Maybe we're doing this one first. Oh, okay. Determines the kinetic energy determinmine the car before it starts to it determine workouoh workouokay a half M V square, not point five. Times 1800 times 20 squared squared don't forget. You know, I forgot of that. 36 not not not not. Jeles. Do you agree? Wait. Yeah, I agree. Okay. Use your result from this bit to determine the mean breaking force during deceleration. During deceleration, mean breaking. The MaaS W equal to F S good. So. F times s. Equals 36 north not not not. So it needs 40m so we know s so. 36 zero not not not divided by 40. Will give us our force, okay? What do you get? I get this answer divided by 40. 9000. It sounds right. So that is 9009 killer usions. Let's write it like that for space, despite the main energy transfers that take place while the car is decelerating. One of the cis decelerating. Main energy transfer fers main energy transfer, it's decelerating. So kinetic energy to kinetic energy turn to the thermal energy Yeah. You could maybe get a bit of sound energy as well, okay? A rubber ball is dropped from a height of 1m onto a horizontal surface, and then it bces several times. His velocity time graph is reproduced below. Air resistance can be neglected. The bounce is determined by the kinetic energy of the ball as it leaves the surface. How does the graph show that some kinetic energy is lost as a result of each bounce? How does the graph show? So looking at the graph, kinetic energy Yeah a result of leash bounds. So this is a rubber ball dropped from a height of one meters onto your horizontal surface. Plus several times vertical time growth is spbelow. Every season can be neglected. So it bounces many times at each of them. And each of them. Yeah each of has less velocity. Good. So if it's less velocity, it's got less kinetic energy. Why does the bounce height decrease after each bounce decreasless velocity at the same time? So there is the less distance. So why does the heist reduce? It gradually gets less high. We know the kinetic energy is decreasing. We can see that from the graph. So if the velocity is decreasing, the kinetic energy is decreasing. So if the kinetic energy is decreasing, what else is decreasing? Think of this, what is your kinetic energy converted to as the ball bounces? The kinetic energy maybe is a transfer is a transfer from inefrom Oh sorry, gravitational potential energy to kinetic energy Yeah so if the kinetic energy is decreasing. Kinetic energy is less so. Gravitational potential energy is also less. So. Less hide. Reached. What percentage of the bo's kinetic energy is lost as a result of the first bounce? What percentage of the bulls kinetic energy is lost as a result of the first balance? Yeah, the result. So firstly, the result is 2s. So and firstly, that is the velocity energy age and basically, okay, so that is not 0.48. In the first, not one for 8s and the second is. How for a second? 1.6 no no no not 1.6. So this philosophy is 4.5. The first impact. And this is one, two, 3.5. So this is 3.5. Meters per second, so we can say 3.5. Squared. Over 4.5. Velcity. So if we get 3.5 squared divided by 4.5 squared. That's 0.6. So we get 60% decrease after the first bounce. We could do a half mv squared over a half mv squared, but all we just need to do the v's, the ratio of the v's. Does that make sense, Jackson? Yeah Yeah Yeah. Okay. Conservation of energy. So this is a fundamental law of physics. In a closed system that means not affected by external sources in which nothing can enter or leave, the total energy remains constant, although it can be transferred within the system. So if we have a gear or a mosystem driving a car, wheel, etc, a motor pulley system lifting a load is an example of energy conversion. The electrical energy is converted to kinetic energy, and the kinetic energy is used to lift something up. The energy is converted from one form to another, but the total amount of energy in the electrical form is fully accounted for. None is lost through, although some may be converted to heat energy and sound energy. So in a closed system, the amount of energy is the same. So if we have an object of two kilogram MaaS raised to a height of 30m above the ground and then dropped, describe the energy changes that take place from the moment the object is released until it comes to rest on the ground. So gravitational potential energy to kinetic energy on impact, some of the energy is converted to thermal energy, sound, heat, deformation of the ground. Work is done with the impact on the ground. Use the principle of conservation of energy to calculate the speed with which it hits the ground. So again, gpe lost. If we're dropping us from the height of 30m, I is that one gp lose. So cakind will increase. Yes. Basically, if you have a thousand joules of gpe itbe converted to a thousand joules of kinetic energy. Again, we can cancel the masses that and that change in height. We know in this time it's 30m height. It was dropped from a half. We can work out the velocity squared. So v is square root of 30 times g times two, two times g times the height, 30. So square rooting as we get the speed on impact. So conservation of energy. If we have a train truck hitting a buffer, and the buffer spring is compressed by 12.5 cm, when the truck is bought to rest, it was moving at that speed. What is the kinetic energy of the moving truck? We know the MaaS. We know the speed. So a half times the MaaS, times the speed squared, so we know it's 310 joules to two significant figures. What is the average force exerted by the buffer? Assuming all the kinetic energy is converted to stored energy in the buffer spring? So when that bit hits that this will compress, move that way. So the work done in compressing the spring is force times distance. If we know the distance that moves, we can work out the force times the distance is the work done which we know from before. So the force is 2500 nutions. Okay, let's try this question. Okay, how much work must be done? The store m. A stone of MaaS 50 grams is thrown with the philosopcity of so to angle of 53 to the horizontal. So that is cosine cosine 50 degrees. No point 64. Accate how much work will be done on the stone? So it's a half times. Note or note five. Times 50 squared. So 50 squared times or not ught five times 0.5 62.5 joules. So 62.5 joules work done. State the type of energy the stone has at the top of its flight. Okay, so types of energies don't hat the trip. The type of energy grataational potential energy and the kinetic energy mothere may be therenergy because of the air resistance. The bottom is that is knowledgeable. It might still have a bit of kinetic energy because it doesn't totally stop. It might still be spinning or something. Calculate how high the stone will travel. Oh yes, the stone will travel how high? So we know that J equal to 9.81Newton kilograms. And the stone MaaS MaaS is through with a velocity of an angle to the horizontal through with height ocity. So with our velocity to this, so the velocity is zero. Initial velocity is 50m per second about us. So we can use Yeah you have to take into account the angle at which the velocity is working this time. So 50m at an angle of 53 degrees. So this time. Mgh. Equals a half times m times 50. Times sign. 53 degrees because it's working at an angle. So we can ignore the m's. So if we work out 50 times sine 53. Sine 53 times 50. So that's. A half times. Third to nine. Point 93. I forgot to swear it, so I have 39.93 squared is equal to 9.81. Times H so we can find H then we have enough information to find H but this time, as you can see here, the velocity we have to take into account that the velocity is. At an angle to the vertical. 53 degrees to the horizontal. So H will be equal to. One over two times 9.81. Times 39.93 squroot. 1594. Divided by two times, answer is one point sorry, one, five, nine, 42 times 9.81. So the final answer is 81.2. Yeah, 81.2, 81.3. Okay, hour, then power is the rate at which work is done. In physics, work is defined as force times, distance, power is work done or energy converted per unit time. So it's joules per second. So Yeah. I think you are happy with this. At the London Olympics in 2012, Yun Colum of North Korea lifted 168 kilograms to equal the then world record in his class. In this particular lift, the clean and jerk, that's where you don't bend your arms, I think. Are you lifted there? And then you put it over your head. The weight is first lifted to shoulder height and kept there until the second stage yet. So you bring it to here and then you lift it over your head. The jerk that completes the lift by raising the weight above the head, the weight, the weight, 1.5m times 168 kilograms. 1.5 times. That's the MaaS. Isn't it so times 9.81 force times distance. So that should give us the work done. So 9.81 times. Each. 2472 joules. The total time taken for the lift was 5s determinmine the average power required to perform the lift. So 2472 divided by five. So that gives us 49. For what unit do we use for power, Jackson? Explain why his peak output power must have been significantly greater than your answer to part b, but lifting the hand a total the time though. So the is that the lift lifting it to the higher way? It's a two stage lift. When you're doing this, you lift it to your shoulders and then you lift it over your head. So his peak output power, he's doing the clean first, then he waits and then he jerks it up over his head. So his actual speed of lifting must be much more rapid because he does it. I think in competition, you have a total of 5s to do the the whole to hear and then to hear. So. He does the left in two stages, so each stage is done more rapidly. Four, nine, four lots. Thanks. Do you understand? It's not like you're you're lifting it constantly. You lift it to here and then you lift it over your head. A lift in a skyscraper is powered by a 90 kilowatt motor and has a total load capacity of 18 hundred kilograms. Determine the vertical speed that the lift would attain when traveling from the ground floor to the top floor, a total of 300m. So they nicely give us the power. We know the power. We know we can work out mg, we know H. We can work at the time. We know age, we know the power and we know the M G, so we can know the t. We want to know about Yeah, we don't know the tea. We don't know tea. Divide by t. So we can work out t, we know the distance, we'll find out the time and then we can work out the speed. So that's okay. 18 hundred times 9.8, one times 300, divide by 90000. So t equals. 58.8s. So the speed equals. 5.1m per second. So we work out the time. Okay, a bullet of MaaS s eight grams leaves the barrel of the rifle at 700m per second. The bullet accelerates uniformly along a barrel of length. This distance determine the kinetic energy of the bullet. Energy, so kinetic energy equal to a half mv square. So m equal to eight times ten to the minus three and the genequal to 700m per second. Yes, square square n times I have. So that is 700 square times zero point. No, eight times ten to the minus three times half. 1960. Jews. Oops. Determine the average power record to accelerate the bullet down the barrel. The power of this barrel. So we know that the kinetic energy is 1960s and average power. Average power, therefore, fall at a barrel of length of 2m power. So the speed of the bullet, it's going from zero two. To 700m per second. So the average speed, we can say is 300. And so the average. Velocity is 350m per second. So speed. Distance. So the distance it has to travel down the barrel is so distance over speed will give us time. So distance 0.52m. Power il length divided by average speed down the barrel. Will give us the time. So that's 1.449 by ten to the -3s. So very quick as youexpect. So average power. So we know the time for this amount of work to be done. So 196 najews of work done divided by 1.49 by ten to minus three. Yeah. Give us. 196. One, two, three, four, five. So that gives us 13 by ten to the mto, ten to the five watts. One, one, two, three, four, five, 13, by ten to the five watts quite big power. Yeah at the. It's 350, but why is that? Is 350. So. If you think of your gun, your rifle, okay, so initially the bullish is here with zero velocity, zero meters per second. And then by the time it leaves, the gun is traveling at 700m per second, okay? So you take an average velocity. From zero to 700m per second, zero meters per second. To 700m per second. So the average velocity is 350m or second. And what that is. Leisure per second, okay. Okay, Jackson, so. You let me know in advance, if you can, what youlike to study next time. Otherwise we'll finish efficiency. We'll do some exam style questions on this topic. But two, you're pretty good at materials now. You're pretty good at this work, energy and power. Have you exams coming up soon? Next Friday? Yeah. Are they just tests or what? Yeah, just the test. But there's maybe there's serious exam in February mox. Mock exokay Jackson, so enjoy the rest of your weekend. I'll say goodbye, bye bye, ace.