1. Chapter 6
Work, Energy, and Power

2. 6.1 Work: The Scientific Definition
- Many different forms of energy
- Energy can be defined as the ability to do work, in some circumstances
not all energy is available to do work.
Work done on a system  product of the component
of the force in the direction of motions times the distance. F is the net force.

3. A person holding a briefcase,
does no work on it, since there
is no motion.

4. No work is done on the briefcase, and no energy is transferred it, when it is carried horizontally at a constant speed.

5. Energy is transferred to the briefcase.

6. Energy is transferred out the briefcase
and into the electric generator. Here the work done on the briefcase by the generator is negative, since F and d are in opposite direction.
F . d < 0

7. Example
Suppose the player in your game must push a secret stone wall at least 2m to gain access to a bonus level. If the player’s character can push with a horizontal force of 1500N and kinetic frictional force of 50N is working against him, how much work total will it take to move the wall out of the way?
Solution
W = FDX = ( )*2 = 2900 J

8. Example
Suppose you are coding a top-down game where the object being moved is free to move in both the x and y directions. A net force of is applied to the object, but due to constraints, the object moves How much work is done to move the object?
Solution
Use Dot product of two vectors
W = F.d = (2000cos 30)*3 = 5196 J F
300 Dx

9. 6.2 Kinetic energy and the work-energy theorem

11. - The package is accerlated from v0 to v by a net force F
Net W = (Net F) d
- The package is accerlated from v0 to v by a net force F
Net F = ma, a=constant
v2 = v02 + 2ad
Net W = ½ m (v2 - v02) = DKE
Work-Energy Theorem
the net work on a system = the changes in the kinetic energy

12. Example
Suppose you are programming a baseball game, and the shortstop need to throw the ball to the first baseman. It takes some work for him to get the ball moving . If he can produce a force of 400N for a distance of 0.75m (from behind his head to the point where he lets go of the ball), and the baseball has a mass of 145g, what speed should the ball have when he lets go?
Solution
W = DKE
400*0.75 = 0.5*0.145*(vf2 - 0)
vf = 64.3 m/s

13. 6.3 Gravitational Potential Energy
A mass is lifted up at constant speed
W = Fd = mgh
Define change in gravitational PE
PEg(h)-PEg(0)= DPEg = mgh
(Gravitational Potential energy, PEg)
- if we release the mass, gravity will do an amount of work mgh on it
Drop the mass, by the Work-energy Theorem
Increasing the mass KE
conversion of PE to KE
PEg = 0

14. The change of gravitational PE
(DPEg = mgh) between point A and B
is independent of path.
Gravity is a conservative force (守恒力)

15. Example
v0 =0,  mgd = DKE = ½ mvf2  vf = 19.8 m/s
v0 = 5m/s  vf = 20.4 m/s
Mass cancel when friction is neglible
Speed depend on the initial position and final postion NOT on the
PATH taken or mass (when friction is neglible)
- Final speed =0.6 m/s << 5m/s

16. Example
Suppose you are programming an Indiana Jones game and you get to the part where he jumps into the mining cart and rides the track up and down a series of gills. If the cart is at a height of 100m when Indy jumps in, and together he and the cart weigh Kg, how fast should they be going when they reach the bottom of the first hill (ground level)?
Solution
½ mvi2 + mgyi = ½ mvf2 + mgyf
*9.8*100 = 0.5*136.18*vf2 + 0
vf2 = m/s
100 m

17. 6.4 Conservative Forces and Potential Energy

18. 6.4 Conservative Forces and Potential Energy
- Define the potential energy of a spring,
- PEs = ½ kx2 (a general expression, such as atomic vibration)
- It represents the work done on the spring and the energy stored
in it as a result of strectching it a distance of x.

19. 6.4 Conservative Forces and Potential Energy
Work is done to deform the guitar string
PE  KE  PE as the string oscillate back and forth
A small fraction is dissipated as sound energy
A nonconservative process

20. Conservation of Mechanical Energy (PE+KE)
6.4 Conservative Forces and Potential Energy
The net work done by all forces acting on a system equals to DKE,
Net W = DKE (Work-energy Theorem )
If only conservative force involved
 Wc = DKE
conservative force (gravity or a spring force) does work  system loses PE  Wc = mgh = - DPEg
DKE + DPE = 0
KEf – KEi + PEf – PEi = 0
PEi + KEi= PEf + KEf
Conservation of Mechanical Energy (PE+KE) h mg

21. 1/2mvi2+mghi+1/2kxi2 = 1/2mvf2+mghf+1/2kxf2
Example 6.7
A 100g toy car is propelled by a compressed spring, as shown in the figure. It follows a track that rises 0.18m above the starting point. The spring si compressed 4 cm and has a force constant of 250 N/m. Assuming no friction, find (a) how fast the car isi going before it starts up the slope and (b) how fast it is going at the tip of the slope,
PEi + KEi= PEf + KEf
1/2mvi2+mghi+1/2kxi2 = 1/2mvf2+mghf+1/2kxf2
1/2kxi2 = 1/2mvf2  vf = (250/0.1)0.5 *0.04 = 2.0 m/s
1/2kxi2 = 1/2mvf2+mghf
 vf = [(250/0.1) *0.042 – 2*9.8*0.18]0.5 = 0.69 m/s

22. 6.5 Nonconservative Forces : Open systems
Frictional force, most of the energy goes into thermal energy
Open system  energy may enter or leave it
Taken into account of both conservative and non-conservative forces
Net W = Wc + Wnc = DKE
-DPE + Wnc = DKE
or Wnc = DPE + DKE (change in Mechanical Energy)
 PEi + KEi + Wnc = PEf + KEf
Path taken by the frictional force is longer More energy lost

23. Conservative Forces : Close systems
Wnc = 0

24. 6.5 Nonconservative Forces : Open systems

25. 6.5 Nonconservative Forces : Open systems
PEi + KEi + Wnc = PEf + KEf
Work done by frictional force

26. 6.6 Conservation of Energy
PEi + KEi + Wnc + OEi = PEf + KEf
OE = other energy  fuel, electrical energy, chemical fuels, solar energy

27. 6.7 Power
P = W/t
1 hp = 746 Watt

28. W = DKE + DPE, if we set h=0, then both KE and PE are at the bottom.
Example 6.9
What is the power output of mechanical energy for a 60 Kg woman who runs up a 3.0 m high flight of stairs in 2.5s, starting from rest but having a final speed of 2.0 m/s ?
Solution
W = DKE + DPE, if we set h=0, then both KE and PE are at the bottom.
Thus W = KE + PEg = ½ mv2 + mgh
P = W/t = [(0.5*60* *9.8*3.0)]/2.5 = 754 W

29. Eff = Wout/Ein or Eout/Ein
6.8 Work, Energy, and Power in Humans; Efficiency
Eff = Wout/Ein or Eout/Ein

30. Eff = Wout/Ein = 120*9.8*2.3/(8.0*4186) = 8.33%
Example 6.11
Find the efficiency of the person in the figure who metabolized 8.0 kcal of food energy while lifting 120 kg to a height of 2.3 m above its starting point
Solution
The work output goes into increasing the gravitational PE of the weights, thus
Wout = mgh
Eff = Wout/Ein = 120*9.8*2.3/(8.0*4186) = 8.33%
Table 6.3
Efficiency of he human body and mechanical devices
Table 6.4
Energy and oxygen consumption rates (power)