Energy Conservation
1. Mechanical energy conservation
E  K  U  0
E  K U
For closed isolated system
2. Open system
K  U  Eint  Wext
U  Wcons
Eint  Wnoncons
3. Conservative and nonconservative forces
•Forces such as gravity or the elastic force, for which the work dose not
depend on the path taken but only on the initial and final position, are called
conservative forces
•For conservative forces the work done on a closed path (a lop) is equal to zero
•Friction is a nonconservative force
Example: A block is moved from rest at point A to rest at point B.
Which path requires the most work to be done on the object?
A) The table is leveled and friction is present.
Path 1
Path 2
Path 3
All the same
B) The table is tilted and frictionless.
Path 1
Path 2
Path 3
All the same
B
1
2
A
3
Example: A hammer slides along 10 m down a 30 inclined roof and off
into the yard, which is 7 m below the roof edge. Right before it hits the
ground, its speed is 14.5 m/s. What is the coefficient of kinetic friction
between the hammer and the roof?
Δx = 10 m
h’
30
This can be solved using Newton’s
laws and kinematics, but it’s
looooooooooooooooooooooooong.
h=7m
v = 14.5 m/s
1
mv 2  mg x sin   k  mg cos   x  mgh
2
k 
2 g h  x sin    v 2
2 g x cos 
2(9.8)  7  10 sin30   14.52

 1.5
2(9.8)10cos30
Example: In the system below, a 200 g box is pushed 4 cm against a
spring with k = 250 N/m and released. The box slides along a frictionless
horizontal surface and then up an incline which makes an angle of 30
with respect to the horizontal. The coefficient of kinetic friction between
the box and the incline is 0.2. How far along the incline is the box when
its speed is half its maximum speed?
1) In the first part of the motion, mechanical energy is conserved.
Einitial
E final
1 2
 0  kx
2
1
 mv 2  0
2
Einitial  E final
vMAX/2
d?
μk = 0.2
θ = 30
v (MAX)  x
vMAX
k
m
 (0.04 cm)
250 N/m
0.2 kg
 1.4 m/s
Compression:
x = 4 cm
m = 200 g
k = 250 N/m
2) For the whole process, mechanical energy E = K + Ug + Uelastic is not
conserved due to friction: ΔE = Wfriction
Einitial
1
 0  0  kx 2
2 2
1  vMAX 
E final  m 
  mgh  0
2  2 
Wfriction  fkd
v (MAX)  x
k
m
2
1  vMAX 
1
2
m

mgh

kx
 fk d

2  2 
2
h  d sin 
fk  k mg cos
1
1
2
kx  mgd sin   kx 2   k mgd cos 
8
2
3(250 N/m)(0.04 cm)
3kx 2

d 
2
8mg  sin   k cos  8(0.2 kg)(9.8 m/s )  sin30  0.2cos30 
2
 0.114 m  11.4 cm
Relation between U and F (conservative force)
1D)
final
 F dx
U  W  
x
initial
dU
Fx  
dx
U    Fx dx  const
Examples:
U  mgy
U 
1
kx 2
2


dU
Fy  
 mg
dy
Fx  
dU
 kx
dx
The force is minus
the slope of the
U (x) curve
Spring
F x  kx
U  x   12 kx 2
x
x = 0, F = 0
x
x < 0, F > 0
U
U
dU/dx = 0
dU/dx < 0
x
x
x > 0, F < 0
x
x
x > 0, F < 0 and larger in
magnitude
U
U
dU/dx > 0
and steeper
than before
dU/dx > 0
x
The force always points “downhill”!!!
x
Example: Which of the force versus position graphs matches the
potential energy function U(x)?
U
Slope:
A
B
Force = − slope !
C
0
+
0
−
0
x
F
x
F
F
x
x
Relation between U and F (conservative force)
1D)
U    Fx dx  const
2 and 3D)
  
U    F (r )dr  const
Fx  
U
U
U
, Fy  
, Fz  
x
y
z
Fr  
U
r
dU
Fx  
dx


F  U
The force is minus
the slope of the
U (x) curve
The force is minus
the gradient of the
function U (x,y,z)
in cartesian coordinates
for the radial component in spherical coordinates
Visualization of a gradient in 2D
Think of a hilly terrain where U is the altitude.
The negative gradient of U is a vector whose:
• Direction points down the hill in the direction water would flow from
that location (i.e., in the steepest direction).
• Magnitude is the slope of the hill in that direction
U
U
x
y
U
y
x
Example: Find the force exerted at point P (0,1,2) m if the potential
energy associated with the force is:
U (r )  3xy  4x 2  yz 3
U
  3y  8x 
x
U
Fy  
  3x  z 3 
y
U
Fz  
   3yz 2 
z
Fx  

Fx ,P  (3  0)  3
Fy ,P  (0  8)  8
Fz ,P  ( 3  1  4)  12


F r   8 x  3 y iˆ  z 3  3x ˆj  3 yz 2
FP  3iˆ  8 jˆ  12kˆ
Equilibrium
Whenever F = 0 (ie, dU/dx = 0), we have equilibrium.
U
xS, xU and xN are points of
equilibrium
xS
xU
x
xN
Stable/unstable/neutral equilibrium
What happens if the particle moves some small dx away from the
equilibrium point?
The force pulls it away from
U
unstable the equilibrium point.
stable
neutral
x
The force brings it back
to the equilibrium point.
The force remains zero,
so the particle stays at
the new position, which
is also an equilibrium
position.
• Force points “downhill”
• Maximum = unstable equilibrium
• Minimum = stable equilibrium
• Turning points : E =U (so K = 0)
Energy Diagrams
Example 1: A box attached to a spring on a horizontal, frictionless table is
released at x = x0 from rest.
2
1
U
E  K  U  0  2 kx0
UMAX ( = E )
KE = 0, turn-around
points
1
E  kxo2
2
UMIN
KEMAX
–x0
Forbidden region
(KE < 0)
x0
Particle
moves here
x
Forbidden region
(KE < 0)
Example 2: The box is brought to x = x0 and pushed, so its initial
velocity is v0.
1
1 2
2
E  KE  U  mvo  kxo
2
2
U
New turn-around points.
E 
1
1
mvo2  kxo2 (now)
2
2
1 2
E  kxo (before)
2
–xt
x0 xt
x
U
U=0
U = UMAX = E
KE = KEMAX = E
KE = 0
KE
E
KE
U
–xt
U
xt
x
How much kinetic/potential energy does the
system have at every point?
U
Example: Potential with two pits.
A particle is subjected to the force associated with this
potential. No other forces are exerted on the particle.
Describe the motion of the particle in the following cases.
x
1. The particle is released from rest at point A.
E  K U
From the initial conditions, E  0  UA
At M1, U is minimum, so K (and speed) is maximum
At xB, U = E, so K (and speed) is zero
→ turn around point
The particle is forbidden from x < xA or x > xB (K < 0)
U
The particle oscillates between A and B.
A
M1 B
x
UA
forbidden OK
E
OK
forbidden
Direction of force F
2. The particle is released at point A with a small* initial velocity v0.
1
mv02  UA  UA (*but not too much larger)
2
At M1, U is minimum, so K (and speed) is maximum
From the initial conditions, E 
The turn-around points are defined by K = 0, so U = E : points C and D.
U
The particle oscillates between C and D.
CA
M1
D
x
E
UA
forbidden
OK
OK
forbidden
Direction of force F
3. The particle is released from rest at point G.
From the initial conditions, E  UG
At M1, U is minimum, so K (and speed) is maximum
The particle keeps moving in the +x direction (no
oscillations).
U
UG
E
M1
x
G
forbidden
OK
OK
OK
Direction of force F
4. The particle is released from rest at point H. The particle
has maximum speed at point:
A. M1
B. M2
C. M3
The particle oscillates
between H and J.
U
J M3 H
M1
M2
forbidden
x
E
OK
forbidden
Direction of force F
5. The particle is released from rest at point K.
From initial conditions, E = UK
The particle oscillates between K and L.
U
E
L
forbidden
x
K
OK
forbidden
6. The particle is released from rest at point M1.
From initial conditions, E = UM1
U
→
Equilibriu
m
If someone pushes the particle
Stable
slightly away from M1, the force
equilibriu
pushes it back.
m
Force = 0 with v = 0
M1
x
Direction of force F
E
7. The particle is released from rest at point M2.
From initial conditions, E = UM2
Force = 0 with v = 0
U
→
Equilibrium
If someone pushes the particle slightly
away from M1, the force pushes it further
away.
Unstable equilibrium
E
x
M2
Direction of force F