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Work and Energy

Work and Energy. Chapter 5. Work. Work is defined in physics as the product of the magnitudes of the component of a force along the direction of displacement and the displacement. Work = force ·distance W = F · d

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Work and Energy

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  1. Work and Energy Chapter 5

  2. Work • Work is defined in physics as the product of the magnitudes of the component of a force along the direction of displacement and the displacement. Work = force ·distance W = F·d • Work is not done on an object unless the object is moved due to the action of a force. • Work is done only when components of a force are parallel to a displacement. • Components of the force perpendicular to a displacement do no work. • Work has dimensions of force times length. • In the SI system work is N·m =Joules

  3. Energy • Energy is the most central concept underlying all of science. • Energy is spent when we lift a load against Earth’s gravity. The heavier the load or the higher we lift, the more work we do. • Work is a type of energy that is defined as force x distance. • Two things enter in every case where work is done 1. The application of a force 2. The movement of something by that force. W = f d • The unit of measurement of work is Nm = Joule • One Joule of work is done when a force of 1N is exerted over a distance of 1 meter.

  4. Work F ө d • Imagine that you push a crate along the ground. If the force that you exert on the crate is horizontal, all of your effort moves the crate. • If your force is other than horizontal, only the horizontal component of your applied force causes a displacement and does work. • If the angle between the force and the direction of the displacement is ө, work can be written as follows: W = F cos ө· d Net work = net force ·cosine of the angle·displacement

  5. WorkExample 1: Panchito is raised 6 m above a platform by Kelly using a conveyor belt. Panchito’s mass is 70 kg. How much work is done on Panchito? • Given: m = 70 kg , d = 6 m a =-9.81 m/s² • Unknowns: F, W • Solution: F= m·a Fg = (70 kg) (-9.81 m/s²) Fg = -686.7 kg·m/s² F = 687 N W = F·d W = (687 N) (6 m) W = 4,122 N·m W = 4,122 J

  6. Example #2 Find the work done on a box that was pushed 5 m by a Force of 50 N at an angle of 30° below the horizontal. The mass of the box is 5 Kg. Fn 30 ° 50 N Fg ∆ x = d = 5 m

  7. Example #2Find the work done on a box that was pushed 5 m by a Force of 50 N at an angle of 30° below the horizontal. The mass of the box is 5 Kg. • Given: F = 50 N m = 5 kg d = 5 m • ө = 30° • Solution: W = F d cos ө W = (50 N) (5m) (cos 30°) W = 216.51 Joules 30° 60º 30º d

  8. Physics Problem Set # 19 October 31, 2007Student Name : _________________________ Class Period: ___ 1. A flight attendant pulls her 70 N flight bag a distance of 253 m along a level airport floor at a constant speed. The force she exerts is 40 N at an angle of 53° above the horizontal. A) Find the work that she does on the flight bag Answer: __________ B) Find the work done by the force of friction on the bag Answer: __________ . - 2. Yogi Berra, from the New York Yankees was the best catcher that ever played the game. As he often caught the baseball, he would gave in so that his glove was displaced 10 cm with a force of 525 N from the pitcher. How much work is done by the ball? Answer: __________ 3. How much work is done on a vacuum cleaner pulled 5 m by a force of 75 N? Answer: __________ 40 N 53°

  9. Bonus Points • P/1 If ½ is ¾ of 4/5 of a certain number. • What is the number?

  10. Work and Energy

  11. Energy • We study primarily two different forms of energy: Potential Energy Kinetic Energy Potential Energy: • is also known as stored energy • It describes an object that has to move because of its position with respect to some other location

  12. Potential Energy • Is also known as stored energy • The SI unit for Potential Energy is the Joule • It describes an object that has the potential to move because of its position with respect to some other location. • The energy associated with an object due to the object’s position relative to a gravitational source is called gravitational potential energy PEg = m g h

  13. Potential Energy Example # 1 Juan went to the top of the roof at his High School. Juan has a mass of 41 kg.If the height of the roof is 15 m, calculate Juan’s Potential Gravitational Energy to the ground. • Given: h = 15 m g = 9.81 m/s² m= 41 kg • Unknown: PEg = ? • PEg = mgh • PEg = (41 kg) ( 9.81 m/s²) ( 15 m) • PEg = 6,033.15 J

  14. Kinetic Energy • Is energy associated with an object in motion • Kinetic energy depends on the speed of the object. • Kinetic energy depends on the mass of the objects. (bowling ball vs volleyball going at the same speed) KE = ½ mv²

  15. Kinetic Energy:Example # 1: A 6 kg bowling ball moves at 4 m/s.a) How much kinetic energy does the bowling ball have?b) How fast must a 2.5 kg tennis ball move in order to have the same kinetic energy as the bowling ball? • Given: mb= 6 kg vb = 4 m/s mt = 2.5 kg • Unknown: KE = ? • Solution: KE = ½ mb vb² KE = ½ (6 kg) (4 m/s)² KE = 48 J KE = ½ mt vt² vt = √2 KE/mt = 6.20 m/s

  16. Conservation of EnergyPower Chapter 5

  17. Conservation of Energy • When we say that something is conserved it means that it remains constant, it doesn’t mean that the quantity can not change form during that time, but it will always have the same amount. • Conservation of Mechanical Energy: MEi = MEf • initial mechanical energy = final mechanical energy

  18. Conservation of Energy • If the only force acting on an object is the force of gravity: • KEi + PEi = KEf + PEf • ½ mvi² + mghi = ½ mvf² + mghf

  19. Conservation of Mechanical EnergyExample # 1:Kelly zooms down a frictionless slide with an initial height of 3 m. Kelly’s mass is 25 kg. What is her speed at the bottom of the slide? 3 m

  20. Conservation of Mechanical EnergyExample # 1:Kelly zooms down a frictionless slide with an initial height of 3 m. Kelly’s mass is 25 kg. What is her speed at the bottom of the slide? • Given: hi = 3 m m = 25 kg vi = 0 m/s • hf = 0 m

  21. Power Chapter 5

  22. Power • Power is the rate at which work is done. • Power is the rate of energy transfer by any method. • The SI unit of power is the watt, W • 1 watt = 1 Joule/s • 1000 watts = 1 kW

  23. Power Power =Work/ time Work= force · distance Power = force·distance/time Power = force·velocity P = F·v

  24. Power:Example 1: A 200 kg curtain needs to be raised 8m. in as close to 5 s as possible. You need to decide among three motors to buy for this, each motor cost a little more the bigger the power rating.The power rating for the three motors are listed as 1.0 kw, 3.5 kw and 5.5 kw. Which motor is the best for the job? • Given: m = 200 kg d = 8 m ∆t = 5s • Unknown: Power and work • Solution: Find the work done first and then divide by the time to get the power. W = F·d W = m·g·d W = (200 kg)·(9.81 m/s²)·(8 m) W = 15,696 Joules P = W/∆t P = 15,696 J / 5 s P = 3,139 watts

  25. Work-Kinetic Energy Theorem • The net work done on an object is equal to the change in the kinetic energy of the object. Wnet = ΔKE

  26. PowerExample # 2A 1,200 kg elevator carries a maximum load of 900 kg. A constant frictional force of 400 N retards the elevator’s motion upward.What minimum power, in kilowatts must the motor deliver to lift the fully loaded elevator at a constant speed of 4 m/s?

  27. PowerExample # 3A 1,500 kg car accelerates uniformly from rest to 10 m/s in 3 sa) What is the work done on the car in this time interval?b) What is the power delivered by the engine in this time interval?

  28. Physics Problem Set # 21 Nov/27/07Student Name: _________________________________ Class Period: ________

  29. Physics Problem Set # 21 Nov/01/05Student Name: ________________________ Class Period: ________ 1. Alyssa’s car accelerates uniformly from rest to 10 m/s in 3 s. Allyssa and her car have a combined mass is 1600 kg. • What is the work done on the car in this time interval? Given: vi = 0 , vf = 10 m/s , ∆t = 3 s m = 1,600 kg Unknowns: W, F, a, ∆x Solution: a =∆v/ ∆t = vf-vi/ ∆t = 10m/s – 0 m/s / 3s = 3.33 m/s² F = m∙a = (1,600 kg) (3.33 m/s²) = 5,328 N W = F ∙ d d = ? → ∆x = vi(∆t) + ½ a (∆t)² ∆x = 0 +½ (3.33 m/s²)(3s)² = 14.99 m W = F ∙ d = (5,328 N) (14.99 m) = 79,867 J B) what is the power delivered by the engine in this time interval? P = W/t = 79,867 J / 3s = 26622 Watts

  30. 2. Erika’s car accelerates uniformly from rest to 13 m/s in 2 s. Erika and her car have a combined mass is 1,550 kg. • What is the work done on the car in this time interval? F= (1,550 kg) (13 m/s / 2s) = 10,075 N ∆x = ½ (6.5 m/s²)(2s)² = 13 m W = F∙d = (10,075 N) (13m) = 130,975 J B) what is the power delivered by the engine in this time interval? P = W/∆t = 130,975 J/ 2s = 65,488 Watts 3. A 1000 kg elevator carries a maximum load of 800 kg. A constant frictional force of 4000 N retards the elevator’s motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s ? Given: m = 1,000 kg + 800 kg = 1,800 kg Fk = 4,000 N v = 3 m/s Unknown: P Solution: P = F∙ v = (Fg + Fk) v = (mg + Fk) v P = {(1,800 kg)(9.81 m/s²) + 4,000 N }(3 m/s) P = 64, 974 Watts

  31. 4. A rain cloud contains 2,660,000 kg of water vapor. How long would it take for a 2 kW pump to raise the same amount of water to the cloud altitude of 2 km? Given: m = 2,660,000 kg P = 2 kW → P = 2,000 Watts d = 2 km → d = 2,000 m Unknown: ∆t, W Solution: W = F∙d = (mg)∙d = (2,600,000 Kg) (9.81 m/s²) (2,000 kg) W = 51,012,000,000 J P = W/∆t → ∆t = W/P ∆t = 51,012,000,000 J/ 2,000 Watts ∆t = 25,506,000 s (about 8.27 years)

  32. PhysicsProblem Set # 13Monday Dec. 5, 2005Student name: ___________________________________ Class Period: _____

  33. Physics Problem Set # 14 Tuesday Dec.6, 2005 Student Name: _________________________ Class Period: _____ 1 A 5.60 kg rubber ball has a speed of 3 m/s at point A and a kinetic energy of 7.5 J at point B. Determine the following: a) The ball kinetic energy at point A ________ b) The ball speed at point B ________ c) The total work done on the ball as it moves from A to B ________ • Erica threw a ball that has a mass of 3.50 kg. The initial speed of the ball was 12 m/s at point A and a kinetic energy of 19.5 J at point B. Determine the following: a) The ball kinetic energy at point A ________ b) The ball speed at point B ________ c) The total work done on the ball as it moves from A to B ________ • Starting from rest, a 20 kg suitcase slides 5 m. down a frictionless ramp inclined at a 40° angle from the floor. The suitcase then slides an additional 8 m. until it comes to a stop. Determine the following; • The speed of the suitcase at the bottom of the ramp. ________ • The coefficient of kinetic friction between the suitcase and the floor ________ • The mechanical energy lost due to friction ________ 4 A skier of mass 60 kg is pulled up a slope by a motor-driven cable. How much work is required to pull the skier 60 m up a 30° slope (assumed to be frictionless) at a constant speed of 3 m/s? ________

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