Download notes6b

WNC  E f  E0
 If no net non-conservative forces
WNC  0  E f  E0  E
 Then, conservation of mechanical energy holds
KE  PE
Crate on Incline Revisited
FN
fk
h
mg
fk

s
x
FBD
FN

mg

s
WN  0
Wg  mg sin  s  mgh  WC
W f   k mg cos s  WNC
 The crate starts from rest, v0=0
KE0  0, E0  PE0  mgh  Wg
WNC  E f  E0
E f  E0  WNC  mgh   K mg cos s
E f  E0
 Some energy, WNC is loss from the system
 In this case it is due to the non-conservative
friction force  energy loss in the form of heat
 Because of friction, the final speed is only 9.3
m/s as we found earlier
 If the incline is frictionless, the final speed
would be:
E f  E0 , since KE0  0, PE f  0
2
1
2 mv f  mgh  Wg
2Wg
2(7510 J)
vf 

 12.3 ms
m
100 kg
 Because of the loss of energy, due to friction,
the final velocity is reduced. It seems that energy
is not conserved
Conservation of Energy
 There is an overall principle of conservation of
energy
 Unlike the principle of conservation of mechanical
energy, which can be broken, this principle can not
 It says: ``The total energy of the universe is, has
always been, and always will be constant. Energy can
neither be created nor destroyed, only converted
from one form to another.’’
 So far, we have only been concerned with
mechanical energy
 There are other forms of energy: heat,
electromagnetic, chemical, nuclear, rest mass
(Em=mc2)
E
E
others
E
E
E
 Ef
others
E
E
E
 Ef
E
E
W
 WNC  E
E
Q
total
0
mech
0
mech
f
mech
f
total
f
others
mech
0
f
mech
others
0
0
mech
0
NC
others
others
0
f
 Q (WNC) is the energy lost (or gained) by the
mechanical system
 The electrical utility industry does not produce
energy, but merely converts energy
PE=mgh
Lake
Light,
heat
Power
Average power:
electricity
River
Hydro-power
plant
W
P
t
F cos  s
P
 F cos  v
t
h
KE=mv2/2
Units of J/s=Watt (W)
Measures the rate at
which work is being
done
Example
A car accelerates uniformly from rest to 27 m/s in
7.0 s along a level stretch of road. Ignoring friction,
determine the average power required to
accelerate the car if (a) the weight of the car is
1.2x104 N, and (b) the weight of the car is 1.6x104
N.
Solution:
Given: v0=0, vf=27 m/s, t=7.0 s,
mg= 1.2x104 N, (b) mg= 1.6x104 N
Method: determine the acceleration
(a)

s

F
W F cos  s
P

t
t
 We don’t know the displacement s
 The car’s motor provides the force F to
accelerate the car – F and s point in same direction
Fs mas s
P

t
t
Need ass
v  v  2a s s  a s s 
2
f
2
0
(v  v )
2
f
2
0
2
2
2
2


(
v

v
)


m f
m
27

0
0


  52m
P
 (7.0 s)
t 
2
 2 

4
52mg 52(1.2 x10 )
4
P

 6.4x10 W
g
9.80
4
52(1.6 x10 )
4
P
 8.5x10 W (b)
9.80
Or from work-energy theorem
W  KE  mv
2
W mv f
P

t 2t
1
2
2
f
Same as on
previous slide
(a)
Example: Problem 6.77
A particle starting from point A, is projected down
the curved runway. Upon leaving the runway at
point B, the particle is traveling straight upward
and reaches a height of 4.00 m above the floor
before falling back down. Ignoring friction and air
resistance, find the speed of the particle at point A.

v0
4.00 m
A
3.00 m