Transcript Lecture 1

“He had the odd feeling of being like
a man in the act of adultery who is
surprised when the woman's
husband wanders into the room,
changes his trousers, passes a few
idle remarks about the weather and
leaves again.”
The Hitchhiker's Guide to the Galaxy
Next Test will cover:
Chapter 10: Springs and elasticity
Chapter 11: Fluids and pressure
Chapter 16: Waves and sound
How do you think you scored on Test 2?
A) 13 or more
B) 10-12
C) 7-9
D) 4-6
E) <4 (really?!)
Chapter 10: Springs. Please READ THE ENTIRE
CHAPTER.
Springs have an equilibrium length. To compress or
stretch a spring requires force.
Chapter 10: Springs.
springs have an equilibrium length. To compress or
stretch a spring requires force.
Once stretched or compressed, a spring tries to
return to the equilibrium length/position.
Chapter 10: Springs.
springs have an equilibrium length. To compress or
stretch a spring requires force.
The amount of force required to stretch/compress a
spring is directly proportional to how far from
equilibrium it has moved, and the stiffness of the
spring. This is called Hooke's Law.
Chapter 10: Springs.
springs have an equilibrium length. To compress or
stretch a spring requires force.
Fx = -kx where k is called the spring constant and is
a measure of a spring's stiffness.
Chapter 10: Springs.
springs have an equilibrium length. To compress or
stretch a spring requires force.
Fx = -kx where k is called the spring constant and is
a measure of a spring's stiffness.
Notice that the force always acts opposite to how the
spring is moved.
Chapter 10: Springs.
Fx = -kx where k is called the spring constant.
Example: A spring with k=3.67 N/m is stretched by
0.4 m. How much force does it take to hold the
spring in this position?
A) 0.4N
B) -0.4N
C) 1.5N
D) -1.5N
Chapter 10: Springs.
Fx = -kx where k is called the spring constant.
Example: A spring with k=3.67 N/m is stretched by
0.4 m. How much force does it take to hold the
spring in this position?
Fx = -(-kx) = 3.67(0.4) = 1.47 N which is C)1.5N
Why wasn't it -1.5N?
Chapter 10: Springs.
Fx = -kx where k is called the spring constant.
Example: A spring with k=3.67 N/m is stretched by
0.4 m. How much force does it take to hold the
spring in this position?
Fx = -(-kx) = 3.67(0.4) = 1.47 N which is C)1.5N
Why wasn't it -1.5N?
Spring
Me
holding
the
spring in
place.
Chapter 10: Springs.
Fx = -kx where k is called the spring constant.
Example: a spring with a 0.1N weight is at
equilibrium. When an additional 0.1N weight is
added, the spring stretches by 4cm. What is the
spring constant?
How would you do this
problem?
Chapter 10: Springs.
Fx = -kx where k is called the spring constant.
Example: a spring with a 0.1N weight is at
equilibrium. When an additional 0.1N weight is
added, the spring stretches by 4cm. What is the
spring constant?
k = |F/x| (spring constants are
always positive)
k = 0.1 / 0.04 = 2.5 N/m
Springs: Potential Energy
When a mousetrap is
set, the energy within
the spring is just
waiting to be released,
converting its energy to
kinetic energy of the
wire in the trap. That
amount of energy is
PE = ½kx2 where
again x is how far
displaced from
equilibrium the spring
is.
Springs: Potential Energy
PE = ½kx2 What is the potential
energy stored in a typical
mousetrap if the radius of the
moving part is 0.0814 m (3.2
inches), and the spring constant
(k) is 96.5 N/m?
Can you do this problem?
(Hint: Think about what X is)
A) 0.320J
C) 3.16J
B) 0.639J
D) 12.35J
Springs: Potential Energy
PE = ½kx2 What is the potential
energy stored in a typical
mousetrap if the radius of the
moving part is 0.0814 m (3.2
inches), and the spring constant
(k) is 96.5 N/m?
Can you do this problem?
PE = ½ kx2 in this case 'x' is the arc length the trap
moves in its motion. It covers p radians, so X = s =
rq
= 0.0814(p) = 0.256 m.
Springs: Potential Energy
PE = ½kx2 What is the potential
energy stored in a typical
mousetrap if the radius of the
moving part is 0.0814 m (3.2
inches), and the spring constant
(k) is 96.5 N/m?
Can you do this problem?
PE = ½ kx2 with x = s = rq = 0.256 m.
PE = ½(96.5)(0.2562) = 3.16 J This is C) 3.16J
Springs: Potential Energy
PE = ½kx2 Example problem: A child's
toy is made to shoot a ping-pong ball
(m=2.7 g) out of a tube. When the ball is
loaded, the spring (k=18N/m) is
stretched 9.5cm inside the tube. If the
toy is shot straight up (neglecting any
friction/air resistance), how high will the
ball go?
Can you do this problem?
A) Yes
B) No
Springs: Potential Energy
PE = ½kx2 m=2.7x10-3 kg x= 0.095m
k=18N/m friction = 0, ho= 0, hf=?
This is an energy
conservation/conversion problem.
Converting the spring's potential energy
into gravitational potential energy:
PES= PEg
A) 1.2m
B) 2.1m
C) 3.0m
D) 4.5m
Springs: Potential Energy
PE = ½kx2, m=2.7x10-3, kg x= 0.095m,
k=18N/m, friction = 0, ho= 0, hf=?
Energy conservation.
Converting the spring's potential energy
into gravitational potential energy:
PES= PEg
½ kx2= mgh Did you get this far?
Springs: Potential Energy
PE = ½kx2, m=2.7x10-3, kg x= 0.095m,
k=18N/m, friction = 0, ho= 0, hf=?
Can you do this problem? Converting the
spring's potential energy into
gravitational potential energy: PES= PEg
½ kx2= mgh
½ (18)(0.0952) = (2.7x10-3)(9.8)hf
0.0812 = 0.0265hf → hf= 3.06 m C) 3.0m
A problem for you tonight:
A block (m=5.0kg) is released from point A
and it slides down the incline where
the coefficient of kinetic friction
is 0.3. It hits a spring with a
spring constant of 500 N/m.
While it is being acted upon by
the spring, assume it is on a frictionless surface.
a) How far does the block go up the plane on the
rebound from the spring?
b) How far is the spring compressed?
Answer revealed on Monday after spring break.
Since W(ork) = DE, if we
can change the energy of a
spring system, we have
done work.
W = ½ k(xf - xo)2
W = ½ k(xf – xo)2
How much work does it take
to stretch a spring 20cm if it
has a spring constant of
545N/m?
W = ½ k(xf – xo)2
How much work does it take
to stretch a spring 20cm if it
has a spring constant of
545N/m?
W= ½ (545)(0.2-0.0)2
W= 10.9J
If you take a block attached
to a spring on a frictionless
surface and you compress
it, what's going to happen?
If you take a block attached
to a spring on a frictionless
surface and you compress
it, what's going to happen?
The block moves back to
the equilibrium point, but will
it stop when it gets there?
If you take a block attached
to a spring on a frictionless
surface and you compress
it, what's going to happen?
The block moves back to
the equilibrium point, but will
it stop when it gets there?
No, it will move past that
point, after which the spring
force will act in the opposite
direction.
If you take a block attached
to a spring on a frictionless
surface and you compress
it, what's going to happen?
The block moves back to
the equilibrium point, but will
it stop when it gets there?
No, it will move past that
point, after which the spring
force will act in the opposite
direction. If the surface is
truly frictionless, it will
continue moving back-andforth forever!
Simple Harmonic Motion
Simple Harmonic Motion: the mass bounces back
and forth. If you plot its position with time, you get
the graph shown.
Simple Harmonic Motion: the mass bounces back
and forth. If you plot its position with time, you get
the graph shown. A is the amplitude: this is the
maximum distance (displacement) away from
equilibrium the mass will move.
Simple Harmonic Motion: the mass bounces back
and forth. If you plot its position with time, you get
the graph shown. A = amplitude. T is the time (in
seconds) it takes to complete one period (down, up
and back to the equilibrium position). This is the
same as 1/frequency (units are Hertz, which is
1/seconds). 10 sec = 0.1Hz
Simple Harmonic Motion: the mass bounces back
and forth. If you plot its position with time, you get
the graph shown. A = amplitude. T = time
f = frequency. w is the angular frequency (not the
angular velocity we had before (the relation to what
we had before is wf= 2p.wv )
Simple Harmonic Motion: the mass bounces back and
forth. If you plot its position with time, you get the
graph shown. A = amplitude. T = time
f = frequency. w = angular frequency. The position at
any time can also be determined.
And there
are
relations to
the spring
constant
and mass.
Example: A ball (m=0.2kg) is bouncing (A=6cm) on a
spring (k=15N/m) and crosses equilibrium at t=0.
Where is it 12.3 seconds later? Can you solve this
problem?
Example: A ball (m=0.2kg) is bouncing (A=6cm) on a
spring (k=15N/m) and crosses equilibrium at t=0.
Where is it 12.3 seconds later? Can you solve this
problem? Use the second part of the y equation.
Example: A ball (m=0.2kg) is bouncing (A=6cm) on a
spring (k=15N/m) and crosses equilibrium at t=0.
Where is it 12.3 seconds later? Can you solve this
problem? Use the second part of the y equation.
y = A sin[(k/m)1/2]t = 0.06 sin[(15/0.2)1/212.3]
Example: A ball (m=0.2kg) is bouncing (A=6cm) on a
spring (k=15N/m) and crosses equilibrium at t=0.
Where is it 12.3 seconds later? Can you solve this
problem? Use the second part of the y equation.
y = A sin[(k/m)1/2t] = 0.06 sin[(15/0.2)1/212.3] =
0.06*sin(8.66*12.3) = 0.06(-0.29)
= -0.0175 m