Transcript ½kx 2

Chapter 8C - Conservation of Energy
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
A waterfall in
Yellowstone Park
provides an
example of
energy in
nature. The
potential energy
of the water at
the top is
converted into
kinetic energy at
the bottom.
Objectives: After completing this
module, you should be able to:
• Define and give examples of conservative
and nonconservative forces.
• Define and apply the concept of
conservation of mechanical energy for
conservative forces.
• Define and apply the concept of
conservation of mechanical energy
accounting for friction losses.
Potential Energy
Potential Energy is the ability to do work
by virtue of position or condition.
h
m
Example: A mass held a
distance h above the earth.
mg
If released, the earth can
do work on the mass:
Earth
Work = mgh
Is this Positive!
work + or - ?
Gravitational Potential Energy
Gravitational Potential Energy U is equal to
the work that can be done BY gravity due to
height above a specified point.
U = mgh
Gravitational P. E.
Example: What is the potential energy when
a 10 kg block is held 20 m above the street?
U = mgh = (10 kg)(9.8 m/s2)(20 m)
U = 1960 J
The Origin of Potential Energy
Potential energy is a property of the Earthbody system. Neither has potential energy
without the other.
F
mg
h
Work done by
lifting force F
provides positive
potential energy,
mgh, for earthbody system.
Only external forces can add or remove energy.
Conservative Forces
A conservative force is one that does
zero work during a round trip.
F
mg
h
Weight is conservative.
Work done by earth
on the way up is
negative, - mgh
Work on return is
positive, +mgh
Net Work = - mgh + mgh = 0
The Spring Force
The force exerted by a spring
is also conservative.
F
x
m
When stretched, the spring
does negative work, - ½kx2.
On release, the spring does
positive work, + ½kx2
Net work = 0
(conservative)
x
m
F
Independence of Path
Work done by conservative forces is
independent of the path.
C
A
B
Force
due to
gravity
mg
C
A
B
Because
vertical
of
Work
(A only
C) =the
Work
(A B component
C)
Why?
the weight does work against gravity.
Nonconservative Forces
Work done by nonconservative forces
cannot be restored. Energy is lost and
cannot be regained. It is path-dependent!
B
A
f
A
B
m
f
Friction forces are nonconservative forces.
Work of Conservative Forces
is Independent of Path:
For gravitational force:
B
C
(Work)AB= -(Work)BCA
Zero net work
For friction force:
A
(Work)AB -(Work)BCA
The work done against friction is greater
for the longer path (BCD).
Stored Potential Energy
Work done by a conservative force is stored in
the system as potential energy.
m
x
xo
The potential energy is
equal to the work done in
compressing the spring:
F(x) = kx to compress
Potential energy of
compressed spring:
Displacement is x
U  Work  kx
1
2
2
Conservation of Energy
(Conservative forces)
In the absence of friction, the sum of the
potential and kinetic energies is a constant,
provided no energy is added to system.
h
y
v=0
v
mg
At top: Uo = mgh; Ko = 0
At y: Uo = mgy; Ko = ½mv2
At y=0: Uo = 0; Ko = ½mvf 2
0
vf
E = U + K = Constant
Constant Total Energy
for a Falling Body
TOP: E = U + K = mgh
h
K=0
y
At any y: E = mgh + ½mv2
v
Bottom: E = ½mv2
mgh = mgy + ½mv2 = ½mvf2
Total E is same at any point.
(Neglecting Air Friction)
0
U=0
vf
Example 1: A 2-kg ball is released from
a height of 20 m. What is its velocity
when its height has decreased to 5 m?
20m
Total Etop = Total E at 5 m
v=0
mgh = mgy + ½mv2
2gh = 2gy +
v2
5m
v2
= 2g(h - y) = 2(9.8)(20 - 5)
v=
(2)(9.8)(15)
0
v
v = 17.1 m/s
Example 2: A roller coaster boasts a
maximum height of 100 ft. What is the
speed when it reaches its lowest point?
Assume zero friction:
At top: U + K = mgh + 0
Bottom: U + K = 0 + ½mv2
Total energy is conserved
mgh = ½mv2
v=
(2)(32 ft/s2)(100 ft)
v = 2gh
v = 80 ft/s
Conservation of Energy
in Absence of Friction Forces
The total energy is constant for a conservative
system, such as with gravity or a spring.
Begin: (U + K)o = End: (U + K)f
Height?
mgho
Spring?
½kxo2
Velocity?
½mvo2
=
mghf
Height?
½kxf2
Spring?
½mvf2
Velocity?
Example 3. Water at the bottom of a falls has
a velocity of 30 m/s after falling 35 ft.
What is the water speed
at the top of the falls?
ho = 35 m; vf = 30 m/s2
First look at beginning point—top of falls.
Assume y = 0 at bottom for reference point.
Height? Yes (35 m)
mgho
Spring? No
½kxo2
Velocity? Yes (vo)
½mvo2
Example 3 (Cont.) Water at the bottom of falls
has a velocity of 30 m/s after falling 35 ft.
What is the water speed
at the top of the falls?
ho = 35 m; vf = 30 m/s2
Next choose END point at bottom of falls:
Height? No (0 m)
mghf
Spring? No
½kxf2
Velocity? Yes (vf)
½mvf2
Example 3 (Cont.) Water at the bottom of falls
has a velocity of 30 m/s after falling 35 ft.
ho = 35 m; vf = 30 m/s2
What is the water speed
at the top of the falls?
Total energy at top = Total energy at bottom
mgh  mv  0  mv
1
2
2
0
1
2
2
f
2 gh  v  v
2
0
2
f
v02  v2f  2 gh  (25.8 m/s)2  2(9.8 m/s2 )(33.2 m)
v0  14.9 m /s
2
2
vo = 3.86 m/s
Example 4. A bicycle with initial velocity 10
m/s coasts to a net height of 4 m. What is
the velocity at the top, neglecting friction?
E(Top) = E(Bottom)
vf = ?
4m
1
2
Etop = mgh + ½mv2
vo = 10 m/s
EBot = 0 + ½mvo2
mv  mgh  mv
2
f
1
2
2
0
1
2
v  v  gh
2
f
1
2
2
0
v  v  2gh  (10 m/s)  2(9.8 m/s )(4 m)
2
f
2
0
v f  21.6 m2 /s2
2
2
vf = 4.65 m/s
Example 5: How far up the 30o-incline
will the 2-kg block move after release?
The spring constant is 2000 N/m and it
is compressed by 8 cm.
mgho
½kxo
2
½mvo2
=
mghf
End
Begin
½kxf2
s
30o
h
½mvf2
Conservation of Energy:
kx02 (2000 N/m)(0.08m) 2
h

2
2mg
2(2 kg)(9.8 m/s )
½kxo2 = mghf
h = = 0.327 m
Example (Cont.): How far up the 30oincline will the 2-kg block move after
release? The spring constant is 2000
N/m and it is compressed by 8 cm.
Continued:
h = 0.327 m = 32.7 cm
sin 30o =
s=
h
sin 30o
End
Begin
s
30o
h
s
=
32.7 cm
Sin 30o
s = 65.3 cm
h
Energy Conservation and
Nonconservative Forces.
f
Work against friction
forces must be accounted
for. Energy is still
conserved, but not
reversible.
Conservation of Mechanical Energy
(U + K)o = (U + K)f + Losses
Problem Solving Strategies
1. Read the problem; draw and label a sketch.
2. Determine the reference points for gravitational and/or spring potential energies.
3. Select a beginning point and an ending
point and ask three questions at each point:
a. Do I have height?
U = mgh
b. Do I have velocity?
K = ½mv2
c. Do I have a spring?
U = ½kx2
Problem Solving (Continued)
4. Apply the rule for Conservation of Energy.
mgho
½kxo2
½mvo2
=
mghf
½kxf2
½mvf2
+
Work
against
friction:
fk x
5. Remember to use the absolute (+) value
of the work of friction. (Loss of energy)
Example 6: A mass m is connected to a cord
of length L and held horizontally as shown.
What will be the velocity at point B? (d = 12 m,
L = 20 m)
A
1. Draw & label.
L v
B
c
2. Begin A and end B.
d
r
3. Reference U = 0.
0
(U + K)o =(U + K)f + loss
mgL + 0 = mg(2r) + ½mvc2
2gL - 4gr = vc2
U=0
(Multiply by 2, simplify)
Next find r from figure.
Example (Cont.): A mass m is connected to a
cord of length L and held horizontally as
shown. What will be the velocity at point B?
(d = 12 m, L = 20 m)
A
2gL - 4gr = vc2
L
vc
r=L-d
r = 20 m - 12 m = 8 m
vc2 =2gL - 4gr = 2g(L - 2r)
B
r
d
U=0
vc2 = 2(9.8 m/s2)[20 m - (2)(8 m)]
vc =
2(9.8 m/s2)(4 m)
vc = 8.85 m/s
Example 7: A 2-kg mass m located 10 m above
the ground compresses a spring 6 cm. The
spring constant is 40,000 N/m and mk = 0.4.
What is the speed when it reaches the bottom?
2 kg
h
s
30o
n
f
Begin
mg Cos
End
mg Sin 30o
30o
30o
mg
Conservation: mgh + ½kx2 = ½mv2 + fkx
(Work)f = (mkn) x = mk(mg Cos 30o) x
Continued . . .
Example (Cont.): A 2-kg mass m located 10 m
above the ground compresses a spring 6 cm.
The spring constant is 40,000 N/m and mk = 0.4.
What is the speed when it reaches the bottom?
mgh + ½kx2 = ½mv2 + fkx
2 kg
h
10 m
x
30o
x=
10 m
= 20 m
Sin 30o
fkx = mk(mg Cos 30o) x
fkx = (0.4)(2 kg)(9.8 m/s2)(0.866)(20 m) = 136 J
mgh = (2 kg)(9.8 m/s2)(10 m) = 196 J
½kx2 = ½(40,000 N/m)(0.06 m)2 = 72.0 J
Example (Cont.): A 2-kg mass m located 10 m
above the ground compresses a spring 6 cm.
The spring constant is 40,000 N/m and mk = 0.4.
What is the speed when it reaches the bottom?
mgh + ½kx2 = ½mv2 + fkx
2 kg
h
10 m
x
30o
mgh = 196 J
½kx2 = 72.0 J
fkx = 136 J
½mv2 = mgh + ½kx2 - fkx
½(2 kg) v2 = 196 J + 72 J - 136 J = 132 J
v =11.4 m/s
Summary:
Energy Gains or Losses:
Gravitational Potential Energy
U = mgh
Spring Potential Energy
U  kx
Kinetic Energy
K  mv
Work Against Friction
1
2
1
2
2
2
Work = fx
Summary:
Conservation of Energy
The basic rule for conservation of energy:
mgho
½kxo2
½mvo2
=
mghf
½kxf2
½mvf2
+
Work
against
friction:
fk x
Remember to use the absolute (+) value of
the work of friction. (Loss of energy)
CONCLUSION: Chapter 8C
Conservation of Energy