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Chapter 7 Internal Energy
Work done by a spring
❑ We know that work equals force times
displacement
Fs = −kx
Hooke’s Law for the restoring
force of an ideal spring. (It is
a conservative force.)
s = x f − xi
But the force is not constant
xi
xf
Fs ,i = −kxi ,
Fs , f = −kx f
Take the average force
Fs,avg
x=0
Fs,avg
Fs,i + Fs, f
=
2
= − 12 k(x f + x i )
Then the work done by the spring is
!
W s = Fs,avg cos φ s = Fs,avg cos0 s
2
2
1
1
= − 2 k(x f€+ x i )(x f − x i ) = − 2 k(x f − x i )
2
2
1
1
W s = 2 kx i − 2 kx f =-ΔU
2
i
2
f
Ws = kx − kx = U s ,i − U s , f
1
2
1
2
U elastic = U s = kx
1
2
2
Units of N/m m2 =
Nm=J
❑ Total potential energy is
U total = U g + U s = mgy + kx
1
2
2
Example Problem
A block (m = 1.7 kg) and a spring (k = 310 N/m)
are on a frictionless incline (θ = 30°). The spring is
compressed by xi = 0.31 m relative to its
unstretched position at x = 0 and then released.
What is the speed of the block when the spring is
still compressed by xf = 0.14 m?
x=0
xi
xf
x=0
θ
θ
Given: m=1.7 kg, k=310 N/m, θ=30°, xi=0.31 m,
xf=0.14 m, frictionless
Method: no friction, so we can use conservation of
energy
Initially
2
E = mv + mgh + kx
v i = 0, hi = xi sin θ
1
2
1
2
2
i
Ei = mgxi sin θ + kx
1
2
2
Finally
h f = x f sin θ , find v f
2
2
1
1
E f = 2 mv f + mgx f sin θ + 2 kx f
E f = Ei
2
2
1
1
2 mv f + mgx f sin θ + 2 kx f
2
1
= mgxi sin θ + 2 kxi
2
2
2
1
1
2 mv f = mg ( xi − x f ) sin θ + 2 k ( xi − x f )
k 2
2
v f = 2 g ( xi − x f ) sin θ + ( xi − x f )
m
310
2
2
v f = 2(9.8)(0.31 − 0.14) sin 30 +
(0.31 − 0.14 )
1.7
v f = 1.666 + 13.95 = 3.95 ms
!
Interesting to plot the potential energies
Energy
K
➔
Ug
xf
E=K+U
Utotal
Us
xi
xf
xi
xf
xi
Molecular potentials (two atoms)
Molecule AB
Harmonic oscillator
potential
A+B
0
re
UM
1
= ks (r
2
re ) 2
EM
Morse potential
Better approximation, but
need EM, re, α, where
↵=
p
ks /2EM
But at large r, Morse
potential does poorly
0
UM
Morse
UM
= EM [1
exp[ ↵(r
C6
! 6
r
re )]]
2
Multidimensional potentials
C
H
Map
H
O
180
600
500
40
35
30
25
20
15
10
5
0
−5
−10
−15
−20
−25
−30
−35
−40
−45
−50
−55
−60
−65
160
400
300
140
200
100
120
0
100
100
θ2
Interaction Potential (cm−1 )
H2-CO, U(R,r1,r2,θ1,θ2,Φ)
7
8
9
R(a0 )
10
11
12
360
300
240
180
120
60
0
θ1 (degree)
r1,r2,θ2,Φ fixed
80
60
R,r1,r2,Φ fixed
40
20
0
0
50
100
150
200
θ1
250
300
350
Example problem
If it takes 4.00 J of work to stretch a
Hooke’s law spring 10.0 cm from its
unstretched length, determine the extra
work required to stretch it an additional
10.0 cm.
Thermal energy
Modeling a solid as
a collection of
spring-masses
Internal energy due
to K of atoms and U
of “springs”
Conservative and Non-conservative Forces
❑ Conservative Force: a force for which the work
it does on an object does not depend on the
path. Gravity is an example.
❑ We know we can obtain the work with the
xf
work integral.
W = ∫ Fx dx = Wc
xi
❑ If the force is conservative, then W=Wc and
this work can be related to the change in
xf
potential energy W =
F dx = −ΔU
c
∫
xi
x
y
m
mg
s
A
A
B
C
h
s
s
mg
B
θ
mg
C
W = F cos φ s = mgh
h
W = (mg sin θ ) s = (mg sin θ )
= mgh
sin θ
W = mgh
- Non-conservative Force - a force for which the
work done depends on the path
- friction
- air resistance
If the force is conservative, we can find the
potential energy due to the force
xf
U f ( x) = − ∫ Fx dx + U i ( x) it is
xi
usually convenient to take Ui(x)=0
Or if we know U(x) and the force is
conservative, we can obtain F
dU ( x)
dU ( x) = − Fx dx ⇒ Fx = −
dx
The x-component of a conservative force equals
the negative derivative of the potential energy
with respect to x
If both conservative and non-conservative forces
act on an object, the work-energy theorem is
modified
Wtotal = WC + WNC = K f − K i
WNC = K f − K i − WC
For the case of gravity
Wg = WC = mg ( yi − y f )
WNC = K f − K i − mg ( yi − y f )
WNC = K f − K i − mgyi + mgy f
WNC = ΔK + ΔU
= K f − Ki + U f − U i
= (K f + U f ) − (Ki + U i )
WNC = E f − Ei
=
Wsurr
❑ If no net non-conservative forces
WNC = 0 ⇒ E f = Ei = E
❑ Then, conservation of mechanical energy holds
ΔK = −ΔU
Crate on Incline Revisited
FN
fk
fk
FN
s
h
mg
x
θ
FBD
s
θ
θ
mg
WN = 0
Wg = mg sin θ s = mgh = WC
W f = − µ k mg cosθ s = WNC
❑ The crate starts from rest, vi=0
K i = 0, Ei = U i = mgh = Wg
WNC = E f − Ei
E f = Ei + WNC = mgh − µ K mg cosθ s
E f < Ei
❑ 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 = Ei , since K i = 0,U 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)
Ei = E f
mech
others
mech
others
Ei + Ei
= Ef + Ef
mech
mech
others
others
E f = Ei + Ei − E f
mech
mech
E f = Ei + WNC
others
others
⇒ WNC = Ei − E f = Q
total
total
❑ Q (WNC) is the energy lost (or gained) by the
mechanical system
❑ The electrical utility industry does not produce
U=mgh
energy, but merely converts energy
Lake
electricity
Light,
heat
River
Hydro-power plant
h
K=mv2/2
Example Problem
A ball is dropped from rest at the top of a 6.10-m tall
building, falls straight downward, collides inelastically
with the ground, and bounces back. The ball loses
10.0% of its kinetic energy every
time it collides with the ground. How many bounces
can the ball make and still reach a window sill that is
2.44 m above the ground?
Solution:
Method: since the ball bounces on the ground, there
is an external force. Therefore, we can not use
conservation of linear momentum.
An inelastic collision means the total energy is not
conserved, but we know by how much it is not
conserved. On every bounce 10% of K is lost:
Qi = 0.1K i , i = 1,2,3,..., n n is the number of bounces
Given: h0 = 6.10 m, hf = 2.44 m
Eo = U o = mgho
E1 = K1 = Eo
o
3
6
Since energy is conserved from
point 0 to point 1.
However, between point 1 and 2,
energy is lost
Q2 = 0.1K1
WNC = −Q = E final − Einitial
− Q2 = E2 − E1
− 0.1K1 = E2 − E1
− 0.1Eo = E2 − Eo
E2 = 0.9 Eo = E3 = E4
1 2 4
5
Total energy after
one bounce
By the same reasoning
E5 = 0.9 E4 = 0.9 E2 = 0.9(0.9 Eo )
2
E5 = 0.9 Eo Total energy after two bounces
The total energy after n bounces is then
n
log(h f / ho )
E f = 0.9 Eo
n
=
n
log(0.9)
mgh f = 0.9 mgho
log(2.44 / 6.10)
n
n=
h f = 0.9 ho
log(
0
.
9
)
n
0.9 = h f / ho
n = 8.7
n
log(0.9) = log(h f / ho ) Answer is 8 bounces
n log(0.9) = log(h f / ho )
Power
Average power:
Pavg
W
=
Δt
or
Units of J/s=Watt (W)
Measures the rate at
which work is done
F cos φ s
Pavg =
= F cos φ v avg
Δt
€
Instantaneous power:
W dW
P = lim
=
Δt →0 Δt
dt
! !
! !
F ⋅ ds
P=
= F ⋅v
dt
W can also be replaced by the
total energy E. So that power
would correspond to the rate
of energy transfer
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: vi=0, vf=27 m/s, Δt=7.0 s,
(a) mg= 1.2x104 N, (b) mg= 1.6x104 N
Method: determine the acceleration
Pavg
W F cos φ s
=
=
Δt
Δt
!
s
!
F
❑ 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 mass Need as and s
Pavg =
=
Δt
Δt
2
2
(v
−
v
f
i)
2
2
v f = v i + 2ass ⇒ ass =
2
2
2 (
%
m (v f − v i )
m % 27 2 − 0 (
Pavg = '
*=
'
* = 52m
Δt &
2
) (7.0 s) & 2 )
4
Pavg
Pavg
52mg 52(1.2x10 )
4
=
=
= 6.4x10 W
g
9.80
4
52(1.6x10 )
4
=
= 8.5x10 W (b)
9.80
Or from work-energy theorem
1
2
2
f
2
f
W = ΔK = mv
W mv
Pavg =
=
Δt 2Δt
Same as on
previous slide
(a)