Work and Energy

1
Work and Energy

• An Introduction

2
Work

• Work tells us how much a force or combination of
  forces changes the energy of a system.
• Work is the bridge between force (a vector) and
  energy (a scalar).
• W F Dr cos ?
• F force (N)
• Dr displacement (m)
• ? angle between force and displacement

3
Units of Work

• SI System Joule (N m)
• 1 Joule of work is done when 1 N acts on a body
  moving it a distance of 1 meter
• British System foot-pound
• (not used in Phys B)
• cgs System erg (dyne-cm)
• (not used in Phys B)
• Atomic Level electron-Volt (eV)

4
Force and direction of motion both matter in
defining work!

• There is no work done by a force if it causes no
  displacement.
• Forces can do positive, negative, or zero work.
  When an box is pushed on a flat floor, for
  example
• The normal force and gravity do no work, since
  they are perpendicular to the direction of
  motion.
• The person pushing the box does positive work,
  since she is pushing in the direction of motion.
• Friction does negative work, since it points
  opposite the direction of motion.

5
Conceptual Checkpoint

• Question If a man holds a 50 kg box at arms
  length for 2 hours as he stands still, how much
  work does he do on the box?
• Answer NONE AT ALL

6
Conceptual Checkpoint

• Question If a man holds a 50 kg box at arms
  length for 2 hours as he walks 1 km forward, how
  much work does he do on the box?
• Answer NONE AT ALL

7
Conceptual Checkpoint

• Question If a man lifts a 50 kg box 2.0 meters,
  how much work does he do on the box?
• Answer
• W FDr cos q
• (50 kg)(10 m/s2)(2.0 m)(cos 0o)
• 1,000 kg m2/s2
• 1,000 J

8
Work and Energy

• Work changes mechanical energy!
• If an applied force does positive work on a
  system, it increases mechanical energy.
• If an applied force does negative work, it
  decreases mechanical energy.
• The two forms of mechanical energy are called
  potential and kinetic energy.

9
Sample problem

• Jane uses a vine wrapped around a pulley to lift
  a 70-kg Tarzan to a tree house 9.0 meters above
  the ground.
• How much work does Jane do when she lifts Tarzan?
• How much work does gravity do when Jane lifts
  Tarzan?

10
Solution

• WJ F Dr cosq mg Dr cos0o
• WJ (70 kg)(9.8 m/s2)(9.0 m)(1)
• WJ 6174 kg m2/s2 6174 J
• (Jane does positive work because the force
  she exerts is upward when it gets to Tarzan, and
  his displacement is also upward.)
• WG -6174 J
• (Gravity pulls down but Tarzan goes up,
  so the work done by gravity is negative.)

11
Sample problem

• Joe pushes a 10-kg box and slides it across the
  floor at constant velocity of 3.0 m/s. The
  coefficient of kinetic friction between the box
  and floor is 0.50.
• How much work does Joe do if he pushes the box
  for 15 meters?
• How much work does friction do as Joe pushes the
  box?

12
Solution

• WJ F Dr cos0o
• ?kNDr ?kmgDr
• (The force Joe exerts must be equal in magnitude
  to the friction for there to be constant
  velocity.)
• WJ (0.5)(10 kg)(9.8 m/s2)(15.0 m)
• WJ 735 J
• Wf -735 J
• (The work done by friction is opposite to
  the work done by Joe.)

13
Sample problem

• A father pulls his child in a little red wagon
  with constant speed. If the father pulls with a
  force of 16 N for 10.0 m, and the handle of the
  wagon is inclined at an angle of 60o above the
  horizontal, how much work does the father do on
  the wagon?

14
Solution

• WJ FDr cos ?
• (16.0 N)(10.0 m) cos 60o
• 80 J

15
Kinetic Energy

• Energy due to motion
• K ½ m v2
• K Kinetic Energy
• m mass in kg
• v speed in m/s
• Unit Joules

16
Sample problem

• A 10.0 g bullet has a speed of 1.2 km/s.
• What is the kinetic energy of the bullet?
• What is the bullets kinetic energy if the speed
  is halved?
• What is the bullets kinetic energy if the speed
  is doubled?

17
Solution

• K ½ m v2
• ½ (0.010 kg) (1,200 m/s)2
• 7200 J
• K ½ m ( ½ v) 2 ¼ Ka
• ¼ (7200 J) 1800 J
• K ½ m ( 2 v) 2 4 Ka
• 4 (7200 J) 28800 J

18
The Work-Energy Theorem

• The net work due to all forces equals the change
  in the kinetic energy of a system.
• Wnet DK
• Wnet work due to all forces acting on an object
• DK change in kinetic energy (Kf Ki)

19
Sample problem (like 7.16 7.17)

• An 80-g acorn falls from a tree and lands on the
  ground 10.0 m below with a speed of 11.0 m/s.
• What would the speed of the acorn have been if
  there had been no air resistance?
• Did air resistance do positive, negative or zero
  work on the acorn? Why?
• How much work was done by air resistance?
• What was the average force of air resistance?

20
Solution

• v2 vo2 2g?y
• v -(2)(9.8 m/s2)(-10.0 m) 1/2
• v 14 m/s
• Air resistance did negative work on the acorn,
  since it was in the direction opposite the
  displacement.
• Wnet ?K
• WG WD ½ m v2
• WD ½ m v2 WG ½ m v2 - FG?r
• WD ½ m v2 - mg ?r
• WD ½ (0.080kg) (11 m/s)2 -
  (0.080kg)(9.8m/s2)(10.0m)
• WD 4.84 J - 7.84 J 3.00 J
• WD F ?r cos q -F ?r
• F -WD /?r 3.00 J /10.0 m 0.30 N

21
Constant force and work
F(x)

• The force shown is a constant force.
• W FDr can be used to calculate the work done by
  this force when it moves an object from xa to xb.
• The area under the curve from xa to xb can also
  be used to calculate the work done by the force
  when it moves an object from xa to xb

x
xa
xb
22
Variable force and work
F(x)

• The force shown is a variable force.
• W FDr CANNOT be used to calculate the work done
  by this force!
• The area under the curve from xa to xb can STILL
  be used to calculate the work done by the force
  when it moves an object from xa to xb

x
23
Springs

• When a spring is stretched or compressed from its
  equilibrium position, it does negative work,
  since the spring pulls opposite the direction of
  motion.
• Ws - ½ k x2
• Ws work done by spring (J)
• k force constant of spring (N/m)
• x displacement from equilibrium (m)
• The force doing the stretching does positive work
  equal to the magnitude of the work done by the
  spring.
• Wapp - Ws ½ k x2

24
Springs stretching
0
Ws negative area - ½ kx2
Fs -kx (Hookes Law)
25
Sample problem (7.23)

• A spring with force constant 2.5 x 104 N/m is
  initially at its equilibrium length.
• How much work must you do to stretch the spring
  0.050 m?
• How much work must you do to compress it 0.050 m?

26
Solution

• Wapp ½ k x2
• Wapp ½ (2.5 x 104 N/m) (0.050 m)2
• Wapp 31.25 J
• The same amount of work is needed to compress the
  spring as to stretch it.

27
Sample problem (7.29)

• It takes 130 J of work to compress a certain
  spring 0.10 m.
• What is the force constant of the spring?
• To compress the spring an additional 0.10 m, does
  it take 130 J, more than 130 J, or less than 130
  J? Verify your answer with a calculation.

28
Solution

• Wapp ½ k x2
• 130 J ½ k (0.10 m)2
• k 2 (130 J) / (0.10 m)2
• k 26,000 N/m
• To go from 0.10 m to 0.20 m, you need more than
  130 J, since you need more and more force to
  compress the spring the more you compress it.
• Wapp ½ k xf2 130 J
• Wapp ½ (26,000 N/m)(0.20m)2 130 J
• Wapp 520 J 130 J 390 J

29
Sample Problem

• How much work is done by the force shown when it
  acts on an object and pushes it from x 0.25 m
  to x 0.75 m?

30
Solution

• W (0.40 N)(0.25 m) (0.80 N)(0.25 m)
• W 0.30 J

31
Sample Problem

• How much work is done by the force shown when it
  acts on an object and pushes it from x 2.0 m to
  x 4.0 m?

32
Solution

• W ½ (base)(height)
• W ½ (2 m)(0.60 N) 0.60 J

33
Power

• Power is the rate of which work is done.
• P W/Dt
• W work in Joules
• Dt elapsed time in seconds
• When we run upstairs, t is small so P is big.
• When we walk upstairs, t is large so P is small.

34
Unit of Power

• SI unit for Power is the Watt.
• 1 Watt 1 Joule/s
• Named after the Scottish engineer James Watt
  (1776-1819) who perfected the steam engine.
• British system
• horsepower
• 1 hp 746 W

35
How We Buy Energy

• The kilowatt-hour is a commonly used unit by the
  electrical power company.
• Power companies charge you by the kilowatt-hour
  (kWh), but this not power, it is really energy
  consumed.
• 1 kW 1000 W
• 1 h 3600 s
• 1 kWh 1000J/s 3600s 3.6 x 106J

36
Sample problem (7.34)

• A record was set for stair climbing when a man
  ran up the 1600 steps of the Empire State
  Building in 10 minutes and 59 seconds. If the
  height gain of each step was 0.20 m, and the
  mans mass was 70.0 kg, what was his average
  power output during the climb? Give your answer
  in both watts and horsepower.

37
Solution

• P W/t
• F ?r/t
• m g ?r/t
• (70.0 kg)(9.8 m/s2)(0.20 m)(1600) / (659
  s)
• 333 W
• and in horsepower
• 333 W / 746
• 0.45 hp

38
Sample problem (7.36)

• Calculate the power output of a 1.0 g fly as it
  walks straight up a window pane at 2.5 cm/s.

39
Solution

• P W / t
• Choose 1 second as the time frame.
• At a velocity of 2.5 cm/s, the fly will go
  0.025 m in 1 second.
• P F?r/t
• mg ?r/t
• (0.001 kg)(9.8 m/s2)(0.025) / 1 s
• 0.000245 W

40
Force types

• Forces acting on a system can be divided into two
  types according to how they affect potential
  energy.
• Conservative forces can be related to potential
  energy changes.
• Non-conservative forces cannot be related to
  potential energy changes.
• So, how exactly do we distinguish between these
  two types of forces?

41
Conservative forces

• Work is path independent.
• Work can be calculated from the starting and
  ending points only.
• The actual path is ignored in calculations.
• Work along a closed path is zero.
• If the starting and ending points are the same,
  no work is done by the force.
• Work changes potential energy.
• Examples
• Gravity
• Spring force
• Conservation of mechanical energy holds!

42
Non-conservative forces

• Work is path dependent.
• Knowing the starting and ending points is not
  sufficient to calculate the work.
• Work along a closed path is NOT zero.
• Work changes mechanical energy.
• Examples
• Friction
• Drag (air resistance)
• Conservation of mechanical energy does not hold!

43
Potential energy

• Energy of position or configuration
• Stored energy
• For gravity Ug mgh
• m mass
• g acceleration due to gravity
• h height above the zero point
• For springs Us ½ k x2
• k spring force constant
• x displacement from equilibrium position

44
Conservative forces and Potential energy

• Wc -?U
• If a conservative force does positive work on a
  system, potential energy is lost.
• If a conservative force does negative work,
  potential energy is gained.
• For gravity
• Wg -?Ug -(mghf mghi)
• For springs
• Ws -?Us -(½ k xf2 ½ k xi2)

45
More on paths and conservative forces.

• Q Assume a conservative force moves an object
  along the various paths. Which two works are
  equal?
• A W2 W3
• (path independence)
• Q Which two works, when added together, give a
  sum of zero?
• A W1 W2 0
• or
• W1 W3 0
• (work along a closed path is zero)

46
Sample problem

• A box is moved in the closed path shown.
• How much work is done by gravity when the box is
  moved along the path A-gtB-gtC?
• How much work is done by gravity when the box is
  moved along the path A-gtB-gtC-gtD-gtA?

47
Solution

• WG 0 FDr
• WG 0 mgh -mgh
• WG 0 -mgh 0 mgh
• 0
• The work in b) is zero because work along a
  closed path is zero for any conservative force.

48
Sample problem

• A box is moved in the closed path shown.
• How much work would be done by friction if the
  box were moved along the path A-gtB-gtC?
• How much work is done by friction when the box is
  moved along the path A-gtB-gtC-gtD-gtA?

49
Solution

• Wf -mkmgd - mkmgd
• Wf -2mkmgd
• Wf -mkmgd - mkmgd - mkmgd - mkmgd
• -4 mkmgd
• Because friction is a nonconservative force, work
  along the closed path in b) is not zero.

50
Sample problem (8.6)

• As an Acapulco cliff diver drops to the water
  from a height of 40.0 m, his gravitational
  potential energy decreases by 25,000 J. How much
  does the diver weigh?

51
Solution

• DUG mghf - mghi
• DUG mg(hf - hi)
• mg DUG / (hf - hi)
• -25,000 J / 40.0 m
• 625 N

52
Sample problem (like 8.9)

• If 60.0 J of work are required to stretch a
  spring from a 2.00 cm elongation to a 5.00 cm
  elongation, how much work is needed to stretch it
  from a 5.00 cm elongation to a 8.00 cm elongation?

53
Solution

• We need to find k for the spring first.
• DUS ½ kxf2 - ½ kxi2
• DUS ½ k(xf2 - xi2)
• k 2DUS / (xf2 - xi2)
• 2(60) / (0.052 0.0222) 57,000 N/m
• Now that we know k, we can calculate DUS.
• DUS ½ k(xf2 - xi2)
• ½ (57,000)(.082 .052)
• 111 J

54
Law of Conservation of Energy

• In any isolated system, the total energy remains
  constant.
• Energy can neither be created nor destroyed, but
  can only be transformed from one type of energy
  to another.

55
Law of Conservation of Mechanical Energy

• E K U Constant
• K Kinetic Energy (1/2 mv2)
• U Potential Energy (gravity or spring)
• ?E ?U ?K 0
• ?K Change in kinetic energy
• ?U Change in gravitational or spring potential
  energy

56
Roller Coaster Simulation
Roller Coaster Physics Simulation (demonstrates
gravitational potential energy and kinetic energy)
57
Pendulums and Energy Conservation

• Energy goes back and forth between K and U.
• At highest point, all energy is U.
• As it drops, U goes to K.
• At the bottom , energy is all K.

58
Pendulum Energy
½mvmax2 mgh For minimum and maximum points of
swing
K1 U1 K2 U2 For any points 1 and 2.
59
Springs and Energy Conservation

• Transforms energy back and forth between K and U.
• When fully stretched or extended, all energy is
  U.
• When passing through equilibrium, all its energy
  is K.
• At other points in its cycle, the energy is a
  mixture of U and K.

60
Spring Energy
0
K1 U1 K2 U2 E For any two points 1 and 2
½kxmax2 ½mvmax2 For maximum and minimum
displacements from equilibrium
61
Spring Simulation
Spring Physics Simulation
62
Sample problem

• What is the speed of the pendulum bob at point B
  if it is released from rest at point A?

63
Solution

• Ui Ki Uf Kf
• mgh 0 ½ mv2 0
• v ?2gh
• v ?2(9.8)1.5(1-cos40o)
• v 2.62 m/s

h 1.5 1.5 cos 40o h 1.5 (1 - cos 40o)
64
Sample problem (8.15)

• A 0.21 kg apple falls from a tree to the ground,
  4.0 m below. Ignoring air resistance, determine
  the apples gravitational potential energy, U,
  kinetic energy, K, and total mechanical energy,
  E, when its height above the ground is each of
  the following 4.0 m, 2.0 m, and 0.0 m. Take
  ground level to be the point of zero potential
  energy.

65
Solution (8.15) for 4.0 m

• E U K
• mgh 0 (dropped from rest)
• (0.21 kg)(9.8 m/s2)(4.0 m) 8.2 J
• Therefore
• E 8.2 J (should be 8.2 J for entire problem)
• U 8.2 J(maximum value)
• K 0 (minimum value)

66
Solution (8.15) for 2.0 m

• E U K
• mgh K
• 8.2 J (0.21 kg)(9.8 m/s2)(2.0 m) K
• 8.2 J 4.1 J K
• K 4.1 J
• Therefore
• E 8.2 J
• U 4.1 J
• K 4.1 J

67
Solution (8.15) for 0.0 m

• E U K
• mgh K
• 8.2 J 0 K
• 8.2 J K
• Therefore
• E 8.2 J
• U 0
• K 8.2 J

68
Sample problem (8.18)

• A 1.60 kg block slides with a speed of 0.950 m/s
  on a frictionless, horizontal surface until it
  encounters a spring with a force constant of 902
  N/m. The block comes to rest after compressing
  the spring 4.00 cm. Find the spring potential
  energy, U, the kinetic energy of the block, K,
  and the total mechanical energy of the system, E,
  for the following compressions 0 cm, 2.00 cm,
  4.00 cm.

69
Solution (8.18) for 0 cm

• The spring has no energy, so all energy is
  kinetic energy of the block.
• E K
• ½ m v2 ½ (1.60 kg)(.950 m/s)2
• 0.722 J
• Therefore
• E 0.722 J(should be 0.722 J for entire problem)
• U 0 (minimum value)
• K 0.722 J (maximum value)

70
Solution (8.18) for 4.00 cm

• The block has stopped, so all energy is spring
  potential energy!
• Therefore
• E 0.722 J
• U 0.722 J
• K 0

71
Solution (8.18) for 2.00 cm

• The energy is a mixture of K and U, but the total
  energy is known. The spring potential energy can
  be calculated.
• E U K ½ k x2 ½ m v2
• 0.722 J ½ (902 N/m) (.0200 m)2 K
• 0.722 J 0.180 J K
• K 0.722 J 0.180 J 0.540 J
• Therefore
• E 0.722 J
• U 0.1804 J
• K 0.540 J

72
Law of Conservation of Energy

• E U K Eint Constant
• Eint is thermal energy.
• ?U ?K ? Eint 0
• Mechanical energy may be converted to and from
  heat.

73
Work done by non-conservative forces

• Wnet Wc Wnc
• Net work is done by conservative and
  non-conservative forces
• Wc -?U
• Potential energy is related to conservative
  forces only!
• Wnet ?K
• Kinetic energy is related to net force
  (work-energy theorem)
• ?K -?U Wnc
• From substitution
• Wnc ?U ?K ?E
• Nonconservative forces change mechanical energy.
  If nonconservative work is negative, as it often
  is, the mechanical energy of the system will drop.

74
Sample problem (8.22)

• Catching a wave, a 72-kg surfer starts with a
  speed of 1.3 m/s, drops through a height of 1.75
  m, and ends with a speed of 8.2 m/s. How much
  non-conservative work was done on the surfer?

75
Solution (8.22)

• Wnc ?U ?K
• Uf Ui Kf Ki
• mghf mghi ½ mvf2 ½ m vi2
• mg(hf hi) ½ (vf2 vi2)
• 72(9.8)(0 - 1.75) ½ (8.22 1.32)
• 1125 J

76
Sample problem (8.29)

• A 1.75-kg rock is released from rest at the
  surface of a pond 1.00 m deep. As the rock falls,
  a constant upward force of 4.10 N is exerted on
  it by water resistance. Calculate the
  nonconservative work, Wnc, done by the water
  resistance on the rock, the gravitational
  potential energy of the system, U, the kinetic
  energy of the rock, K, and the total mechanical
  energy of the system, E, for the following depths
  below the waters surface d 0.00 m, d 0.500
  m, d 1.00 m. Let potential energy be zero at
  the bottom of the pond.

77
Solution (8.29) for 0.00 m

• Wnc F?r 0
• E U K
• mgh 0 mgh
• (1.75 kg)(9.8 m/s2)(1.00 m) 17.15 J
• Therefore
• Wnc 0 (the rock hasnt moved yet)
• E 17.15 J(will be reduced by the drag force of
  water)
• U 17.15 J (maximum value)
• K 0 (minimum value)

78
Solution (8.29) for 0.50 m

• Wnc F?r (-4.10 N)(0.50 m) -2.05 J ?E
• E 17.15 J ?E 17.15 J - 2.05 J 15.1 J
• E U K mgh K
• 15.1 J (1.75 kg)(9.8 m/s2)(0.50 m) K
• 15.1 J 8.6 J K
• K 15.1 8.6 J 6.5 J
• Therefore
• Wnc -2.05 J
• E 15.1 J(reduced by the drag force of water)
• U 8.6 J (determined by height)
• K 6.5 J (reduced by the drag force of water)

79
Solution (8.29) for 1.00 m

• Wnc F?r (-4.10 N)(1.00 m) -4.10 J ?E
• E 17.15 J ?E 17.15 J - 4.10 J 13.05 J
• E U K 0 K K
• 13.05 J K
• Therefore
• Wnc -4.10 J
• E 13.05 J (reduced by the drag force of water)
• U 0 (lowest point in problem)
• K 13.05 J (maximum value)

80
Announcements
81
Pendulum lab

• Figure out how to demonstrate conservation of
  energy with a pendulum using the equipment
  provided.
• The photogates must be set up in gate mode this
  time.
• The width of the pendulum bob is an important
  number. To get it accurately, use the caliper.
• Turn in just your calculations, which must
  clearly show the speed you predict for the
  pendulum bob from conservation of energy, the
  speed you measure using the caliper and photogate
  data, and a difference for the two.