The relationship between force and distance has a name. WORK

Transcription

1 EWP

2 The relationship between force and distance has a name. WORK

3 WORK - the product of the net force and distance through which an object moves in the direction of the net force. W=F d where W=work joules F=force kg * m/s 2 ( Newtons) d=distance m (through which the force moves) The SI unit for work is the joule (J) aka a Newton-Meter The American Unit is the Ft-lb

4 Ex 1: You discover a box of rocks in middle of your room. They have a mass of 20 kg. You apply an effort force of 65N to push them 2 meters so they are against the wall. How much work have you done? W = Fd W = 65N x 2m W = 130 J 20 kg

5 You now have a box of rocks against the wall of your room. You decide to lift them up to a height of 1.5 m. Now how much work have you done? What is your total work since discovering this box of rocks? W = Fd F in this case is equal and opposite the F w F w = mg W = (20 kg)(9.8m/s 2 ) x 1.5m W = 294 J WT = 130J + 294J =424J 20 kg

6 Ex 2 Artie pulls a 500-kg boulder a distance of 4m at a constant velocity along a horizontal surface by applying a 250-N force along a rope that is parallel with the horizontal. He then stoops down and lifts the boulder above his head to a height of 2 m. Artie carries the boulder at a constant velocity for a distance of 5 m. How much work does he do to the boulder? BONUS: Where does Artie get his strength from? Answer: J

7 One of the oddest characters in The Adventures of Pete & Pete is Artie, the Strongest Man in the World. In trying to figure out the mystery of Artie, the Strongest Man in the World, we have put together this list of known facts: Artie, the Strongest Man in the World, is the strongest man in the world. He is used in several episodes for his super-human strength. [Various episodes] Artie, the Strongest Man in the World, derives his super powers from the red and blue underwear that is his costume, a 40/60 blend of titanium and cotton. [One of the early shorts]

8 Ex 2 Part I: Drags W = Fd W = (250N)(4m) = 1000J Part II: Lifts W = F w d W = (500kg)(9.8 m/s 2 )(2m) = 9800J Part III: Carrries No work

9 CHALLENGE Example 3 A reindeer pulls a sled a distance of 3.0 m across a frictionless surface. He applies a force of 50.0 N on the rope to the horizontal. How much work was done on the sled? 50 N W = FD Work = (50N)(3m) = 150 J

10 What happens if force is applied at an angle? We have to consider the Horizontal Component of the applied force. How do we calculate this? F F vert θ F hor

11 We have to consider the Horizontal Component of the applied force. How do we calculate this? 50N F vert 30 F hor Fcosθ

12 Example 3 cont. A reindeer pulls a sled a distance of 3.0 m across a frictionless surface. He applies a force of 50.0 N on the rope at an angle of 30.0 with the horizontal. Calculate the work done on the sled. 50N 30 Fcosθ

13 Example 3 A reindeer pulls a sled a distance of 3.0 m across a frictionless surface. He applies a force of 50.0 N on the rope at an angle of 30.0 with the horizontal. Calculate the work done on the sled. 50N 30 W=Fdcos. 130J Fcosθ

14 Example 3 The work on the sled decreased by applying the force at an angle. The Force of effort was the same. 50N 30 Fcosθ

15 Who is doing the most work? A. Mary holds a 40 N bag of groceries for 15 minutes B. Veronica slides a bag a 40 N bag of groceries from the edge of a counter and then carries it 10 meters across the room C. Phillip lifts a 40 N bag from the floor to a shelf 2 meters high

16 Who is doing the most work? A. Briton holds a 40 N bag of groceries for 15 minutes B. Teagan slides a bag a 40 N bag of groceries from the edge of a counter and then carries it 10 meters across the room C. Anna lifts a 40 N bag from the floor to a shelf 2 meters high

17 KEY IDEA Work Force & Direction Phillip is the only one doing work Why? When calculating work, the applied force of effort is parallel and in the direction of motion.

18 What is the difference? Ricky pushes a notebook 2 m across a table with a force of 10 N in 4 seconds. Christian pushes his notebook 5 meters across the table with a with a force of 4 N in 10 seconds.

19 Power The rate at which work is done P = W/t Power is measured in Watts. (W)

20 W t P power (Watts) P W work (Joules) t time (seconds) Fd F Force ( Newtons) P t d distance (m)

21 Ex 4 One watt is one joule of work performed in one second. This is very small: calculate the power when lifting a 5N plate of apples 0.25m in 1 second. P = W/t P = (5N)(0.25m)/1sec 1.25 Watt Since it is so small a unit, power is often described in kilowatts.

22 How is this related to light bulbs? In terms of electromagnetism, one watt is the rate at which work is done when one ampere (A) of current flows through an electrical potential difference of one volt (V).

23 Ex 5 Who has a higher power rating? Stephanie pushes a 20 kg box with a net force of 60N. She moves it 30 m across the floor in 5 seconds. Dustyn pushes a 20 kg box with a net force of 72N. He moves it 25 m across the floor in 4 seconds.

24 Who has a higher power rating? Stephanie pushes a 20 kg box with a net force of 60N. She moves it 30 m across the floor in 5 seconds. 360 W Dustyn pushes a 20 kg box with a net force of 72N. He moves it 25 m across the floor in 4 seconds. 450 W

25 What is horse power? HP is a term used in the US for power. 1 HP = 746 watts US unit for work = foot-pound US unit for power = foot-pound/s 1 hp = 550 foot-pound/s

26 Ex 6: A winch lifts a maximum load of 600 kg at a constant velocity for a distance of 4 meters during a time period of 2 seconds. What power in watts did this require? What would be the horsepower? What power in kilowatts did this require? P = W/t = Fd/t P = (600kg)(9.8m/s 2 )(4m) /2sec P = Watts P = (11760 Watts) )(1 hp / 746 Watts) = hp P = (11760 Watts) (1 kw /1000W) = kw

27 energy How many examples can you think of

28 What is energy? The capacity to do work

29 Define mechanical energy

30 Mechanical Energy can be classified as Potential Energy or Kinetic Energy

31 How would you define Potential Energy?

32 Potential Energy: stored energy due to an object s position or condition in a field of force.

33 What factors do you think effect E P? What would be the units?

34 Gravitational Potential Energy dependent on mass, height, & gravitational acceleration E p = mgh m is mass (kg) g is gravity (9.8m/s 2 ) h is the height (m) of the object E p is Joules (J) one J= kg*m 2 /s 2 SI Unit for Energy is the Joule, also referred to as the Newton-Meter

35 How would you define kinetic energy? Energy of motion Types? Besides just moving: thermal, sound

36 Kinetic energy energy of motion. This is dependent upon mass of moving object and the square of it s velocity. E k =.5mv 2

37 Kinetic energy energy of motion. This is dependent upon mass of moving object and the square of it s velocity. E k =.5mv 2 m is mass (kg) v is velocity (m/s) E k is Joules (J) one J= kg*m 2 /s 2

38 Example 7: A panda lifts a barbell with a mass of 85 kg above his head. The barbell is 243 cm above the ground. How much gravitational potential energy does the barbell have with respect to the ground? (85 kg)(9.8 m/s 2 )(2.43 m) = 2020 J

39 Ex 8 An object with a mass of 2.0 kg is moving horizontally with a velocity of 0.5 m/sec. Calculate the kinetic energy possessed by the object. (.5)(2.0kg)(0.5 m/s) 2 = 0.25 J

40 Example 9: A 1400-kg car has a net forward force of 4500 N applied to it. The car starts from rest and travels down a horizontal highway. What are its kinetic energy and speed after it has traveled 124 m? What would you solve for first? Acceleration then it is a dvvat problem! a= 4500 N/1400 kg = 3.21 m/s 2 v f = v i + 2ad = 28.2 m/s KE = 557,000 J

41 Example 10: A kitchen pot has a weight of 180 N. (about 40 pounds!!!) a. How much work is required to lift the pot to a height of 35 cm above the ground with nothing but the hair on your face?? b. What was the increase in GPE of the kitchen pot at this height? 63 J

42 If the pot fell from a height of 35 cm, what would be its final velocity? How would you determine this? v f2 = v i2 + 2ad v f = v i + 2ad v f = 2.62 m/s What would be its kinetic energy? We have the velocity, now determine mass 18.4 kg Solving for KE= 63J!!!

43 Work Energy Theorem W = ΔE Work is the transfer of energy by means of forces Work equals the change in energy in a system

44 Ex 11: How would you solve this? A man pushes a kg cart horizontally across the floor starting at m/s and accelerating to 3.70 m/s. How much work is done on the cart? Find the change in KE W = ΔKE ΔKE =.5 m (v f2 v i2 ) or determine them separately and find difference KE f Ke i Ans: 79.8 J

45 How do you determine force when calculating work done dragging an item at an angle? You must determine F HOR (F T cosθ)

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47 Pendulum(swing) PE PE KE

48 The Law of Conservation of Mechanical Energy M E = E P + E K Mechanical Energy is equal to the sum of potential and kinetic energy Energy cannot be created nor destroyed, but is only changed from one form to another The amount of energy in the universe remains constant.

49 EX 12 A 60-kg skier is at the top of a ski slope. At this highest point the skier is 10 m vertically above the chalet. What is the skier s gravitational potential energy at the peak? (60 kg)(9.8 m/s 2 )(10 m) = 5880 J What is the skier s gravitational potential energy at the chalet? 0 J What is the skier s gravitational potential energy at a point directly between the peak and the chalet? (60 kg)(9.8m/s 2 )(5 m) = 2940 J

50 A 60-kg skier is at the top of a ski slope. At this highest point the skier is 10 m vertically above the chalet. What is the skier s kinetic energy at this very point? (directly between the peak and the chalet) Solving using kinematics: v f = v i + 2ad = 9.90 m/s KE = (.5)(60 kg)(9.90 m/s) J

51 A 60-kg skier is at the top of a ski slope. At this highest point the skier is 10 m vertically above the chalet. You can solve this problem very easily using your knowledge of energy:

52 Alpha = peak PE = 5880 J Now you can use KE to solve for velocities at any point! PE = KE = 2940 J PE = 0 J gamma = Chalet What is the KE at the base of the hill?

53 Example 13 A rocket with a mass of 7.8 kg is launched vertically with a kinetic energy of 1420 J. How high does it rise? KE i = PE f PE f = mgh f PE f /mg = h f h f =1420 J /[(7.8 kg)(9.8m/s 2 )] h f = 18.6 m

54 The Law of Conservation of Mechanical Energy (or total energy) E M = E P + E K This can be expanded

55 EXTREME Law of Conservation of Mechanical Energy

56 The Law of Conservation of Mechanical Energy This can be expanded E M = E P + E K or E Mi = E Mf or E Pi + E Ki = E Pf + E Kf or mgh i +.5mv 2 i = mgh f +.5mv 2 f

57 Ex 14 A rocket is launched vertically with a velocity of 29 m/s. How high does it rise? PE i + KE i = PE f + KE f mgh i +.5mv i2 = mgh f +.5mv 2 f.5v i2 = gh f.5 v i2 / g = h f.5(29 m/s) 2 / 9.80 m/s 2 = h f 42.9 m = h f

58 Example 15: A dog is riding a sled. She and the sled weigh 844 N. They move down a frictionless hill through a vertical distance of 12.8 m. Use the conservation of mechanical energy to find the speed of the dog and sled at the bottom of the hill, assuming they are pushed off with an speed of 5.29 m/s. Dog on sled

59 PE i + KE i = PE f + KE f mgh i +.5mv i2 = mgh f +.5mv 2 f gh i +.5v i2 =.5v 2 f (gh i +.5v i2 )/.5 = v 2 f [(9.80)(12.8 m) + [.5(5.29 m/s) 2 ]/.5 = v f m/s = v f

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