#### Transcript Work-Energy Theorem

```Chapter 6
Work, Energy, and Power
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.
A person holding a briefcase,
does no work on it, since there
is no motion.
No work is done on the
briefcase, and no energy is
transferred it, when it is
carried horizontally at a
constant speed.
Energy is transferred to the briefcase.
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
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 = (1500-50)*2 = 2900 J
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 2000N@600 is applied to
the object, but due to constraints, the object moves 3m@300. 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
6.2 Kinetic energy and the work-energy theorem
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
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
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
The change of gravitational PE
(DPEg = mgh) between point A and B
is independent of path.
Gravity is a conservative force (守恒力)
Example
(a) v0 =0,  mgd = DKE = ½ mvf2  vf = 19.8 m/s
(b) 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 20.4 - 19.8=0.6 m/s << 5m/s
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 136.18 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
• 0 + 136.18*9.8*100 = 0.5*136.18*vf2 + 0
• vf2 = 44.27 m/s
100 m
6.4 Conservative Forces and Potential Energy
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.
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
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
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
(a) 1/2kxi2 = 1/2mvf2  vf = (250/0.1)0.5 *0.04 = 2.0 m/s
(b) 1/2kxi2 = 1/2mvf2+mghf
 vf = [(250/0.1) *0.042 – 2*9.8*0.18]0.5 = 0.69 m/s
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
Conservative Forces : Close systems
Wnc = 0
6.5 Nonconservative Forces : Open systems
6.5 Nonconservative Forces : Open systems
PEi + KEi + Wnc = PEf + KEf
Work done by frictional force
6.6 Conservation of Energy
• PEi + KEi + Wnc + OEi = PEf + KEf
• OE = other energy  fuel, electrical energy,
chemical fuels, solar energy
6.7 Power
P = W/t
1 hp = 746 Watt
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*2.02 + 60*9.8*3.0)]/2.5 = 754 W
6.8 Work, Energy, and Power in Humans; Efficiency
Eff = Wout/Ein or Eout/Ein
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)
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