Reminder (from last lecture)
Kinetic energy:
B r
r
Potential energy difference: ΔU = U − U 0 = − ∫ F ⋅ dr
K = 12 mv2
A
Work – kinetic energy theorem:
The net work, W, done by the net force on an
object equals the change, ΔK, in its kinetic energy.
LECTURE 12: CONSERVATION
OF ENERGY
W = K 2 − K1 = ΔK
Prof. Flera Rizatdinova
Reminder (from last lecture)
Conservation of Energy
A conservative force is one for which W = 0
along a closed path. And this is independent of
the path taken.
Energy can be neither created nor destroyed
For a closed system
the change in energy
is zero
If, along a closed path, the work done by a
force is nonzero, then the force is called
nonconservative.
Closed System
ΔEsys = 0
Open System
E in
ΔEsys ≠ 0
E out
4
Conservation of Mechanical Energy
5
Conservation of Mechanical Energy
6
According to the work-kinetic energy theorem
ΔK = Wnet.
The net work can be split into the work Wc done
by conservative forces and work Wnc done by
nonconservative forces
ΔK = Wc + Wnc
ΔK = –ΔU + Wnc
or
ΔK + ΔU = Wnc
2/17/2009
ΔK + ΔU = Wnc
We see that the change in kinetic energy plus
the change in potential energy is equal to the
work done by nonconservative forces.
If all the forces are conservative, then Wnc = 0.
In this case, K – K0 + U – U0 = 0, that is,
K + U = K0 + U 0
2/17/2009
Conservation of Mechanical Energy
Clicker Question 1
7
8
The sum K + U of the kinetic and potential
energy is called the mechanical energy.
The rule
1)
K + U = K0 + U 0
2)
3)
expresses the
law of conservation of mechanical energy
4)
5)
A ball is held at a height H above a floor. It is then released
and falls to the floor. If air resistance can be ignored, which of
the five graphs below correctly gives the mechanical energy of
the Earth-ball system as a function of the altitude y of the ball?
A
B
C
D
E
2/17/2009
Example
2/17/2009
(1)
Example
9
(2)
10
How high does the block go?
Since the forces are conservative the
mechanical energy is conserved
Initial mechanical energy
of system
Ei = 12 kx
1
2
2
kx 2 = mgh
Height reached
Final mechanical energy
of system
h=
E f = mgh
kx 2
2mg
2/17/2009
11
2/17/2009
Example –
Pushing a box up a slope (1)
12
To push a box of mass m a distance d up a slope,
the amount of work you do is W. What is the
coefficient of friction, assuming you move the
box at constant velocity?
y
If F is the force you exert,
then the work you do is
B r
r
W = ∫ F ⋅ dr
A
r
f
θ
If the force you exert is constant, then we can
write
r
r
r r
B
B r
r
W = ∫ F ⋅ dr = F ⋅ ∫ dr = F ⋅ d
A
x
r
N
r
w
2/17/2009
r
F
Example –
Pushing a box up a slope (2)
How do we compute F? We
use the fact that the box does
not accelerate. In this case,
r r r r r
F + w+ N + f = 0
so
r
r r r
F = −( w + N + f )
A
y
x
r
N
r
f
θ
r
w
2/17/2009
r
F
13
Example –
Pushing a box up a slope (3)
14
The work you do is therefore
…putting together the pieces
r r
r r r r
W = F ⋅ d = −( w + N + f ) ⋅ d
r r r r r r
= −w ⋅ d − N ⋅ d − f ⋅ d
= − wd cos(π / 2 + θ ) − 0 − fd cos(π )
r
= wd sin(θ ) + fd
N
The magnitude of the
friction force is
f = μk N = μk w cos(θ )
Example –
Pushing a box up a slope (4)
r
f
θ
r
w
W = wd sin(θ ) + μk w cos(θ )d
r
d
x
W − wd sin(θ )
μk =
wd cos(θ )
r
F
Note, when θ = 0, we
get μk = W/wd, as expected
15
r
w
How far up a slope does the box go, if its initial
velocity is v and the coefficient of friction
is μk?
2/17/2009
Clicker question 2
Objects A and B interact with each other via both conservative
and nonconservative forces. Let K1 and K2 be the kinetic energies,
U the potential energy, and Eint be the internal energy. If no
external agent does work on the object then:
r
F
Example –
Sliding a box up a slope
which can be written as
ΔK + ΔU = Wnc
is conserved
is conserved
is conserved
is consvered
is conserved
θ
r
d
Start with
K + U = K0 + U0 + Wnc
K + U = K0 + U0 + Wnc
K1 + U
K1 + U + K2 + Eint
K1 + U + K2
K1 + U + Eint
U + K2
r
f
2/17/2009
16
The effect of nonconservative forces, such as
friction, is to remove energy from a system.
Energy, of course, is still conserved if we
account for the work done Wnc by these forces:
x
r
N
2/17/2009
Nonconservative Forces
1)
2)
3)
4)
5)
r
d
y
we get
x
r
N
r
f
θ
r
w
2/17/2009
Example
A block slides along a track from one level to a higher level
after passing through an intermediate valley. The track is
frictionless until the block reaches the higher level. There a
frictional force stops the block in a distance d. The block’s
initial speed is 6 m/s, the height difference h is 1.1 m and μk is
0.6. Find d.
K i + U i = K f + U f − W f ⇒ W f = ΔK + ΔU ; K f = 0
ΔK = − K i = − 12 mv02 ; ΔU = mgΔh; W f = − μ k mg ⋅ d
d=
1
μk g
(
1
2
)
v02 − gΔh = 1.2m
Potential-energy curve
Potential-energy curve: plot of potential energy
versus position
Force and Potential Energy
Fx = −
U
E
x
Force is the slope of the potential-energy curve
Potential well – path between two turning points
dU
dx
On peaks and valleys, the force is equal to 0.
Peaks are non-stable equilibrium positions; bottom
of valleys are stable equilibrium positions.