Download Force = -kx Springs

10/17/2011
Announcements
Clicker Question
• CAPA Set #9 due Friday at 10 pm
• This week in Section: Lab #3 – Momentum
Room Frequency BA
Two paths lead to the top of a big hill.
Path #1 is steep and direct and Path #2 is twice as long but less steep.
Both are rough paths and you push a box up each.
How much more potential energy is gained if you take the longer path?
• Reading – Chapter 7 on Momentum
A) none
B) twice as much
C) four times as much
D) half as much
d
h
#1
2d
h
#2
θ
d 2  h2
φ
4d 2  h2
PE  mgh in both cases.
Clicker Question
Room Frequency BA
Springs
Consider the same two paths as in the previous problem.
Is the amount of work done the same along each path?
A) Yes
B) No, more work is done on the steeper path.
C) No, more work is done on the longer path.
D) Probably not, but the details are needed to know which
is larger.
d
h
#1
d h
h
#2
  mg
φ
4d 2  h2
2
| W f |  Ff d   Nd   (mg cos )d
d 2  h2
d   mg d 2  h 2
d
Force
Compressed spring
pushes back
Force
Stretched spring
pulls back
2d
θ
2
Spring in “relaxed” position
exerts no force
| W f |  Ff d   N(2d)   (mg cos  )(2d)
  mg
Hooke’s “Law”
Force = -kx
(2d)2  h 2
2d   mg 4d 2  h 2
2d
Clicker Question
Stored Energy in Springs
Compressed spring has elastic potential energy
Stretched spring also has elastic potential energy
If no change in any other energy, then PE = Wext =Fspring x d
However Fspring = -kx changes as
we stretch the spring
Use average Fspring = -½ kx
Elastic potential energy:
k = “spring constant”
units = N/m
x = displacement of spring
PE = ½ kx2
Room Frequency BA
A force of magnitude F stretches a spring through a
displacement x. The force is then increased so that
displacement is doubled.
F = -kx, PE = ½kx2
What is the magnitude of force needed to double the
displacement and what will be the effect on the
potential energy of doubling displacement?
A)
B)
C)
D)
E)
Force doubles, PE doubles.
Force doubles, PE quadruples.
Force quadruples, PE doubles.
Force doubles, PE halves.
Force quadruples, PE quarters.
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10/17/2011
Clicker Question
Room Frequency BA
Which of the following types of potential energy can
be negative?
A) Gravitational Potential Energy
PEgravity = mgy
B) Elastic Potential Energy
PEelastic 
1 2
kx  0
2
C) Both of the above
D) Neither of the above
A block of mass m is released from rest at height H
on a frictionless ramp. It strikes a spring with spring
constant k at the end of the ramp.
How far will the spring compress (i.e. x)?
0
0
KEi + PEi + Wfrict + Wexternal = KEf + PEf
0 + mgH + 0 +
x
2mgH
k
= 0 + (0 + ½ kx2)
0
gravitational
elastic
Elastic Potential Energy – Does Not Have to Look Like a Spring
What is Power?
Power is a rate of energy per time
P  average power =
work
time
Units are Joules / second = Watts
The average power can be written in terms of the
force and the average velocity:
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10/17/2011
Raise a box of mass 2 kg at constant velocity by a
height of 6 meters over a time interval of 2 seconds.
How much energy is required to do
the work?
Wext = PE = mgh
6m
Also Wext = Fext x d  Fext = 117J/6m = 19.6 N
How much power is required to do
the work in that amount of time?
Power = Wext / time = 117 J / 2 s = 59 Watts
Also Power = Fext v = 19.6 N x (6m/2s) = 59 Watts
Collisions
m1
Approximately how many 100 Watt lightbulbs would
need to be left on all the time to amount to that energy
consumption?
xx
A) 1 lightbulb
B) 10 lightbulbs
C) 100 lightbulbs
D) 1000 lightbulbs
Energy = Power x time
= (10 x 100 W) x (30x24x60x60 s)
E) 10,000 lightbulbs
≈ 2.6 x 109 Joules
Elastic Collision
Before the collision:
v2i
v1i
Room Frequency BA
A kiloWatt-hour is a unit of energy (Power x time).
1 kW-hour = 3.6 x 106 Joules
The average household in the USA uses 2.7 billion Joules
of electrical energy each month.
since KE=0
Wext = 2 kg x 9.8 m/s x 6 m = 117 Joules
2 kg
Clicker Question
m2
Total kinetic energy remains constant
during the collision
KE is conserved or ΔKE = 0
After the collision:
v1f
m1
v2f
No energy converted to thermal or potential energy
m2
v1f = final velocity of mass 1
v2f = final velocity of mass 2
These collisions should obey Newton’s Laws
Demonstration
KE1i  KE2i  KE1 f  KE2 f
1
1
1
1
m1v1i 2  m2 v2i 2  m1v1 f 2  m2 v2 f 2
2
2
2
2
New Conserved Quantity  Momentum
momentum
p  mv
(a vector)
Kind of like kinetic energy (associated with motion).
Macroscopic objects never have perfectly elastic
collisions but it’s often a good approximation.
Inelastic Collision: KE is not constant;
some is converted to another kind of energy;
typically heat or deformation.
Totally Inelastic Collision: KE is not conserved, and the
objects stick together after collision.
However, not in units of Joules (instead kg m/s)
and so not an energy.
In addition to total energy being conserved in every collision,
the total momentum vector is also conserved!
p1i  p2i  p1 f  p2 f
mv1i  mv2i  mv1 f  mv2 f
Conservation of Momentum
(in one-dimension)
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