Work. Work is the measure of energy transferred. Energy: the capacity to do work. W = F X d

Transcription

1 ENERGY CHAPTER 11

2 Work Work is the measure of energy transferred. Energy: the capacity to do work. W = F X d Units = Joules Work and energy transferred are equivalent in ideal systems.

3 Two Types of Energy Energy WRT position is called: Potential Energy or ( P.E. ). When an object is in motion, the mass has: Kinetic Energy or ( K.E. ).

4 Potential Energy and Base Levels A change in P.E. is always measured WRT some arbitrary position where the P.E. is defined as 0 J. An object in a high position can do work if it falls to a lower position.

5 The object has stored energy. Gravitational P.E. PE = W = Fd PE = PE F -PE I PE F -PE I = Fd Since the object is falling, it is changing heights. d = h PE F -PE I = F h

6 PE F -PE I = mg h = mg( h f -h i ) W = PE F -PE I = mg( h f -h i ) If PE I is assumed to be 0 J at h i, then: PE = mgh

7 (IC#1) A 10 kg block is lifted at a constant velocity from the floor to a shelf that is 3.0 meters off the ground by a forklift. A. What is the change in PE?

8 PE F 3 m PE 10 kg PE I PE F -PE I = mg( h f -h i ) PE = 10 kg ( 9.8 m ) 3 m PE = PE F = mgh f s 2 = 294 J

9 B. How much energy would be released if the block fell off the shelf? 10 kg PE I 3 m PE F PE = mg( h f -h i )

10 PE = mg( h f -h i ) PE = 10 kg ( 9.8 m ) - 3 m s 2 PE = -294 J Means energy was given off. 294 J of work could be done.

11 Kinetic Energy K.E. is the energy an object possesses by virtue of its motion.

12 F f F A ΣF X = F net = F A + F f If µ = 0, then F f = 0 N and F net = F A F net cause accelerations. F net = ma

13 If F net acts over a given distance: F net d = mad V f2 = V i2 + 2Ad Ad = V f2 -V2 i 2 F net d = m (V f2 -V i2 ) 2 W = F net d = 1mV 2-1mV f i 2 2 2

14 KE at any given time: KE = 1mV 2 2 KE depends on two things: Mass in kg Velocity in m/s W = F net d = KE f -KE i = KE

15 (IC#2) A 10 kg block is sliding at 50 m/s. A. What is the KE of the block? KE = 1mV 2 2 KE = 10 kg ( 50 m/s ) 2 2 KE = J

16 B. If the block slows to 30 m/s, what is the change in KE? KE = KE f -KE i KE = 1mV f 2-1mV i = 10 kg (30 m/s) 2-10 kg (50 m/s) Lost KE KE = J

17 C. If the block was slowed over a 35 meter distance, what was the force of friction? KE = F net d J = F net (35 m) F net = N Opposes motion

18 F f F A F X = F net = F A + F f F net = - F f N = - F f F f = N

19 D. How much work could be done by allowing the block to slow to a stop? W = KE = KE f -KE i = 10 kg (0 m/s) 2-10 kg (50 m/s) KE = J Energy lost translates to work that can be done.

20 (IC#2**) A 1400 kg car, starting from rest, has a net force of 4500 N applied to it. The car travels 100 meters. A. How much work was done on the car? W = F net d W = 4500 N ( 100 m) W = 4.5 X 10 5 J

21 B. What is the KE of the car at the 100 meter mark? W = KE = KE f -KE i W = KE 100m = 4.5 X 10 5 J

22 C. What is the velocity of the car at the 100 meter mark? KE = 1mV 2 2 V = 2 KE m V f = 2 (4.5 X 10 5 J ) 1400 kg V f = 25.4 m/s

23 Conservation of Energy Energy of an isolated system can t change unless work is done on it by an outside force... or the system does work on something outside the system. Energy can t be created or destroyed.

24 PE i + KE i = PE f + KE f mgh i + (1/2)mV i2 =mgh f + (1/2)mV f 2 As an object rises or falls, a tradeoff between PE and KE occurs. This energy tradeoff can be utilized to do work.

25 Energy Energy conserved. leaves system.

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27 (IC#3) A 100 kg diver jumps off a 30 meter platform during the Olympic games. A. What is the PE and KE at this point?

28 PE = mgh PE i = 100 kg ( 9.8 m ) 30 m s 2 PE i = J KE i = 0 J Diver not moving.

29 B. The diver falls freely for 10 meters. What is the PE and KE at this point? PE i + KE i = PE f + KE f

30 PE f PE f = mgh f = 100 kg ( 9.8 m ) 20 m s 2 PE f = J J = J + KE f KE f = 9800 J

31 Note: How fast was the diver falling at this point? KE f = 9800 J KE = 1mV 2 2 V f = 14 m/s V f = 2 ( 9800 J ) 100 kg

32 C. The diver falls freely for another 5.0 m. What is the PE and KE at this point? PE i + KE i = PE f + KE f

33 PE f PE f = mgh f = 100 kg ( 9.8 m ) 15 m s 2 PE f = J J = J + KE f KE f = J

34 D. What is the PE and KE when the diver hits the water? PE i + KE i = PE f + KE f PE f = mgh f = 0 J

35 PE i = KE f J = KE f E. How fast is the diver moving as he hits the water? V f = 2 ( J ) 100 kg V f = 24.2 m/s

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37 F PE i + KE i = PE f + KE f (IC#4) A 0.5 kg baseball is thrown straight up. A. If 500 J of work was done on the ball, how high will the ball go? KE i = PE f I

38 What provided the KE I? The 500 J of work that was done on the ball. 500 J = PE f 500 J = mgh f 500 J =.5 kg ( 9.8 m ) h f s 2 h f = 102 m

39 B. What was the velocity of the ball as it left the throwers hand? 500 J = KE KE = 1mV2 i J = (.5 kg )V i 2 2 V i = 44.7 m/s

40 Non Conservative and Conservative Forces If an object changes KE, it is changing velocity. Therefore, it is accelerating. The object had to have a F net exerted on it over a distance. F net d = KE

41 Looking closer at F net : F f F A F X = F net = F A + F f

42 In the case of an airplane, F net d can also cause a PE. F net d = KE + PE ( F A -F f ) d = KE + PE F A d - F f d = KE + PE W = F d W = W f + KE + PE

43 (IC#6) An airplane has a mass of kg and is moving at 60 m/s. The pilot revs the engine so that the forward thrust becomes N. If the force of air resistance on the plane is N, what is the speed of the plane after it has traveled 500 meters?

44 W = W f + KE + PE F A d = F f d +.5 m ( V f2 -V i2 ) + No change in elevation. mg( h f - h i ) 7.5 x 10 4 N ( 500 m ) = 4 x 10 4 N ( 500 m ) +.5 ( 1.5 x 10 4 kg ) [ V f2 -( 60 m/s ) 2 ] V f = 77 m/s

45 (IC#8) A N car is traveling at 100 m/s due east. Suddenly, the car slams into a N truck traveling at 50 m/s due west. A. What is the KE of the car before the collision? KE = 1mV 2 2

46 KE car = 1 ( N )( 100 m ) m s 2 s KE car = 1.02 x 10 7 J

47 B. What is the KE of the truck before the collision? KE truck = 1 ( N )( - 50 m ) m s 2 s Opposite direction KE truck = 2.55 x 10 7 J

48 C. What is the total KE before the collision? KE car = 1.02 x 10 7 J + KE truck = 2.55 x 10 7 J KE total = 3.57 x 10 7 J

49 D. If the collision is inelastic, what is the velocity of the car after the collision? P car + P truck = P car&truck 2041 kg (100 m/s) kg (-50m/s) = kg ( V car&truck ) V car&truck = m/s

50 E. What is the KE after the collision? KE total = 1 ( kg )( m ) 2 2 s KE total = x 10 7 J

51 F. What is the change in KE? Why is there a change? KE = x 10 7 J x 10 7 J KE = x 10 7 J Most energy is converted to heat and a small amount in the form of sound.

52 (IC#10) When a long barreled rifle is fired, the force of the expanding gases acts on the bullet through a longer distance. What effect does this have on the velocity of the emerging bullet? 1. More work is done on the bullet because the force acts over a longer distance.

53 2. Since more work is done on the bullet, there is a greater change in KE. 3. Since the mass of the bullet is constant, the change in KE is caused by an increased velocity.

54 The End