Transcript PHYS 218 - Texas A&M University

PHYS 218
sec. 517-520
Review
Chap. 7
Potential Energy and
Energy Conservation
Potential energy
• Energy associated with the position of bodies in a system
• A measure of the potential or possibility for work to be done
• The work done on the object to change its position is stored in the object in
the form of an energy
Gravitational potential energy
Work done by gravity = Wgr = Fs = w( y1 - y2 ) = mgy1 - mgy2
We can define the gravitational potential energy as
U gr = mgy
y2
Then
Wgr = U gr,1 - U gr,2 = - (U gr,2 - U gr,1 ) = - D U gr
y1
When the body moves up, the work done by the gravitational
force is negative and the potential energy increases.
Potential energy
The potential energy is a relative quantity. You have to specify the reference
point when you define the potential energy. For example, the gravitational
potential energy is mgy and the point where y = 0 should be specified, which is
the reference point when you define the potential energy.
What is physically meaningful is the change of the potential energy. The
absolute value does not have physical meaning. Note that the work done by a
force is equal to the negative of the potential energy change.
W = - DU
Conservation of mechanical energy
r
Consider the gravitational force only so that Fother = 0
Work-Energy theorem gives Wtot = D K = K 2 - K1
The definition of the potential energy gives Wtot = - D U gr = U gr,1 - U gr,2
Putting these together, D K = - D U gr or K 2 - K1 = U gr,1 - U gr,2
K1 + Ugr,1 = K2 + Ugr,2
This defines the total mechanical energy of the system.
E = K + U gr =
1 2
mv + mgy
2
Conservation of mechanical energy
E = K + U gr =
1 2
mv + mgy = constant
2
So, K + U is conserved
Ex 7.1
Height of a baseball from energy conservation
Energy conservation is very useful to obtain speed or position, in particular, when it its
very difficult to use Newton’s laws of motion.
The maximum height can be obtained easily by using energy conservation.
y = y2
v1
y = y1 = 0
The initial total energy is
1
Eini = K1 + U gr,1 = mv12 + 0 (Q y1 = 0)
2
The final total energy is
E fin = K 2 + U gr,2 = 0 + (Q v2 = 0)
Then
Eini = E fin
1 2
Þ
mv1 = mgy2 Þ
2
v12
y2 =
2g
When forces other than gravity do work
In this case, Wtot = Wgr + Wother .
The Work-Energy theorem gives Wtot = K 2 - K1
Wgr can be related to the potential energy as Wgr = U gr,1 - U gr,2
Putting altogether,
K 2 - K1 = U gr,1 - U gr,2 + Wother
Þ K1 + U gr,1 + Wother = K 2 + U gr,2
Wother = E2 - E1
Therefore, for example, if there is friction force, the total mechanical energy
is not conserved. Instead, we have E1 + Wother = E2
Ex 7.2
Work and energy in throwing a baseball
y2 = 0.5 m, v2 = 20 m/s, what is F ?
What is its speed when y3 - y2 = 15 m?
From y = 0 to y1
State 3
y = y3
K1 = 0, U gr,1 = mgy1 = 0
1 2 1
mv2 = (0.145 kg)(20 m/s) 2 = 29 J,
2
2
U gr,2 = mgy2 = (0.145 kg)(9.8 m/s 2 )(0.5 m) = 0.71 J
K2 =
v2
Wother = E2 - E1 = ( K 2 + U gr,2 ) - ( K1 + U gr,1 ) = 29.7 J
State 2
y = y2
\ F=
y3
y = y1 = 0
State 1
F
Wother
29.7 J
=
= 59 N
y2 - y1
0.5m
From y = y1 to y2
K 3 = ( K 2 + U gr,2 ) - U gr,3 = (29.7 J) - (0.145 kg)(9.8 m/s 2 )(15.5 m) = 7.7 J
\
v3 = ±
2 K3
= ± 10 m/s
m
Two solutions; moving up
and moving down
Here we use different choice for the y-axis from the textbook. But the final answers are the same as it should be.
Gravitational potential energy for motion
along a curved path
Fother
r
Work done by the gravitational force during the displacement D s
r r
w ×D s = - mg ˆj ×(D x iˆ + D y ˆj ) = - mgD y
y1
w = mg
y2
Dx
Dy
w = mg
Only Dy contributes
The total work becomes
r
Ds
Wgr = - mg ( y2 - y1 )
The total work done by the gravitational force
depends only on the difference in height
Ex 7.4
Calculating speed along a vertical circle
Point 1
Speed at the bottom of the ramp
1 2
mv2 , U gr,2 = 0
2
Conservation of mechanical energy gives
K1 = 0, U gr,1 = mgR, K 2 =
R
n
K1 + U gr,1 = K 2 + U gr,2
Þ v2 =
Point 2
w
2 gR
What we need is speed
not velocity, so v2 is
positive
Normal force at the bottom of the curve
At the bottom of the curve, it is describing a circular motion. Therefore, it should have centripetal acceleration,
v22 2 gR
which is arad =
=
= 2 g.
R
R
The net force acting on the man at the bottom of the curve is
å
Fy = n - w = n - mg , which should give marad = 2mg .
Þ n - mg = marad = 2mg Þ n = 3mg
Ex 7.5
Vertical circle with friction: same as Ex 7.4 but there is friction force. What
is the work done by the friction force?
By the work energy theorem and potential energy,
K1 + U gr,1 + Wother = K 2 + U gr,2
Þ Wother = K 2 + U gr,2 - (K1 + U gr,1 ) =
1 2
mv2 - mgR
2
Note that v2 in this case is less than 2 gR .
This is natural as it lost energy by friction force.
Compare the values of v2 in Ex. 7.4 and Ex. 7.5 in the textbook.
Ex 7.6
Inclined plane with friction
d1
d2
Point 2
q
Motion of a crate:
Point 1 (speed v1)
g Point 2 (speed v2 = 0)
g Point 3 (speed v3)
equal to Point 1
h2
Magnitude of a constant friction force
Point 1, 3
Consider the motion from Point 1 to Point 2
K1 =
1 2
mv1 , U gr,1 = 0,
2
K 2 = 0, U gr,2 = mgh2 = mgd 2 sin q, W fr = - fd 2
W fr = - fd 2 = (K 2 + U gr,2 )- (K1 + U gr,1 ) = mgd 2 sin q \
f =-
ö
1æ
ççmgd 2 sin q - 1 mv12 ÷
÷
÷
ø
d 2 çè
2
1 2
mv1
2
Speed at Point 3
Consider the motion from Point 1 to Point 3
During the motion from Point 1 to Point 3,
W fr = - f (2d 2 ) = - 2 fd 2
Then
K1 + U gr,1 + W fr = K 3 + U gr,3
Þ
1
K3 = K1 + U gr,1 + W fr - U gr,3 = mv12 - 2 fd 2
{
{
2
0
1 2 1 2
mv3 = mv1 - 2 fd 2
2
2
4 fd 2
\ v3 = v12 m
0
Þ
What we want is speed
not velocity, so v3 is
positive
Be careful:
Speed is always
positive
while velocity
can be negative
Elastic potential energy
Energy stored in an ideal spring.
This is a potential energy that is not gravitational in origin. Although this is not one
of the fundamental forces in nature, its potential energy can be defined.
r
s
x1
x2
r
Fspring = - kx
1 2 1 2
kx2 - kx1
2
2
1
1
Work done by the spring: Wel = kx12 - kx22
2
2
1
Define the elastic potential energy as U el = kx 2
2
Then Wel = U el ,1 - U el ,2 = - D U el
Work done on the spring: W =
If the elastic force is the only force that does work on the body, then
Wtot = Wel = K 2 - K1 by the work-energy theorem, which leads to
K1 + U el ,1 = K 2 + U el ,2
Gravitational potential energy plus Elastic potential energy
The Work-Energy theorem holds for the total work.
If we have both gravitational and elastic forces as well as other forces,
Wtot = Wgr + Wel + Wother = K 2 - K1
and
Wgr = - D U gr , Wel = - D U el , so that
K1 + U gr ,1 + U el ,1 + Wother = K 2 + U gr ,2 + U el ,2
Total potential energy = sum of potential energies
Ex 7.7
Motion with elastic potential energy
m
Point 1
x1
v2
Point 2
m = 0.2 kg, k = 5.0 N/m
v1 = 0, x1 = 0.1 m, x2 = 0.08 m
m
What is v2?
x2
By energy conservation, K1 + U1 = K 2 + U 2 Þ K 2 = K1 + U1 - U 2
From the given conditions,
1
1
1
1
2
2
K1 = mv12 = (0.2 kg )(0) = 0, U1 = kx12 = (5 N/m)(0.1 m) = 0.025 J
2
2
2
2
1
1
2
U 2 = kx22 = (5 N/m)(0.08 m) = 0.016 J
2
2
v2 = + 0.3 m/s is the speed
1
when the object is moving to the right
Þ K 2 = mv22 = K1 + U1 - U 2 = 0.09 J
2
2K2
\ v2 = ±
= ± 0.3 m/s
m
The speed at point 2 is v2 = - 0.3 m/s
Ex 7.8
Motion with elastic potential energy and work
done by other forces: similar to Ex 7.7
F
m
m = 0.2 kg, k = 5.0 N/m, F = 0.61 N
v1 = 0, x1 = 0, x2 = 0.1 m
Then v2 ?
K1 = 0 (Q v1 = 0), U1 = 0 (Q x1 = 0)
m
U2 =
x2
Wother
Since
K1 + U1 + Wother = K 2 + U 2 ,
K 2 = K1 + U1 + Wother - U 2 = 0.036 J
\ v2 =
2K2
=
m
2 (0.036 J )
= 0.6 m/s
0.2 kg
1 2 1
2
kx2 = (5 N/m)(0.1 m) = 0.025 J
2
2
= Fs = Fx2 = (0.61 N)(0.1 m)= 0.061 J
The object is moving to the right. So
the negative value, -0.6 m/s, is not the
solution although it is a mathematical
solution.
Ex 7.9
Motion with gravitational, elastic, and friction forces
f = 17, 000 N
v2 = 0
m=2000 kg
h= 2 m
What is the spring constant k ?
y= 0
Point 1
h
v1 =
Choose Point 1 as y = 0
1 2
mv1 , U gr ,1 = 0 (Q y1 = 0),
2
U el ,1 = 0 (Q the spring is not compressed yet)
K1 =
4 m/s
Point 2
w = mg
Wother = - f h
K 2 = 0 (Q v2 = 0),
U gr ,2 = mgy2 = - mgh (Q y2 = - h), U el ,2 =
Þ K1 + U gr ,1 + U el ,1 + Wother = K 2 + U gr ,2 + U el ,2
Þ U el ,2 = K1 + U gr ,1 + U el ,1 + Wother - ( K 2 + U gr ,2 ) = K1 + Wother - U gr ,2
\ k=
2 (K1 + Wother - U gr ,2 )
h
2
= 1.06´ 104 N/m
1 2
kh
2
Conservative and nonconservative forces
Allows two-way conversion between kinetic and potential
energies
Conservative force
Properties of the work done by a conservative force
1. The potential energy function can be defined.
2. It is reversible.
3. It is independent of the path of the body; depends only on the starting &
ending points
r r
4. When the starting & ending points are the same the total work is zero. ò
ÑF ×dl = 0
f
I
i
II
WI =
ò
f
i along I
r r
F ×dl , WII =
This leads to
r r
f
òÑF ×dl = ò
i along I
= WI - WII = 0
r r
F ×dl +
ò
f
i along II
ò
i
f along II
r r
F ×dl , then WI = WII .
r r
F ×dl =
ò
f
i along I
r r
F ×dl -
ò
f
i along II
r r
F ×dl
Nonconservative
force
Does not allow two-way conversion between kinetic and
potential energies
1. The work done by a nonconservative force cannot be represented by a
potential energy.
2. Under the influence of some nonconservative force, the body looses its
energy. So this is also called a dissipative force.
3. Under the influence of some nonconservative force, the body gets its energy.
Potential energy cannot be defined for nonconservative
forces!
Ex 7.11
Conservative or nonconservative?
r
F = Cxjˆ (C > 0) : This force is in + y direction and depends on the value of x
Is this force conservative?
A conservative force satisfies the condition
r r
òÑF ×dl = 0 for "any" closed path.
y
Leg 3
r
dl
( L, L)
r
dl
r
dl
r
dl
r
r
Along Leg 1, W1 = 0 since F ^ dl
r
( L,L) r
Along Leg 2, W2 = ò
F ×dl
Leg 2
Leg 4
(0, L )
( L ,0)
x
(0, 0)
Leg 1
( L, 0)
ò
y= L
CLdy = CL2 (Q x = L along this path)
y= 0
r
r
Along Leg 3, W3 = 0 since F ^ dl
=
Along Leg 4, W4 = 0 since x = 0 along this path, so F = 0
4
W=
å
Wi = CL2 ¹ 0
i= 1
r
F
\ This force is not conservative.
Force and potential energy
integrate
Force
Potential energy
differentiate
For 3-dim case
For 1-dim case
W = - DU =
ò Fdx
Þ
dU ( x)
dx
F ( x) = -
Fx = -
¶ U ( x, y, z )
¶ U ( x, y, z )
¶ U ( x, y, z )
, Fy = , Fz = ¶x
¶y
¶z
partial derivative
r
F= -
æ¶ U
¶U ˆ ¶U
çç
iˆ +
j+
çè ¶ x
¶y
¶z
r
¶ ˆ ¶ ˆ ¶ ˆ
Ñ=
i+
j+
k
¶x
¶y
¶z
r
ö
÷
ˆ
k ÷=
- Ñ U ( x, y, z ) : gradient of U
÷
÷
ø
Energy diagrams
Energy diagram: a graph of energy vs position
It contains the shape of the potential energy and the total energy is a straight
horizontal line as it is a constant once a body is given an energy.
This allows to know the motion of the body even if the functional form for the
potential energy is not known.
U ( x)
F
F
Slope of the tangent line is positive,
so the force is negative.
Slope of the tangent line is negative,
so the force is positive.
The body is moving always toward
the minimal potential energy point.
maximum of the potential: unstable
equilibrium point
U
Turning point
E1
E0
xa x1
xb
x4
x
Minimum of the potential: stable
equilibrium point
If E = E0 , the body stays at x = x1
If E = E1 , the body is trapped between x = xa and x = xb
in this case, the regions where x < xa and x > xb are unphysical
If E > U ( x4 ), the body can escape to x > x4