PHYSICS PERIODS & 6 UNIT 8 WORK ENERGY WORK MR. LARGO MR. Hill EXPERIMENT ONE You hold a heavy block just above a piece of chalk and then release the block. Outcome: Chalk breaks just a little bit or not at all. System: Block, chalk, Earth External object: Hand Lift the block 30 cm above the chalk and then release the block. Outcome: Chalk is crushed F HonB Direction of displacement of block EXPERIMENT ONE EXPERIMENT TWO The direction of the force exerted on the block, and the direction of the displacement of the block is the same! You push a cart initially at rest. A piece of chalk is taped to the end of the track. F HonB There is an increase in the block s elevation and its ability to break the Direction of chalk displacement of block The fast moving car collides with the chalk. Outcome: Chalk is crushed 1
System: cart, chalk External object: Hand EXPERIMENT TWO The direction of the force exerted on the cart, and the direction of the displacement of the cart is the same! F HonB Direction of displacement of cart F HonC Direction of displacement of cart There is an increase in the cart s speed and its ability to break the chalk EXPERIMENT THREE A piece of chalk rests in the hanging sling of a slingshot. You then pull it back until the slingshot is fully stretched. When released from the stretched sling, the chalk flies along the room. System: Sling, chalk External object: Hand F HonSB Outcome: Chalk is crushed Direction of displacement of the sling EXPERIMENT THREE The direction of the force exerted on the sling, and the direction of the displacement of the sling is the same! F HonS Direction of displacement of the sling There is an increase in the sling s elongation and its ability to break the chalk EXPERIMENT FOUR A heavy box sits on a shag carpet. You pull the box along the carpet to a position several meters from where it started. When you reach the final position you touch the bottom of the box and the carpet. Outcome: They feel slightly warmer
System: box External object: Hand F HonB EXPERIMENT FOUR The direction of the force exerted on the box, and the direction of the displacement of the box is the same! Direction of displacement of the box F HonB Direction of displacement of the box There is an increase in box s temperature and/or structural changes. CONCLUSION OF FOUR EXPERIMENTS In every single case the external force exerted on the object caused a displacement of the object. When a force exerted on an object causes a displacement of the object, Physicists call this: Definition: Symbol: Units: PHYSICAL QUANTITY: WORK Type of PQ: scalar Work is a way to change the energy of a system. W Joules (J) WORK Work is a way to change the energy of a system. The force exerted on the object does work on the object, if the object changes: Elevation Speed Elongation Temperature or structure Eddie pulls the box at rest with a rope. The box starts moving with constant speed. F RonB Direction of displacement of the box The force exerted by the rope on the box does work on the box because the box changed speed!! 3
Force Displacement W = F d Force and displacement are parallel to each other MATHEMATICAL MODEL WORK Work Work [J] Joules Force exerted on object [N] Newtons distance [m] meters Force W F d distance Eddie pulls the box at rest with a rope at an angle. The box starts moving with constant speed. Displacement Force q Force and displacement are at an angle F RonB q W = F d cos(θ) Direction of displacement of the box Vin exerts a force of 176.4 N pulling a 90 kg washing machine along a distance of 50 m The washing machine starts from rest and then moves with constant speed. FORCE DIAGRAM Earth (down) Vin (right) Surface Normal (up) Surface friction (left) F NonW Washing machine F KFonW F RonW Earth F EonW 4
g = - 9.8 m/s 90 kg Mass = F EonW = -88 N F EonW = g m F EonW = 9.8 m 90 kg s F EonW = 88 N F RonW ANGLES Displacement q = 0 F NonW = +88 N (Equilibrium) k = 0. F KFonW = -176.4 N (Equilibrium) +176.4 N F RonW = μ K = F KFonW F NonW μ K = 176.4 N 88 N μ K = 0. F NonW F KFonW F EonW Displacement Displacement Displacement q = 90 q = 180 q = 70 Work done by the force exerted by Vin (Rope) on the washing machine W RonW = F RonW d cos(θ) W RonW = 176.4 N 50 m cos(0 ) W RonW = 880 J The force exerted by Vin does work on the Washing machine (it is not zero), this force CHANGES THE SPEED of the washing machine Work done by the Normal force exerted on the washing machine W NonW = F NonW d cos(θ) W NonW = 88 N 50 m cos(90 ) W NonW = 0 J The Normal force does not do work on the Washing machine (it is zero). Work done by the force of kinetic friction exerted on the washing machine W KFonW = F KFonW d cos(θ) W KFonW = 176.4 N 50 m cos(180 ) W KFonW = 880 J The force of kinetic friction does work on the Washing machine (it is not zero), this force CHANGES THE TEMPERATURE AND STRUCTURE of the washing machine Work done by the force exerted by Earth on the washing machine W EonW = F EonW d cos(θ) W EonW = 88 N 50 m cos(70 ) W EonW = 0 J The force exerted by Earth does not do work on the Washing machine (it is zero). 5
F EonW = -88 N F NonW = +88 N F KFonW = -176.4 N F RonW = +176.4 N W EonW = 0 J W NonW = 0 J W KFonW = -880 J W RonW = 880 J Vin exerts a NEW force of 1.4 N on the washing machine. (all other quantities remain the same) F NET = 0 N W NET = 0 J a = F NET m a = 0 N 90 kg a = 0 m s F EonW = -88 N F NonW = +88 N F KFonW = -176.4 N F RonW = +1.4 N F NET = 45 N W EonW = 0 J W NonW = 0 J W KFonW = -880 J W RonW = 11070 J W NET = 50 J Mr. Largo lifts a 185 pounds (83.91 kg). The distance from his chest to his extended arms is 0.5 m. a = F NET m a = 45 N 90 kg a = 0.5 m s Assume the barbell is lifted at constant speed. Earth (down) Largo (up) FORCE DIAGRAM F LonB Bar g = - 9.8 m/s F EonB = g m 83.91 kg Mass = F EonB = 9.8 m 83.91 kg s F EonB = 8.3 N F EonB = -8.3 N F RonB = +8.3 N (Equilibrium) Earth F EonB 6
ANGLES F LonB Displacement q = 0 Work done by the force exerted by Mr. Largo on the bar W LonB = F LonB d cos(θ) W LonB = 8.3 N 0.5 m cos(0 ) W LonB = 411. J F EonB Displacement q = 180 The force exerted by Mr. Largo does work on the bart(it is not zero), this force CHANGES THE ELEVATION of the bar. Work done by the force exerted by Earth on the bar W EonB = F EonB d cos(θ) W EonB = 8.3 N 0.5 m cos(180 ) W EonB = 411. J F EonB = -8.3 N F RonB = +8.3 N F NET = 0 N W EonB = -411. J W NonB = 411. J 0 J W NET = The force exerted by Earth does work on the bar (it is not zero), this force CHANGES THE ELEVATION of the bar. a = F NET m a = 0 N 83.91 kg a = 0 m s Mr Largo exerts a NEW force of 999.6 N on the bar. (all other quantities remain the same) F EonB = -8.3 N F LonB = +999.6 N W EonB = -411. J W NonB = 499.8 J F NET = 177.3 N 88.6 J W NET = a = F NET m a = 177.6 N 83.91 kg a =.1 m s 7
Justin lifts a 0 kg bucket from a 15 m deep well. Earth (down) Rope (up) FORCE DIAGRAM F RonB The bucket rises at constant speed. Bucket Earth F EonB ANGLES g = - 9.8 m/s F EonB = g m 0 kg Mass = F EonB = -196 N F EonB = 9.8 m 0 kg s F EonB = 196 N F RonW Displacement q = 0 F RonB = +196 N (Equilibrium) F EonW Displacement q = 180 Work done by the force exerted by the rope on the bucket W RonB = F RonB d cos(θ) W RonB = 196 N 15 m cos(0 ) W RonB = 940 J The force exerted by the rope does work on the bucket (it is not zero), this force CHANGES THE ELEVATION of the bucket. Work done by the force exerted by Earth on the bucket W EonB = F EonB d cos(θ) W EonB = 196 N 15 m cos(180 ) W EonB = 940 J The force exerted by Earth does work on the bucket (it is not zero), this force CHANGES THE ELEVATION of the bucket. 8
F EonB = -196 N F RonB = +196 N W EonB = -940 J W NonB = 940 J Justin exerts a NEW force of 00 N on the rope that pulls up the bucket. (all other quantities remain the same) F NET = 0 N 0 J W NET = a = F NET m a = 0 N 0 kg a = 0 m s F EonB = -196 N F RonB = +00 N F NET = 4 N W EonB = -940 J W NonB = 3000 J 60 J W NET = Jordan brings his Lacrosse stick to the second floor of his house. The mass of a Lacrosse stick is 1.9 kg and Jordan went up the stairs 3.0 m. Assume Jordan goes up the stairs at a constant speed. a = F NET m a = 4 N 0 kg a = 0. m s How much work is done Jordan on the Lacrosse stick? Earth (down) Jordan (up) FORCE DIAGRAM Work done by the force exerted by Jordan on the Lacrosse stick W JonL = F JonL d cos(θ) Earth F JonL Lacrosse stick F EonL W JonL = 9.8 1.9 3 m cos(0 ) W JonL = 55.86 J F JonL = F EonL W JonL = g m d cos(θ) The force exerted by Jordan does work on the lacrosse stick (it is not zero), this force CHANGES THE ELEVATION of the lacrosse stick. 9
W SonB = F SonB d cos(θ) Steven does 6.46 J of work lifting his backpack from the floor to the top of the classroom table (0.90 m) What is the mass of Steven s back pack? W SonL = g m d cos(θ) 6.46 = 9.8 m 0.9 cos(0 ) m = 3 kg F SonB = F EonB 6.46 = 8.8 m The force exerted by Steven does work on the backpack (it is not zero), this force CHANGES THE ELEVATION of the backpack. PHYSICAL QUANTITY: POWER Definition: Rate at which work is done. POWER Symbol: P Units: Watt (W) Type of PQ: scalar PRACTICE PROBLEM P = W t Work [J] Joules Power [W] Watts Time [s] seconds 10
Amelia (60 kg) climbs the stairs in 3 s. How much work is done by Amelia? W EonA = F EonA d cos(θ) P = W t W EonA = g m d cos(θ) W EonA = 9.8 60 4 cos(0 ) W EonA = 35 J What is the power output? P = F d t P = W EonA t P = 35 3 P = 784 W P = F v P = F v Your brother got himself a snappy new car. You think it s kind of small, but he claims it has over 100 horsepower. The net force exerted on the car is 5060 N and the car moves with a constant speed of 1 m/s. Power [W] Watts Force [N] Newtons Speed (constant) [m/s] Meters per seconds What is the power developed by the car? P = F v P = 5060 1 P = 60,70 W What is the power developed by the car in Horsepower? 1 Horsepower = 745.7 W P = 60,70 W 745.7 W GRAVITATIONAL POTENTIAL ENERGY P = 81.43 hp 11
Bryan changes a light bulb in the living room of his house. Unfortunately Bryan dipped his hands in butter and the light bulb escaped from his hands and falls to the ground. Initial elevation FORCE DIAGRAM What force changed the elevation of the lightbulb? The force that Earth exerts on the lightbulb changes its elevation Displacement of the light bulb Earth The force exerted by Earth does work on the light bulb. Earth Final elevation F EonLB W EonLB = F EonLB d cos(θ) q = 0 F EonLB = gm cos 0 = 1 gmh i + W EonLB = gmh f W EonLB = gm (h f h i ) cos(0) W EonLB = gm (h f h i ) 1 W EonLB = gmh f gmh i Reorganize the equation. gmh i + W EonLB = gmh f d difference between two elevations NEW COMBINATION OF PHYSICAL QUANTITIES: gmh GRAVITATIONAL POTENTIAL ENERGY U G = gmh GRAVITATIONAL POTENTIAL ENERGY U G = gmh GRAVITATIONAL POTENTIAL ENERGY U G = gmh Energy that a system has due to the relative separation of two objects with mass. Elevation Earth Gravitational Potential Energy [J] Joules Gravitational constant [m/s ] Meters per second squared height (elevation) [m] Meters mass [kg] kilograms 1
gmh i + W EonLB = gmh f U Gi + W = U GF Initial elevation 4 m Bryan drops a light bulb (0.185 kg) from a height 4 m above the ground. The lightbulb falls on a table 0.8 m above the ground. You need to do work to change the elevation of an object Earth Final elevation 0.8 m g = 9.8 m/s F = g = 9.8 m/s g = 9.8 m/s F = g = 9.8 m/s m = 0.185 kg d = m = 0.185 kg m = 0.185 kg d = m = 0.185 kg h i = 4 m h f = 0.8 m h i = 4 m h f = 0.8 m U Gi = 7.5 J W = U GF = 1.4504 J U Gi = 7.5 J W = -5.8016 J U Gf = 1.4504 J U Gi = g m h i U Gi = 9.8 0.185 4 U Gi = 7.5 J U GF = g m h F U GF = 9.8 0.185 0.8 U GF = 1.4504 J U Gi + W = U GF 7.5 + W = 1.4504 W = 5.8016 J g = 9.8 m/s m = 0.185 kg h i = 4 m U Gi = 7.5 J d = h f h i d = 0.8 m 4 m d = 3. m F = 1.813 N d = -3. m W = -5.8016 J W = F d F = W d g = 9.8 m/s m = 0.185 kg h f = 0.8 m U Gf = 1.4504 J F = 5.8016 3. m F = 1.813 N 8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0.5.0 1.5 1.0 0.5 0.0-0.5-1.0-1.5 -.0 -.5-3.0-3.5-4.0-4.5-5.0-5.5-6.0-6.5 WORK ENERGY BAR CHART U Gi = 7.5 J U GF = 1.4504 J UG i W UG F W = -5.8016 J 13
Bryan pushes a cart The speed of the car increases KINETIC ENERGY Initial speed Displacement of the cart Final speed F HonC Bryan pushes a cart The speed of the car increases W HonC = F HonC d cos(θ) q = 0 W HonC = F HonC d 1 cos 0 = 1 W HonC = F HonC d The force that the hand exerts on the cart changes its speed. The force exerted by the hand does work on the cart. Find F using Newton s second law F = m a Find d using independence of time d = v f v i (a) W HonC = F HonC d v f v i W = m a a Acceleration divides out W = mv f mv i W = m v f v i Distribute the mass W = mv f mv i Reorganize the equation mv i + W = mv f 14
mv i + W = mv f NEW COMBINATION OF PHYSICAL QUANTITIES: mv KINETIC ENERGY KE = mv KINETIC ENERGY KE = mv The energy of an object of mass m moving at speed v. KINETIC ENERGY KINETIC ENERGY KE = mv KE = mv Kinetic Energy [J] Joules mass [kg] kilograms speed [m/s] Meters per second Kinetic Energy can t be negative Mass is always positive Any number squared will be positive mv i + W = mv f KINETIC ENERGY A Truck (000 kg) speeds up from 6 m/s to 8 m/s over a displacement of 70 m KE i + W = KE F v i = 6 m/s v F = 8 m/s You need to do work to change the SPEED of an object 15
m = 000 Kg F = m = 000 kg m = 000 Kg F = m = 000 kg v i = 36 m/s d = 70 m v F = 64 m/s v i = 36 m/s d = 70 m v F = 64 m/s Divide by Divide by Divide by Divide by KE i = 36000 J W = KE F = 64000 J KE i = m v i KE F = m v F 000 36 000 64 KE i = KE F = KE i = 36000 J KE F = 64000 J KE i = 36000 J W = 8000 J KE F = 64000 J KE i + W = KE F 36000 + W = 64000 W = 8000 J m = 000 Kg v i = 36 m/s F = 400 J d = 70 m m = 000 kg v F = 64 m/s 64,000.0 WORK ENERGY BAR CHART KE F = 64000 J Divide by Divide by 51,00.0 38,400.0 KE i = 36000 J W = 8000 J KE i = 36000 J W = 8000 J KE F = 64000 J 5,600.0 1,800.0 0.0 W = F d F = 8000 J 70 m -1,800.0-5,600.0-38,400.0 KE i W KE F F = W d F = 400 N -51,00.0-64,000.0 Bryan loads an arrow into a bow and elongated the bowstring back ELASTIC POTENTIAL ENERGY Displacement of the bow string F HonC 16
Bryan loads an arrow into a bow and elongated the bowstring back W AonB = F AonB d cos(θ) W AonB = F AonB d 1 W AonB = F AonB L q = 0 cos 0 = 1 The force that the arrow exerts on the bow changes its elongation. The force exerted by the arrow does work on the bow. Find F using Force of the spring (elastic material ) F SonB = k L d is the same as the elongation d W AonB = F AonB L W AonB = W AonB = Not a constant force (changes with elongation) use average force. k L k L L W = k L f k L i k L i Reorganize the equation + W = k L f k L i + W = k L f NEW COMBINATION OF PHYSICAL QUANTITIES: k L ELASTIC POTENTIAL ENERGY U S = k L U S = k L Energy of a stretched or compressed elastic object (spring, bow, etc). ELASTIC POTENTIAL ENERGY 17
Elastic Potential Energy [J] Joules ELASTIC POTENTIAL ENERGY U S = k L Spring Constant [N/m] Newtons per meter Elongation [m] Meters Elastic potential energy can t be negative ELASTIC POTENTIAL ENERGY U S = k L Spring constant is always positive Any number squared will be positive k L i + W = k L f ELASTIC POTENTIAL ENERGY Bryan loads an arrow into a bow and pull the bowstring back 0.40 m. The spring constant of the bowstring is 900 N/m U Si + W = U SF You need to do work to change the ELONGATION of an object L i = 0.4 m L F = 0 m k = 900 N/m k = 900 N/m k = 900 N/m k = 900 N/m L i = 0.16 L F = 0 L i = 0.16 L F = 0 Divide by Divide by Divide by Divide by U Si = 7 J W = U SF = 0 J U Si = k L i U SF = m L F 900 0.16 U Si = U SF = 900 0 U Si = 7 J KE F = 0 J U Si = 7 J 7 J W = U SF = 0 J U Si + W = U SF 7 + W = 0 W = 7 J 18
7.0 64.0 56.0 48.0 40.0 3.0 4.0 16.0 8.0 0.0-8.0-16.0-4.0-3.0-40.0-48.0-56.0-64.0-7.0 WORK ENERGY BAR CHART US i = 7 J US F = 0 J US i W US F W = -7 J INTERNAL ENERGY Bryan steps abruptly on the brakes to avoid an accident. Imagine that the car s brakes have locked, and the tires are skidding on the roads surface. If you touched the car s tires just after the car came to rest, they would be warm to the touch. You would also notice black skid marks on the road (some of the rubber had been scrapped of the tires). Displacement of the car F KFonC The force of kinetic friction does work on the car (tires) because it changes the temperature and the structure of the car (tires) W KFonC = F KFonC d cos(θ) W KFonC = +F KFonC d q = 180 cos q = -1 W KFonC = F KFonC d 1 W KFonC = +F KFonC d NEW COMBINATION OF PHYSICAL QUANTITIES CHANGE IN INTERNAL ENERGY U int = +F KFonO d 19
CHANGE IN INTERNAL ENERGY U int = +F KFonO d Energy of motion and interaction of the microscopic particles making up the objects in the system. The internal energy of the system changes when the surfaces of the system objects rub against each other. CHANGE IN INTERNAL ENERGY W KFonC = +F KFonC d Use Force of Kinetic friction W KFonC = μ KF F NonC d W KFonC = μ KF g m d F KFonC = μ KF F NonC Normal Force in equilibrium with Force exerted by Earth F NonO = F EonO = gm CHANGE IN INTERNAL ENERGY U int = +F KFonO d Jim is driving a 100 kg sedan 1 m/s and releases his foot from the accelerator pedal. The car slows down to 4 m/s over a displacement of 8 m. The coefficient of friction between the tires and the road is 0.7. U int = +μ KF g m d Jim is driving a 100 kg sedan 1 m/s and releases his foot from the accelerator pedal. The car slows down to 4 m/s over a displacement of 8 m. m = 100 Kg v i = 144 m/s F = d = 8 m m = 100 kg v F = 16 m/s The coefficient of friction between the tires and the road is 0.7. Divide by Divide by KE i + W = KE F W KFonO = U int KE i = 86400 J W = KE F = 9600 J KE i = m v i KE F = m v F 100 144 100 16 KE i = KE F = KE i = 864000 J KE F = 9600 J 0
m = 100 Kg F = m = 100 kg m = 100 Kg F = -9600 J m = 100 kg v i = 144 m/s d = 8 m v F = 16 m/s v i = 144 m/s d = 8 m v F = 16 m/s Divide by Divide by Divide by Divide by KE i = 86400 J W = -76800 J KE F = 9600 J KE i = 86400 J W = -76800 J KE F = 9600 J KE i + W = KE F 86400 + W = 9600 W = 76800 J W = F d F = W d F = 76800 J 8 m F = 3840 N 90,000.0 80,000.0 70,000.0 60,000.0 50,000.0 40,000.0 30,000.0 0,000.0 10,000.0 0.0-10,000.0-0,000.0-30,000.0-40,000.0-50,000.0-60,000.0-70,000.0-80,000.0-90,000.0 WORK ENERGY BAR CHART KE i = 86400 J KE F = 9600 J KE i W KE F W = -76800 J g = -9.8 m/s 100 kg Mass = F EonS = 9.8 m 100 kg s F EonS = 11760 N F EonS = -11760 N F NonS = +11760 N (Equilibrium) F KFonS = μ K F NonS k = 0.7 F KFonS = 0.7 11760 N F KFonS = 83 N F FKonS = 83 N F EonS = g m W KFonS = F KFonS d W KFonS = 83 N 8 m W FKonS = 65856 J W KFonS = 65856 J 70,000.0 60,000.0 50,000.0 40,000.0 30,000.0 WORK ENERGY BAR CHART W = 65856 J U int = 65856 J 0,000.0 10,000.0 0.0-10,000.0-0,000.0-30,000.0-40,000.0-50,000.0-60,000.0-70,000.0-80,000.0-90,000.0 W KF on O DU int Positive work that explains the increase in temperature caused by the Force of Friction TRANSFORMATIONS OF ENERGY 1
TRANSFORMATIONS OF ENERGY Kinetic Energy & Gravitational Potential Elastic Potential & Gravitational Potential Kinetic Energy & internal energy Kinetic & Gravitational Potential KE = U G speed mv = mgh v = gh height v = gh h = v g Elastic & Gravitational Potential U s = U G elongation k L = mgh height Elastic & Gravitational Potential U s = U G Spring constant k L = mgh mass L = mgh k h = k L mg k = mgh L m = k L gh speed Kinetic & Internal Energy KE = U int mv = μ kmgd v = μ k Coefficient v = μ k gd μ k = v gd 10.0 m 3.0 m 0.0 m WATER A diver drops from a board 10.0 m above the water s surface, as shown in the figure. What speed does the diver enter the water with? (Neglect air resistance)
Type of energy transformation: Gravitational potential to kinetic energy Physical quantity I am looking for? speed v = gh v = 9.8 10 v = 14 m s If the diver enters the water with a speed of 1 m/s from what height did he step off the board? Type of energy transformation: Gravitational potential to kinetic energy Physical quantity I am looking for? Height h = v g h =.5 m h = 1 9.8 An 80 kg stuntman drops from a height 10.0 m onto a mattress, as shown in the figure. The mattress has a spring constant 98000 N/m. What is the elongation of the mattress? Type of energy transformation: Gravitational potent. to Elastic Potential Physical quantity I am looking for? Elongation L = mgh k L = L = 0.4 m 80 9.8 10 98000 If the stuntman falls on a NEW mattress, it elongates 0.56 m. What is the spring constant of the new mattress? Type of energy transformation: Gravitational potent. to Elastic Potential Physical quantity I am looking for? Spring Constant k = mgh L k = k = 50000 N m 80 9.8 10 0.56 To avoid a collision while driving, you apply the brakes in your car, leaving 4-m skid marks on the road while stopping. A police officer observes the near collision and gives you a speeding ticket, claiming that you were exceeding the 35-mph speed limit. He estimates your car's mass as 1390 kg and the coefficient of kinetic friction μ k between your tires and this particular road as about 0.70. Do you deserve the speeding ticket? 3
Type of energy transformation: Kinetic Energy to Internal Energy Physical quantity I am looking for? speed v = μ k gd v = 0.7 9.8 4 v = 18.15 m s v = 40.6 miles h A 1900 kg car skids 0 m on a level road while trying to stop before hitting a stopped car in front of it. The two cars barely touch. The car initially riding at 18 m/s slows down to stop. What is the coefficient of kinetic friction between the first car and the road? Type of energy transformation: Kinetic Energy to Internal Energy Physical quantity I am looking for? Coefficient of kinetic friction μ k = v gd μ k = 18 9.8 0 CONSERVATION OF ENERGY μ k = 0.87 Gravitational Potential Energy Kinetic Energy U G = mgh KE = mv height h = U G mg speed v = KE m 4
Elastic Potential Energy elongation U s = k L L = U s k TOTAL ENERGY (TE) TE = U G + KE + U S Total energy remains constant throughout Juanito drops a basket (3 kg) from the roof of the school (3.8 m) h = 3.8 m U G = 111.7 J TE = v = 0 m/s KE = 0 J 111.7 J v = SPEED KE m Floor h = 0 m U G = 0 J TE = v = 8.63 m/s KE = 111.7 J 111.7 J v = 111.7 3 v = 8.63 m s UG KE US TE h v L What is the SPEED of the basket when it is at height h =.3 m Roof 111.7 0 111.7 3.8 0 111.7 111.7 h =.3 m v = 5.4 m/s U G = 67.6 J KE = 44.1 J TE = 111.7 J 0 111.7 111.7 0 8.63 Floor 5
UG KE US TE h v L What is the HEIGHT when it is falling with a speed 7.6 m/s? Roof 111.7 0 111.7 3.8 0 A 67.6 44.1 111.7.3 5.4 111.7 h = 0.85 m v = 7.6 m/s U G = 5.08 J KE = 86.64 J TE = 111.7 J 0 111.7 111.7 0 8.63 Floor UG KE US TE h v L Roof 111.7 0 111.7 3.8 0 A 67.6 44.1 111.7.3 5.4 B 5.08 86.64 111.7 0.85 7.8 0 111.7 111.7 0 8.63 10 100 80 60 40 0 0 ENERGY BAR CHART 111.7 111.7 86.64 67.6 44.1 5.08 0 0 Roof A B Floor UG KE US 6