Ch10.1 – Energy and Work

Ch10.1 – Energy and Work Energy – the ability to produce change.

Ch10.1 – Energy and Work Energy – the ability to produce change. Kinetic Energy – energy of motion KE = ½ m∙v2Units: Ex1) What is the kinetic energy associated with a freight train car travelling 2 m/s with a mass of 1,000 kg? Ex2) What is the kinetic energy associated with a 10 kg bowling ball rolling at 20 m/s ?

Ch10.1 – Energy and Work Energy – the ability to produce change. Kinetic Energy – energy of motion KE = ½ m ∙ v2 Units: kg ∙ m2 = kg ∙ m ∙ m = N∙m = J s2s2 (Joule) Ex1) What is the kinetic energy associated with a freight train car travelling 2 m/s with a mass of 1,000 kg? Ex2) What is the kinetic energy associated with a 10 kg bowling ball rolling at 20 m/s ?

Ch10.1 – Energy and Work Energy – the ability to produce change. Kinetic Energy – energy of motion KE = ½ m ∙ v2 Units: kg ∙ m2 = kg ∙ m ∙ m = N∙m = J s2s2 (Joule) Ex1) What is the kinetic energy associated with a freight train car travelling 2 m/s with a mass of 1,000 kg? KE = ½( 1000 kg )( 2 m/s )2 = 2000 N∙m = 2,000 N.m Ex2) What is the kinetic energy associated with a 10 kg bowling ball rolling at 20 m/s ?

Ch10.1 – Energy and Work Energy – the ability to produce change. Kinetic Energy – energy of motion KE = ½ m ∙ v2 Units: kg ∙ m2 = kg ∙ m ∙ m = N∙m = J s2s2 (Joule) Ex1) What is the kinetic energy associated with a freight train car travelling 2 m/s with a mass of 1,000 kg? KE = ½( 1000 kg )( 2 m/s )2 = 2000 N∙m = 2,000 N.m Ex2) What is the kinetic energy associated with a 10 kg bowling ball rolling at 20 m/s ? KE = ½ ( 10 kg )( 20 m/s )2 = 2000 J

Work – force applied over a distance W = F ∙ d Force and distance must be in the same direction Ex3) How much work is done when a person pushes a shopping cart with a force of 50 N for 500 meters?

Work – force applied over a distance W = F ∙ d Force and distance must be in the same direction Ex3) How much work is done when a person pushes a shopping cart with a force of 50 N for 500 meters? W = F ∙ d = ( 50N )( 500m ) = 25,000 N∙m

Work – force applied over a distance W = F ∙ d Force and distance must be in the same direction Ex3) How much work is done when a person pushes a shopping cart with a force of 50 N for 500 meters? W = F ∙ d = ( 50N )( 500m ) = 25,000 N∙m Work – Energy Theorem W = ∆KE Ex4) A hockey player hits a puck, applying a force of 4,000 N over a distance of 0.5 m, accelerating the puck up to 40 m/s. What is the mass of the puck?

Work – force applied over a distance W = F ∙ d Force and distance must be in the same direction Ex3) How much work is done when a person pushes a shopping cart with a force of 50 N for 500 meters? W = F ∙ d = ( 50N )( 500m ) = 25,000 N∙m Work – Energy Theorem W = ∆KE Ex4) A hockey player hits a puck, applying a force of 4,000 N over a distance of 0.5 m, accelerating the puck up to 40 m/s. What is the mass of the puck? F = 4000N W = ∆KE d = 0.5m vF2 = vi2 + 2ad F ∙ d = KEf – KEi vF = 40 m/s F ∙ d = ½ mvf2 – ½ mvi2 vi = 0 F = m ∙ a ( 4000 )( .5 ) = ½ m( 40 )2 m = ? m = 2.5 kg

1. an object is lifted, work is done overcoming gravity W = (+) 1 2 3 4

1. an object is lifted, work is done F d overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = (–) 1 2 3 4

1. an object is lifted, work is done F d overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = (–) 3. an object is pushed across a surface, F d work is done overcoming friction. 1 2 3 4

1. an object is lifted, work is done F d overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = (–) 3. an object is pushed across a surface, F d F d work is done overcoming friction. 4. an object is carried horizontally, no work is done on the object. 1 2 3 4

1. an object is lifted, work is done F d overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = (–) 3. an object is pushed across a surface, F d F d work is done overcoming friction. 4. an object is carried horizontally, F no work is done on the object. d What about work done in circular motion lab? Top View vel Fc 1 2 3 4

1. an object is lifted, work is done F d overcoming gravity W = (+) 2. an object is dropped, gravity does the work to bring it back down. W = (–) 3. an object is pushed across a surface, F d F d work is done overcoming gravity. 4. an object is carried horizontally, F no work is done on the object. d So what about in between? Force at an angle A tall person pushing a cart: d θ FW = F.d.cosθ 1 2 3 4

Ex5) A sailor pulls a boat 30.0 m along a dock using a rope that makes a 25° angle with the horizontal. How much work does he do if he exerts a force of 255 N on the rope? F = 255 N 25° d = 30 m

Ex5) A sailor pulls a boat 30.0 m along a dock using a rope that makes a 25° angle with the horizontal. How much work does he do if he exerts a force of 255 N on the rope? F = 255 N 25° d = 30 m W = F ∙ d ∙ cosθ = ( 255 N )( 30 m )( cos25° ) = 6933 J

HW #7) An airplane passenger carries a 215 N suitcase up the stairs, a displacement of 4.20 m vertically and 4.60 m horizontally. a) How much work does the passenger do? b) The same passenger carries it back downstairs. How much work now? 4.20m 4.60m Ch10 HW#1 1 – 8

HW #7) An airplane passenger carries a 215 N suitcase up the stairs, a displacement of 4.20 m vertically and 4.60 m horizontally. a) How much work does the passenger do? b) The same passenger carries it back downstairs. How much work now? a) W = F..d 4.20m = 215N.4.20m = 903 J 4.60m b) W = - 903 J Ch10 HW#1 1 – 8

Chapter 10 HW #1 1 – 8 1. A student lifts a box that weighs 185N. The box is lifted 0.800m. How much work? d = .800 m Fg = 185 N 2. Two students exert a force of 825N in pushing a car 35m. How much work do they do on the car? d = 35m F = 825N

Chapter 10 HW #1 1 – 8 1. A student lifts a box that weighs 185N. The box is lifted 0.800m. How much work? FF = Fg d = .800 m Wg = Fg ∙ d = ( 185 N )( .800 m ) = 148 J Fg = 185 N 2. Two students exert a force of 825N in pushing a car 35m. How much work do they do on the car? d = 35m F = 825N

Chapter 10 HW #1 1 – 8 1. A student lifts a box that weighs 185N. The box is lifted 0.800m. How much work? F F = Fg d = .800 m Wg = Fg ∙ d = ( 185 N )( .800 m ) = 148 J Fg = 185 N 2. Two students exert a force of 825N in pushing a car 35m. How much work do they do on the car? d = 35m W = F ∙ d = ( 825 N )( 35 m ) = 28,875 J F = 825N

3) A 0.180kg ball falls 2.5m. How much work does the force of gravity do on the ball? d = 2.5 mFg = 1.8 N 4) A forklift raises a box 1.2m doing 7000J of work on it. What is the mass of the box? F d Fg

3) A 0.180kg ball falls 2.5m. How much work does the force of gravity do on the ball? d = 2.5 mFg = 1.8 N Wg = Fg ∙ d = - 4.4 J 4) A forklift raises a box 1.2m doing 7000J of work on it. What is the mass of the box? F d Fg

3) A 0.180kg ball falls 2.5m. How much work does the force of gravity do on the ball? d = 2.5 mFg = 1.8 N Wg = Fg ∙ d = - 4.4 J 4) A forklift raises a box 1.2m doing 7000J of work on it. What is the mass of the box? F d W = F ∙ d ( W = mgd ) 7000 J = F ∙ 1.2 m F = 5833 N m = 583 kg Fg

5. You and a friend each carry identical boxes to a room one floor above you and down the hall. You carry your box up a set of stairs then down the hallway. Your friend carries a box down the hall then up another stairwell. Who does more work? How much work does the force of gravity do when a 24N object falls a distance of 3.5m? W = F ∙ d d = 3.5 m Fg = 24 N

5. You and a friend each carry identical boxes to a room one floor above you and down the hall. You carry your box up a set of stairs then down the hallway. Your friend carries a box down the hall then up another stairwell. Who does more work? same vertical distance = same work How much work does the force of gravity do when a 24N object falls a distance of 3.5m? W = F ∙ d d = 3.5 m Fg = 24 N

5. You and a friend each carry identical boxes to a room one floor above you and down the hall. You carry your box up a set of stairs then down the hallway. Your friend carries a box down the hall then up another stairwell. Who does more work? same vertical distance = same work How much work does the force of gravity do when a 24N object falls a distance of 3.5m? W = F ∙ d d = 3.5 m = ( 24 N )( 3.5 m ) Fg = 24 N = - 84.0 J

8. A rope is used to pull a metal box 15.0 m across the floor. The rope is held at an angle of 46.0° with the floor, and a force of 628N is used. How much work is done on the box? F = 628 N 46° W = F · d · cosθ Fx d = 15 m

8. A rope is used to pull a metal box 15.0 m across the floor. The rope is held at an angle of 46.0° with the floor, and a force of 628N is used. How much work is done on the box? F = 628 N 46° W = F · d · cosθ Fx= 628 N · 15 m · cos46° d = 15 m = 6544 J

Ch10.2 – Work Under a Varying Force Work = Area under the Force/Distance graph Ex1) A bow string is pulled back 0.5 meters. The force to pull it increases with distance from 0 N to 20 N as shown. How much work is done? 20 15 F (N) 10 5 0.1 0.2 0.3 0.4 0.5 d (m)

Ch10.2 – Work Under a Varying Force Work = Area under the Force – Distance graph Ex1) A bow string is pulled back 0.5 meters. The force to pull it increases with distance from 0 N to 20 N as shown. How much work is done? Work = Area = ½ b · h = ½ ( .5 m )( 20 N ) = 5 J 20 15 F (N) 10 5 0.1 0.2 0.3 0.4 0.5 d (m)

Ex2) How much work is done by this erratic Force? Work = Area → + + 20 15 F (N) 10 5 1 2 3 4 5 d (m)

Ex2) How much work is done by this erratic Force? Work = Area → + + = (½ b·h) + (l·w) + (½ b·h) = ½(2)(10) + (3)(10) + ½(2)(10) = 50 J 10 10 20 15 F (N) 10 5 1 2 3 4 5 d (m)

Ex3) To compress a large coil spring 10 cm requires a force that increases linearly from 10 N to 50 N. How much work is done on the spring? 50 40 30 F (N) 20 10 .02 .04 .06 .08 .10 d (m)

Ex3) To compress a large coil spring 10 cm requires a force that increases linearly from 10 N to 50 N. How much work is done on the spring? Work = Area = 1 + 2 = ½ b·h + l·w 40 = ½(.1)(40) + (.1)(10) = 3 J 10 50 40 30 1 F (N) 20 10 2 .02 .04 .06 .08 .10 d (m)

Power – the rate at which work is done Power = work time Units: Ex1) An electric motor lifts an elevator 9.0 m in 15.0 seconds by exerting a force of 12,000 N. What power does it produce? F d = 9 m Fg = 12,000 N

Power – the rate at which work is done Power = work time P = W or P = F·d or P = F · v t tt Units: J/s → Watt Ex1) An electric motor lifts an elevator 9.0 m in 15.0 seconds by exerting a force of 12,000 N. What power does it produce? F P = F·d = ( 12,000 N )( 9.0 m ) = 7,200 W t ( 15.0 sec ) d = 9 m Fg = 12,000 N

Ex2) Through a set of pulleys, a 10 kg mass is lifted .25 m in 0.5 seconds. How much power was required? F FG Ch10 HW#2 9 – 12

Ex2) Through a set of pulleys, a 10 kg mass is lifted .25 m in 0.5 seconds. How much power was required? F P = F·d = ( mg )·d = ( 100 N )( .25 m ) = 50 W t t ( .5 s ) FG Ch10 HW#2 9 – 12

Lab10.1 Power - due tomorrow - go over Ch10 HW#2 before lab

Ch10 HW#2 9 – 12 9. A 575N box is lifted 20.0m in 10 sec. What is the power required? A 645N rock climber climbs 8.2m in 30 min. a. How much work? b. Power?

Ch10 HW#2 9 – 12 9. A 575N box is lifted 20.0m in 10 sec. What is the power required? P = F·d = (575N)(20.0m) = t (10s) A 645N rock climber climbs 8.2m in 30 min. a. How much work? W = F·d = (645N)(8.2m) = b. Power? P = W = 5280 J = t 1800s

11. An electric motor develops 65 kw of power as it lifts a loaded elevator 17.5 m in 35 seconds. How much force does the motor exert? P = 65,000 WP = W d = 17. 5 m t t = 35 sec P = F·d F = ? t 12. Pushing a stalled car, it takes 210N to get it moving, and the force decreases at a constant rate until it reaches 40N by 15m. How much work is done during this interval?

11. An electric motor develops 65 kw of power as it lifts a loaded elevator 17.5 m in 35 seconds. How much force does the motor exert? P = 65,000 WP = W d = 17. 5 m t t = 35 sec P = F·d F = ? t F = 12. Pushing a stalled car, it takes 210N to get it moving, and the force decreases at a constant rate until it reaches 40N by 15m. How much work is done during this interval? Work = Area = 1 + 2 = ½ b·h + l·w = ½(15m)(170N) + (15m)(40N) = 200 150 1 F (N) 100 50 2 5 10 15 d (m)

13. In the tractor pull competition, the trailer is set up so that as the tractor pulls the trailer, the trailer shifts its mass forward, increasing the drag, and thus increasing the force required to pull. If ur not sure what I’m referring to, Youtube it! A graph of the Force required vs distance is shown: How much work is done by the tractor? 20000 15000 10000 5000 F (N) 10 20 30 40 50 d (m)

13. In the tractor pull competition, the trailer is set up so that as the tractor pulls the trailer, the trailer shifts its mass forward, increasing the drag, and thus increasing the force required to pull. If ur not sure what I’m referring to, Youtube it! A graph of the Force required vs distance is shown: How much work is done by the tractor? Work = Area = 1 + 2 = ½ b·h + l·w = ½(50m)(5000N)+(50m)(15000N) = 20000 15000 10000 5000 1 F (N) 2 10 20 30 40 50 d (m)

Ch10.3 – Machines - make work “feel” easier by changing forces, either magnitude or direction.

Ch10.3 – Machines - make work “feel” easier by changing forces, either magnitude or direction. - you apply an input force, Fin, the machine multiplies the force lifting the object, called the output force, Fout.

Ch10.3 – Machines - make work “feel” easier by changing forces, either magnitude or direction. - you apply an input force, Fin, the machine multiplies the force lifting the object, called the output force, Fout. - this multiplying of forces is at the expense of distance moved. The input distance, din, is greater than the output distance, dout.

Ch10.1 – Energy and Work