Name ANSWERS

Power, Work and Energy Workbook

Year 11 Science

2014

Ruawai College

Work

Work is done when a force moves an object. If the object does not move, no work is done.

Work requires movement against an opposing force. The opposing force is often friction or gravity.

Whenever work is done, energy is changed from one form to another. Energy and work are inter-connected.

Work is calculated from the force used to move an object, and the distance the object moves.

Work = force x distance W = F x d the unit of work is the Joule, J

1 000 J = 1 kJ

1. If you push a lawnmower and it moves, is work done? Yes

2. If you push a lawnmower and it does not move, is work done? No

3. What is the definition of work?

Work is done when a force moves an object.

4. What is the word equation to calculate how much work is done?

Work = Force x distance

5. What is the symbol equation to calculate how much work is done?

W = F x d

6. What is the unit of work? Joules, J

Example:

Calculate the work done when a person lifts a box of mass 6 kg from the floor to a table that is 1.5 m high. The opposing force is the weight of the box = 6 x 10 = 60 N.

W = F x d = 60 x 1.5 = 90 J

7. What is the mass of the box?

6 kg

8. What is the name of opposing force that is needed to be overcome to lift the box?

Gravity

9. What is the amount of the weight force?

60 N

10. What is the height the box is lifted onto the table?

1.5 m

11. What is the work done when the box is lifted onto the table?

90 Joules

12. What is the abbreviation for work? W

13. What is the abbreviation for force? F

14. What is the abbreviation for distance? d

15. Complete the equations using the triangles above:

W = F x d

F = W ÷ d

d = W ÷ F

Exercises:

16. A car is pushed a distance of 3m, from stationary, by a net force of 2 000 N. Calculate the work done.

W = F x d = 2 000 x 3 = 6 000 J

17. Two girls drag a boat 20 m along the beach. They each pull with a force of 50 N. Calculate the total work done by the girls.

W = F x d = (2 x 50 ) x 20 = 2 000 J

18. Describe two factors that must be present for work to be done.

Force and distance

19. For each situation in the following table, say whether work is done or not, and give a reason. The first one has been done for you.

Situation Work done? Reasona. A girl climbs up a ladder. Yes Movement against the force of gravity.

b. A child pulls a toy along the floor.

Yes Movement along floor

c. A crane exerts a force of 10 000 N on a container but the container cannot move.

No Force applied but no movement

d. A rock falls off a cliff into the river.

Yes Movement downwards in response to force of gravity

20. Calculate the work done when a person uses a constant force of 25 N to drag a box for a distance of 15 m.

W = F x d = 25 x 15 = 375 J

21. Calculate the force exerted by a small motor that does 330 J of work to move a toy car 110 m.

F = W ÷ d = 330 ÷ 110 = 3 N

22. If the work done by a 2.4 N force pushing a duster the length of a table is 9.6 J, calculate the length of the table.

d = W ÷ F = 9.6 ÷ 2.4 = 4 m

23. A car is travelling at a steady speed of 20 m s -1 . ( v = d ÷ t )

a. Calculate the distance the car moves in 10 seconds.

d = v x t = 20 x 10 = 200

b. The total resisting force (air resistance and friction) is 900 N. Calculate the work done by the car in 10 s in overcoming this resisting force.

W = F x d = 900 x 200 = 180 000 J = 180 kJ

24. A motorbike of mass 150 kg is lifted 2 m onto a display stand. Calculate the work done to lift the motorbike onto the display stand.

W = F x d = 150 x 2 = 300 J

25. Sarah uses a spring exerciser. When she pulls on the exerciser with an average forced of 400 N, the springs stretch 0.5 m. How much work does Sarah do to the exerciser?

W = F x d = 400 x 0.5 = 200 J

Power

Power, P, is the rate at which work is done - the rate of transferring energy.

Power is calculated by dividing the work done, by the time taken to do the work.

The unit of work is the watt, W.

One Joule per second is one watt.

1 000 watts = 1 kilowatt

26. Complete the equations using the triangle above:

Work = Power x time W = P x t

Power = Work ÷ time P = W ÷ t

time = Work ÷ Power t = W ÷ P

27. What is the unit of power?

Watts, W

28. What is a kilowatt? 1 000 Watts

Example:

A crane lifts a bucket weighing 4 000 N to a height of 6 m in 10 seconds. Calculate the power of the crane.

First calculate the work done -- W = F x d = 4 000 x 6 = 24 000 J

Second, calculate power -- P = W ÷ t = 24 000 ÷ 10 = 2 400 Watts = 2.4 kW

29. What is the weight of the bucket?

4 000 N

30. What height is the bucket lifted?

6 m

31. How much work is done to lift the bucket?

24 000 J

32. How long did it take to lift the bucket?

10 seconds

33. What is the power of the crane?

2 400 W , 2.4 kW

34. How do you work out kilowatts if you know the watts?

1 000 watts = 1 kW , divide watts by 1 000 to get kilowatts

Exercises:

35. An electric motor has a power rating of 6 kW. Convert the power rating to watts.

6 kW = 6 000 W

36. A woman walks along the road. She does 19 800 J of work in 1 minute. Calculate the power the woman develops.

P = W ÷ t = 19 800 ÷ 60 = 330 W

37. A crane lifts a load of 4 000 N through a vertical height of 12 metres in 6 seconds. Calculate the crane’s power in watts and kilowatts.

W = F x d = 4 000 x 12 = 48 000 J

P = W ÷ t = 48 000 ÷ 6 = 8 000 W

38. The work done when a construction worker Ben pushes a wheelbarrow for 3 seconds is 60 J. The work done when Max pushes a wheelbarrow for 5 seconds is 125 J. Which worker has the greater power output? Explain your answer.

Ben - P = W ÷ t = 60 ÷ 3 = 20 W

Max - P = W ÷ t = 125 ÷ 5 = 25 W

Max has greater power output than Ben

39. Jo takes 5 seconds to run upstairs, a vertical height of 4 metres. Jo’s mass is 51 kg. Calculate Jo’s power output. (Hint: 3 calculations needed)

Jo’s weight = 51 x 10 = 510 N

Work = F x d = 510 x 4 = 2 040

Power = W ÷ t = 2 040 ÷ 5 = 408 W

40. A runner jogs for 1 km using a force of 210 N and a power output of 500 W. Calculate how long, in minutes, it takes the runner to complete the 1 km run.

Work = F x d = 210 x 1 000 = 210 000 J

t = W ÷ P = 210 000 ÷ 500 = 4 20 sec = 7 minutes

Energy

Energy is the ability to do work. The amount of work done tells us how much energy has been changed from one form to another. All forms of energy are measured in Joules, J.

41. What is the energy of movement?

Kinetic energy

42. What is the type of energy gained when an object is lifted?

Gravitational potential energy

43. What is the unit of energy?

Joules, J

44. What is the unit of work?

Joules, J

Active forms of energy are kinetic, light, heat and sound.

Potential forms of energy are chemical, elastic, electrical, nuclear and gravitational.

Energy cannot be created or destroyed; it just changes from one form to another.

45. What are some examples of elastic energy?

In spring, in stretched rubber

46. What are some examples of chemical potential energy?

In food, in batteries,

47. How would an object gain gravitational potential energy? By being lifted

Gravitational potential energy, EP , is the potential energy an object has because of its height.

The unit of EP is the Joule, J.

The higher an object is lifted, the greater is its gravitational potential energy.

EP = m x g x h

= 75 x 10 x 4

= 3 000 J

48. What is Ep the abbreviation for? gravitational potential energy

49. What is m the abbreviation for? mass

50. What is the mass of the rock in the example? 75 kg

51. What is g the abbreviation for? gravity

52. What is the value of the strength of gravity? 10

53. What is h the abbreviation for? height

54. What is the height of the rock in the example? 4 m

55. What is the formula to calculate EP? EP = m x g x h

56. What is the unit of EP? Joules

57. How much gravitational energy did the 75 kg rock gain when it was lifted 4 m? 3000 J

Work Done = Energy Gained

58. What was the mass of the box? 1 kg

59. What was the weight of the box? 10 N

60. How was the weight calculated? mass x 10

61. What was the height / distance box lifted? 2 m

62. What work was done by lifting the box? 20 J

63. What formula was used to find the work done? W = F x d

64. What was the gravitational potential energy gained by the box? 20 J

65. What formula was used to find the gravitational energy gained? EP = mgh , EP = W

When an object’s height decreases, the object loses gravitational potential energy.

The same formula Ep = mgh, is used to calculate the loss in gravitational potential energy.

Exercise:

66. A 5 kg object is lifted to a height of 2 m above the floor that it is resting on.

a. Calculate the gain in gravitational potential energy of the object.

EP = mgh = 5 x 10 x 2 = 100 J

b. How much gravitational potential energy does the object lose if it is dropped and falls back down to the floor?

100 J

67. Complete the equations using the triangle opposite:

EP = m g h

m = EP ÷ g h

h = EP ÷ m g

g is gravity = 10 m s-2

68. Robin and her mother go up in a lift.

a. Robin’s mass is 45 kg and she gains 5 400 J of gravitational potential energy as she goes up in the lift. Calculate the height that Robin gains.

h = EP ÷ m g = 5 400 ÷ 450 = 12 m

b. Robin’s mother gains 7 800 J of gravitational potential energy. Calculate the mass of Robin’s mother.

m = EP ÷ g h = 7 800 ÷ 120 = 65 kg

When a pendulum swings, the energy of the mass at the end of the pendulum (called a bob) alternates between kinetic energy and gravitational potential energy.

69. How does the pendulum gain gravitational potential energy?

Is raised above the ground

70. How does the pendulum gain kinetic energy?

Gravitational potential energy is transformed into kinetic energy

Water in dam has a large gravitational potential energy.

Gravitational potential energy changing to kinetic energy along penstock.

Moving water pushes on the turbines, making them spin to generate electrical energy.

gravitational kinetic electricpotential energy energyenergy

71. How does the water gain gravitational potential energy?

Is high up in the dam

72. How does the water gain kinetic energy?

Is moving down the penstock

73. What type of energy is the kinetic energy transformed into in the powerhouse?

Electrical

Kinetic Energy is the energy of moving objects.

The unit of kinetic energy is Joules, J.

The unit of mass is grams, g.

The unit of speed is metres per second.

74. Complete the equations using the triangle above:

EK = ½ m v2

½ m = EK ÷ v2 m = 2 x EK ÷ v2

v2 = EK ÷ 1/2 m v = √ EK ÷ 1/2m

Example: Calculate the kinetic energy of a car of mass 1 100 kg that is travelling at 30 ms-1.

Ek = ½ m v2 = ½ x 1 100 x 302 = 495 000 J (495 kJ)

75. What is the mass of the car? 1 100 kg

76. What is the speed of the car? 30 ms=1

77. What is the value of 302? 900

78. What is the formula to find the kinetic energy of the car? EK = ½ m v2

79. How are joules converted into kilojoules? ÷ 1000

Exercise:

80.

a. Calculate the kinetic energy of an empty truck of mass 1 000 kg moving at 5 ms-1.

EK = ½ m v2 = ½ x 1000 x 52 = 12 500 J

b. When loaded the truck’s mass is 2 000 kg. Calculate the kinetic energy of the truck now moving at the same speed of 5 ms-1.

EK = ½ m v2 = ½ x 2000 x 52 = 25 000 J

c. The loaded truck has twice the mass, so has twice the kinetic energy.

81. Two cars travelling down the motorway both have 300 kJ of kinetic energy.

a. One car is travelling at 25 ms-1. Calculate its mass.

m = 2 x EK ÷ v2 = 2 x 300 000 ÷ 252 = 960 g

b. The other car has mass of 1 000 kg. Calculate its speed.

v = √ EK ÷ 1/2 m = √ 300 000 ÷ 500 = √600 = 24.5 ms-1

82. A forklift moves a crate upwards at a constant speed of 0.28 ms-1. The mass of the crate is 115 kg. Calculate the kinetic energy of the crate.

EK = ½ m v2 = ½ x 115 x 0.282 = 4.5 J

83. Calculate the speed of a car that has 200 kJ of kinetic energy and a mass of 1 000 kg.

v = √ EK ÷ 1/2 m = √ 200 000 ÷ 500 = √500 = 22.4 ms-1

84. You are on a skateboard at the top of a ramp. Your gravitational potential energy is equal to 1 000 J and your mass is 60 kg.

a. Calculate the height of the ramp.

h = EP ÷ m g = 1 000 ÷ 600 = 1.67 m

b. When you skate down the ramp, your potential energy is converted to kinetic energy. At the bottom of the ramp, your kinetic energy will be equal to your potential energy at the top (ignoring energy lost to heat). Calculate your speed at the bottom of the ramp.

v = √ EK ÷ 1/2 m = √ 1 000 ÷ 30 = √33.33 = 5.77 ms-1

85. Extension:

Think back over the calculations in 84 b and devise an experiment to model this in the lab with equipment that we have available.