Transcript Conservation of Energy

Conservation of Energy
If a body is acted upon by a force that does no
total work during any round trip/cycle A, then the
force is conservative. Otherwise, it is nonconservative B!
A) gravity:
UP: W = - mgh
DOWN: W = mgh
Wnet = 0
conservative!
A) spring:
W = +.5kd2
W = - .5kd2
conservative
Return trip is exact
opposite: Wnet = 0
Work done in a conservative system is path
independent– like in the case of gravity!
B) Friction:
The total work done by friction during a
complete cycle is not 0!
Friction is a non-conservative
force!
Potential Energy (U)
Potential Energy is the energy stored within a
system-- also called the energy of configuration.
• it can only be defined in terms of
conservative forces-- not non-conservative
forces!
If we are dealing with a conservative system, then
the change in Potential Energy will be equal to the
work done by the conservative force:
∆U = -W
If the system is conservative, then the total
amount of energy will remain a constant value:
U+K=E
This is the mathematical representation of the
law of conservation of mechanical energy.
Therefore:
∆U + ∆K = 0
This will hold true for more than one
conservative force (more than one force doing
the work) as well!
In the Conservative system: ∆U + ∆K = 0
In the non-conservative (and conservative) part of
the system: W = ∆K
Assume a conservative system unless otherwise
specified!
A spring gun used to horizontally launch a 12 g
ball is compressed 3.2 cm from its relaxed state
and loaded. Use the law of conservation of
energy to find the speed of the ball as it leaves the
gun if k for the spring is 7.5 N/cm.
∆U + ∆K = 0
v=d k
m
(Uf - Ui) + (Kf - Ki) = 0
(0 - .5kd2) + (.5mv2 - 0) = 0
= 8.0 m/s
= .032 m
750 N/m
.012 kg
A 1.93 kg block is placed against a compressed
spring on a frictionless 27.0˚ incline. The spring
force constant is 20.8 N/cm and it is compressed
18.7 cm and released. How far up the incline will
the block go before coming to rest?
m = 1.93 kg
K + Us + Ug = K0 + Us0 + Ug0
ø = 27.0˚
2
2
y
=
.5kx
mgy = .5kx0
0 = 1.92 m
k = 2080 N/m
mg
x0 = .187 m
L
L= y
y
sin Ø
= 4.23 m
Two children are playing a game in which they try to
hit a small box on the floor with a marble fired from a
spring loaded gun that is mounted on a table. The
target box is 2.20 m horizontally from the edge of the
table. Conor compresses the spring 1.10 cm, but the
marble falls 27.0 cm short. How far should Mary
Alice compress the spring to score a hit?
let x = the spring compression
∆Us + ∆K = 0
let d = horizontal distance traveled
.5k(x2 - xo2 ) + .5 m (v2 - vo2) = 0
kxo2 = mv2 = m(d /t)2
k,m,t are constants!
d2 = kt2
d is a constant ratio!
so
x2
m
x
Conservation of Energy in a System of Particles
• When an object interacts with one or more
objects we can choose our system to be as many
or as few of those objects as is convenient.
conservation of energy will always hold if
care is taken to keep track of energies within
the system and transfers in and out of the
system!
system
boundary
system
U,K,Eint
External W
The energy of the system within the boundary
can be changed only when external work is done
to it.
(Internal work-- done by one part within the
boundary acting upon another-- does not
change the total energy of the system,
although it may convert energy from one
form to another, e.g. UK)
Conservation of Energy can then be written:
∆U + ∆K +∆Eint = W
Consider a block attached to a spring sliding
across a table that exerts frictional forces:
A) If the block itself is the system, then:
∆U + ∆K +∆Eint = Wtotal
∆K + ∆Eint = Ws + Wf
B) If the block and the spring are the system:
∆Us + ∆K + ∆Eint = Wf
C) If the block, spring and the table are the
system:
∆U + ∆K + ∆Eint = 0
*the Conservation Laws hold in each case!
A baseball of mass .143 kg is dropped from a
height of 443 m and reaches a terminal speed of
42 m/s. Find the change in internal energy of the
ball and surrounding air during the fall back to the
surface of the earth.
If the system is regarded to be ball, the air
and the earth, then no external forces or
work are present:
∆U + ∆K + ∆Eint = 0
∆Eint = - ∆U - ∆K
∆U = Uf - Ui = mgh - mgho
0 - (.143)(9.8)(443) = - 621 J
∆K = Kf - Ki = .5mv2 - .5mvo2
.5(.143)(42)2 - 0 = 126 J
Eint = - ∆U - ∆K
- (- 621 J) - (126J) = 495 J
The internal energy increase might be observed as a
temperature increase of the air/ball, and/or an
increase in the KE of the air the ball passes through.
A 4.5 kg block is thrust up a 30.0˚ incline with an
initial speed of 5.0 m/s and it travels 1.5 m before
coming to rest. A) How much mechanical energy
does the block lose in this process due to friction?
B) Assuming friction to produce the same loss in
mechanical energy on the return downward
journey, what is the speed of the block as it passes
through its initial location?
If we choose our system to be the block and the
plane, then there are no external forces/work:
∆U + ∆K + ∆Eint = 0
In this case ∆U and ∆K would refer to the block,
and therefore ∆Eint would refer to the energy
converted to friction and thermal energy.
∆E = - ∆U - ∆K
∆U = mgh - mgho = (4.5)(9.8)(1.5 m)(sin 30) - 0
= 33 J
∆K = .5mv2 - .5mv 2
o
= 0 - .5(4.5)(5.0)2 = - 56 J
∆E = - (33 J) - (-56 J) = 23 J
Positive value
would indicate
an overall gain!
Therefore, the block lost 23 J of mech. energy!
Now if we consider the system to be just the
block:
∆U + ∆K + ∆Eint = Wf
The block would experience changes in U and K,
and its loss of energy would be represented by:
∆U + ∆K = Wf - ∆Eint
In this case, ∆U + ∆K = - 23 J, meaning the
energy went out of the system (block). It would
impossible to determine how much energy was
lost due to Wf and how much was due to
internal changes (thermal energy, etc).
B) Again consider the system to be the block,
both going up and then coming back down the
plane:
∆U + ∆K = Wf - ∆Eint
If the block loses 23 J of energy going up (Wf ∆Eint), then it will lose a total of 46 J going up
and down: ∆U + ∆K = - 46 J
Kf - Ki = - 46 J
vf =
2Kf
m
Kf = - 46 J + 56 J = 10 J
= 2.1 m/s
OR: from the top down
∆U = mg(h - ho) = 0 - (4.5)(-9.8)(-1.5 m)(sin
30)
= - 33J
∆U + ∆K + ∆E = 0
∆K = - ∆U - ∆E = - (-33 J) - 23 J = 10 J
vf =
2Kf
m
= 2.1 m/s
A ski run connects to mountains separated by a
valley. A skier starts from rest down the first hill
from a height of 362 m and coasts until topping
the second hill at a height of 131 m at which point
he has a speed of 23 m/s. What is the change in
internal energy of his skis and the snow over
which he traversed?