2/12/2014
Recap: Mechanical Energy E
Last time
The mechanical energy of a body is the sum of the
kinetic energy and the potential energy of the
body.
The total work W done on an object is equal to its
change in kinetic energy
E=K+U
1
1
2
W = mv 2 − mv0 = K − K 0 = ∆K
2
2
• Characterize the instant state of the body
• U is the sum of all potential energy
associated with conservative forces
For a constant force being applied to an object,
• The change of the mechanical energy is:
W = F∆r cosθ
∆E = ∆K + ∆U
Example (text problem 6.84): A roller coaster car is about to roll down a track.
Ignore friction and air resistance.
RECAP
Conservative Force ⇒ Potential energy U
⇒ Conservative Force
Potential Energy
m = 988 kg
40 m
20 m
Mechanical Energy
E=K+U
Conservation of E of a body if only conservative
forces do work.
y=0
(a) At what speed does the car reach the top of the loop?
Ei = E f
∆E = 0
U i + Ki = U f + K f
E2 = E1
1
mgyi + 0 = mgy f + mv 2f
2
v f = 2 g ( yi − y f ) = 19.8 m/s
K 2 + U 2 = K1 + U1
4
Example continued:
Example continued:
(b) What is the force exerted on the car by the track at the top of the loop?
(c) From what minimum height above the bottom of the track can the car be
released so that it does not lose contact with the track at the top of the loop?
FBD for the car:
Using conservation of mechanical energy:
y
Apply Newton’s Second Law:
∑F
y
= − N − w = − mar = −m
x
N
N +w=m
w
N =m
Ei = E f
v2
r
U i + Ki = U f + K f
v2
r
1 2
mgyi + 0 = mgy f + mvmin
2
v2
− mg = 2.9 × 104 N
r
Solve for the starting height
5
yi = y f +
2
vmin
2g
6
1
2/12/2014
Work by Nonconservative force
Example continued:
What is vmin?
What do you do when there are nonconservative
forces? For example, if friction is present
v = vmin when N = 0. This means that
v2
r
v2
w = mg = m min
r
vmin = gr
N +w=m
∆E = E f − Ei = Wfric
The work done by
friction.
The initial height must be
yi = y f +
2
vmin
gr
= 2r +
= 2.5r = 25.0 m
2g
2g
7
Gravitational Potential Energy
Part 2
Why can we take g=const. at earth’s
surface?
Newton says force on mass m a
distance r from center of earth is
Newton says force on mass m a
distance r from center of earth is
But earlier we said
So it must be that
8
When h<<RE
F=mg
g=
GM E
≠ const.
r2
F=mg
The general expression for gravitational potential
energy is:
U (r ) = −
GM 1M 2
r
where U (r = ∞ ) = 0
≈
GM E
2
RE
10
Examples
Example
A planet of mass m has an elliptical orbit around the Sun. The
elliptical nature of the orbit means that the distance between the
planet and Sun varies as the planet follows its orbital path. How
does the speed of a planet vary as it orbits the Sun once. Take
the planet to move counterclockwise from its initial location.
Example: What is the gravitational potential energy of a
body of mass m on the surface of the Earth?
U (r = Re ) = −
GM 1M 2
GM e m
=−
r
Re
Example: Drop a stone from height of h=3 Re . What is the
speed when it hits the Earth?
v=
The mechanical energy of the planet-sun
system is:
B
r
A
1
GmM sun
E = mv 2 −
= constant
2
r
3GM e
2 Re
At the planet moves r decreases and the planet moves faster.
11
As the planet moves past point A r begins to increase and the
planet moves slower.
12
2
2/12/2014
Example: What is the work done by the variable force shown below?
Work by variable force
Fx (N)
F3
In each interval ∆x=xi+1-xi,
F is roughly constant. So
approximate
W=F∆x1+F∆x2+…F∆xN
F
F2
F1
x1
x1… xN
x
W = area under the curve
x2
x3
x (m)
The work done by F1 is
W1 = F1 ( x1 − 0)
The work done by F2 is
W2 = F2 (x2 − x1 )
The work done by F3 is
W3 = F3 (x3 − x2 )
The net work is then W 1+W 2+W 3.
Spring force
14
Work done by spring
F
d
x
Robert Hooke
1600s
F = − kx
Spring Force:
• force direction is opposite to direction of
displacement from the equilibrium point
• force is linearly proportional to displacement
• k – spring constant:
- large k: hard spring
- small k: weak spring
F = −kx
-F
Area under the curve, minus sign!!
 kx 2f kxi2 
W = −
−

 2
2 

x1… xN
x
Q: what work do you do?
As a theorist, Robert Hooke (1635-1703) is mostly remembered for arguing with Isaac Newton
over the nature of light and gravity, a long-running debate that is said to have left both men forever bitter toward each other.
Elastic potential energy
Vertical spring analysis
Lower mass such that it starts
and ends at rest.
1
− k (∆x )2
2
The work done in stretching/compressing a spring
transfers energy to the spring.
Us =
k
x=0
1 2
kx
2
Work done by gravity is
Wg = mg ∆x = m 2 g 2 / k
Work done by spring is
1
Wspr = − k ∆x 2 = − m2 g 2 /(2k )
2
W-KE theorem says
x=∆x
Wtot = Wg + Wspr + Whand = ∆K = 0
so the work
done by your hand is
17
Newton: ∆x=mg/k
Whand = −m 2 g 2 /(2k )
3
2/12/2014
Example 2
0
m
x
Example - spring and gravity
A mass m=2kg is attached to a
hanging spring with force
constant k=10N/m and then
released. The spring
tension is zero at time of
release. a) In resulting
oscillations, what is max
extension L of spring? b)
what is the speed of m when
it is at ½ the max extension?
A block of mass m is dropped from height h
onto a spring. The spring is compressed by a
length d before the body stops. Find the
spring constant k.
For the block-Earth-spring system:
1
E mec 2 = K 2 + U g 2 + U s2 = 0 + mgy 2 + kd 2
2
E mec 1 = E mec 2
1
mgy1 = mgy 2 + kd 2
2
k = 2mg(y1 − y 2 ) d 2
Pav =
P=
d
2
k = 2mg(h + d) d 2
y2
y=0
Example
Example (text problem 6.75): A race car with a
mass of 500.0 kg completes a quarter-mile (402 m)
race in a time of 4.2 s starting from rest. The car’s
final speed is 125 m/s. What is the engine’s
average power output? Neglect friction and air
resistance.
Power is the rate of energy transfer.
Instantaneous Power
y1
1
E mec1 = K1 + U g1 + U s1 = 0 + mgy1 + 0
Power
Average Power
y
∆E
∆t
W F∆r cos θ
=
= Fv cos θ
∆t
∆t
The unit of power is the watt. 1 watt = 1 J/s = 1 W.
21
∆E ∆U + ∆K
=
∆t
∆t
1 2
mv f
∆K 2
=
=
= 9.3 ×105 watts
∆t
∆t
Pav =
22
4