Gravitational potential energy. Conservation of energy
2
h
Fg  mg
Definition:
Wg  mgh  mg (h2  h1 )
1
mv22 mv12

 W  U  mgh  mg(h2  h1 )
2
2
K 2  K 1  (U 2  U 1 )
U  W
U g  mgh
K 2  U 2  K 1  U 1  const
Example: An object of unknown mass is projected with an initial
speed, v0 = 10 m/s at an unknown angle above the horizontal. If air
resistance could be neglected, what would be the speed of the object
at height, h = 3.3 m above the starting point?
v0  10m / s
mv 2
h  3.3m
v ?
v  v02  2 gh 
mv02
 mgh 
 mg 0
2
2
10m / s 2  2  9.8m / s 2  3.3m  6.0m / s
Example: A box of unknown mass and initial speed v0 = 10 m/s moves
up a frictionless incline angled 30°. How high does the box go before it
begins sliding down?
A. 2 m
B. 5 m
C. 10 m
m
30°
Only gravity does work (the
normal is perpendicular to
the motion), so mechanical
energy is conserved.
E  K U
1
E A  mv 02  0
2
EB  0  mgh
1
EA  EB 
mv 02  mgh
2
v 02
(10 m/s)2
 h

5 m
2
2 g 2(10 m/s )
The really nice thing is, we can apply the same thing to any “incline”
The really nice thing is, we can apply the same thing to any “incline”:
Turnaround
point:
where
K=0
E
K
v=0
U
E
K
U
h
E
K
U
Example: Roller Coaster
A roller coaster starts out at the top of a hill of height h. How fast is it
going when it reaches the bottom?
h
Energyinitial  Energy final
U initial  mgh
K final
mv

2
2
mv2
mgh 
2
v  2 gh
Example: Paul and Kathleen start from rest at the same time on
frictionless water slides with different shapes.
a) At the bottom, whose velocity is greater?
The only force doing work is gravity
 mechanical energy is conserved:
Ei = mgh = Ef = 1/2 mv2
gh = 1/2 v2
Since they both start from the
same height, they have the
same velocity at the bottom.
1) Paul
2) Kathleen
3) both the same
b) Who makes it to the bottom first?
Even though they both have the same final velocity, Kathleen is at a lower
height than Paul for most of her ride. Thus she always has a larger
velocity during her ride and therefore arrives earlier!
Example: A cart is released from height h in a roller coaster with a loop
of radius R. What is the minimum h to keep the cart on the track?
E A  EB
vB2
mg  N  m
R
1
mgh  0  mg 2R  mvB2
2
The minimum velocity is fixed by N = 0:
mg  m
A
2
vB,min
R
vB,min 

The minimum height is given by:
gR 
h
vB  2 g (h  2R )
R  2hmin  4R
hmin
5
 R
2
2 g (hmin  2R )
B
mg+N
R
gR
Example: A pebble of mass m sits on top of the perfectly spherical head
of a snow man. When given a very slight push, it starts sliding down.
Where does it leave the snow man’s head?
v2
mg cos  N  m
R
N
θ
As v increases, N decreases. When N = 0,
the pebble flies off.
R
mg cos  max  m
mg
Conservation of energy:
2
mgR  12 mvmax
v
2
max
R
 mgR cos  max
2 equations for vMAX and θMAX
2 gR (1  cosMAX )  Rg cosMAX
cosMAX
2

3

(MAX  48.2 )
Example: Pendulum (Conservation of energy)
Only weight of the pendulum is doing work; weight is a conservative
force, so mechanical energy is conserved:
K  U  const
L
m
K 0
U  max
θ0
θ0
U  mgh
mv 2
K
2
K  max
U  min
K 0
U  max
The angle on the
other side is also θ0.
Example: Pendulum (Conservation of energy)
The pendulum with a mass of 300 g is deviated from the equilibrium
position B to the position A as shown below. Find the speed of the
pendulum at the point B after the pendulum is released.
A. Energy of the pendulum at the point A:
U A  mgh  (0.3kg)(9.8
A
m
)(0.2m)  0.58 J
2
s
KA  0
B
h  20cm
v
B. Energy of the pendulum at the point B:
mv 2
KB 
2
UB  0
C. Conservation of energy:
K  U  const
mv 2
0  mgh 
0
2
v
2K B

m
2(0.58 J )
 1.97 m / s
0.3kg
v  2 gh  2  9.8m / s 2  0.2m  1.97m / s
Example: A 5.00 g block is pushed against a spring with k = 8.00 N/m. The
spring is initially compressed 5.00 cm and then released. The coefficient
of kinetic friction between the block and the table is 0.600. What is the
speed of the block at x = 0?
x = 0, equilibrium
x
k
m
Einitial
1 2
 0  kx
2
v?
μk
E final 
1
mv 2  0
2
1
1 2
2
mv  kx  k mgx
2
2
v 
Wfriction   k mgx
kx 2
v 
 2k gx
m
(8.00 N/m)(0.05 m)2
 2(0.600)(9.81 m/s2 )(0.05 m)  1.85 m/s
0.005 kg
Example: A box sliding on a horizontal frictionless surface runs
into a fixed spring and compresses it to a maximum distance x1
before bouncing back. If the initial speed of box is doubled and its
mass if halved, how far (x2) would the spring be compressed?
x1
A. x2 = x1
Ef =Ei
B. x2 =2 x1
1
2
kx  12 mv
2
C. x2 = 2x1
2
m
x
v
k
x2
m2 v22

x1
m1v12
Example: Spring 1 is stiffer then spring 2; that is k1 > k2. The spring force
F of which spring does more work if the springs compressed by the same
applied force?
A. W1 <W2
B. W1 >W2
C. W1 =W2
F=-kx
x=-F/k
W=½Fx=½F(-F/k)=-½F2/k
Example: A ball is shot by a vertical spring compressed over a distance x1
as shown below. It reaches a height h1 above the initial position. If the
spring is compressed over a distance x2 = 2x1, the maximum height h2 is:
1
Before the shot: E  kx12
2
(vinitial  0)
After the shot: E  mgh1
(v top  0)
1
kx12  mgh1
2
x2 = 2x1
k
h1 
x12
2mg
h2 =4h1
h
x
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 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