Physics 1A
Lecture 5C
"For those who want some proof that
physicists are human, the proof is in the
idiocy of all the different units which they
use for measuring energy.”
--Richard Feynman
Mechanical Energy
When dealing with macroscopic systems
(large objects) it is nice to deal with
mechanical energy:
Emec = KE + PE
In order to change Emec for a system, we need
to perform some kind of work on the system
due to an external force.
Example: By lifting a ball, I have changed the
potential energy of the system (ball and
Earth) and thus the mechanical energy of the
system.
Mechanical Energy
In an isolated system where only conservative
forces cause energy changes, Emec is constant.
This is the principle of conservation of
mechanical energy. In equation form:
If Wnc = 0,
ΔEmec = ΔKE + ΔPE = 0
KE and PE may both change, but the sum is
constant
Thus, if the work done on the system is zero,
then for the two times 1 and 2:
KE1 + PE1 = KE2 + PE2
Mechanical Energy
Example
A block slides across a horizontal,
frictionless floor with an initial velocity of
3.0m/s, it slides up a ramp which makes a
30o angle to the horizontal floor. What
distance will it travel up the ramp?
vo
Δy
30o
Answer
First, you must define a coordinate system.
Let’s choose the upward direction as positive and
make the floor to be y = 0.
Mechanical Energy
Answer
Next, use conservation of mechanical energy:
While on the floor:
Efloor = PE1 + KE1 = 0 + KE1 = (1/2)mvo2
•
While at its maximum height:
•
•
Emax = PE2 + KE2 = PE2 + 0 = mgΔy
Since no energy is taken away from the system
between the initial and final points:
Answer
Mechanical Energy
But this is not the distance up the ramp.
We need to turn to trigonometry to solve this
problem.
Make a triangle to solve:
d
30o
Δy = 0.46m
In Class Question
Blocks A and B, of equal mass, start from rest
and slide down the two frictionless ramps shown
below. Their speeds at the bottom are vA and
vB. Which of the following equations regarding
their speeds is true?
B
A
1m
1m
30o
A) vA > vB
•
B) vA = vB
•
C) vA < vB
60o
Springs
Springs are interesting objects that can store
energy (PE).
When the spring is at some equilibrium
position it will not exert a net force.
But as I displace this spring from equilibrium it
will resist this displacement (Δx).
This resistance (also known as a restoring
force, Fspring) will resist either a compression
or a stretching.
In general, each spring will have a different
resistance to a certain displacement.
Spring Potential Energy
Hooke’s Law describes the restoring force with
respect to displacement:
where k is a constant of proportionality also known as the
spring constant (units of k are [N/m]).
The minus sign in Hooke’s Law is to show that the restoring
force is opposite in direction to the displacement vector.
Please note that the restoring force in Hooke’s Law is not
constant, as we increase displacement the restoring force will
increase linearly.
Spring Potential Energy
The force of the spring will get stronger as you
pull it farther from equilibrium.
This means that you will have to put more and
more energy into the spring for more
displacement.
If you want to calculate the work in order to
compress a spring you can look at:
But force is not constant over the entire distance, it
varies with distance, which value would you use here?
Spring Potential Energy
Use the average value for force, force varies
from 0 to F linearly as you increase
displacement.
The average value for force
will then be (1/2)F.
So, for a spring the work will
be:
Spring Potential Energy
This work that you perform on the spring to
compress it, will be available as potential energy
for possible later use (spring force is
conservative).
The equation for spring potential energy becomes:
This spring potential energy will become a part of
the equation for mechanical energy as:
PE = PEgrav + PEspring
•
Emec = KE + (PEgrav + PEspring)
Spring Potential Energy
Example
A block of mass 12.0kg slides from rest
down a frictionless 35.0o incline and is
stopped by a strong spring with
k=3.00x104N/m. The block slides 3.00m
from the point of release to the point where
it comes to rest against the spring. When
the block comes to rest, how far has the
spring been compressed?
Answer
First, you must define a coordinate system.
Let’s choose the down the incline as the positive
x-direction and let the lowest position be x = 0.
Spring Potential Energy
Answer
Are there any non-conservative forces at work in this problem?
No, the only forces performing work are gravitational force and
spring force. So:
ΔEmec = ΔKE + ΔPE = 0
The block starts at rest and ends at rest such that ΔKE = 0.
This leaves us with:
ΔPE = ΔPEgrav + ΔPEspring = 0
mgΔh + (1/2)k(Δx)2 = 0
-mgΔh = (1/2)k(Δx)2
Spring Potential Energy
Answer
To calculate the change in height of the block use:
•
block
Δh = (3.00m) sin35o
Δh = –1.72m
x=3.00m
Δh
35o
Mechanical Energy
In a system where non-conservative forces can
cause energy changes, Emec may not be constant for
the system.
We change the principle of conservation of
mechanical energy in the following way:
Wnc = ΔEmec = ΔKE + ΔPE
where Wnc is the work done by non-conservative
forces.
When starting energy problems, the first question
you need to ask is: Are non-conservative forces
performing work on the system?
In Class Question
A block slides down an inclined plane which
makes an angle of 25o with respect to the
horizontal. Friction is present. Which of the
following non-conservative forces will perform
work on the block, thereby reducing the block’s
mechanical energy?
A) Force of Gravity.
block
B) Normal Force.
•
C) Force of Friction.
•
D) All of the above.
•
E) Both B and C.
Δx
h
25o
Work and Emec
Example
A 10.0kg block is released from point A in the figure
below. The track is frictionless except for the portion
between points B and C, which has a length of 6.00m.
The block travels down the track, hits a spring (with a
spring constant of 2,250N/m), and compresses the
spring 0.300m from its equilibrium position before
coming to rest momentarily. Determine the coefficient
of kinetic friction, μk, between the block and the rough
surface.
Answer
Work and Emec
Choose up as positive y with the lowest point to be
y = 0.
Are there any non-conservative forces at work in
this problem?
Yes, the frictional force between points B and C. So:
ΔEmec = ΔKE + ΔPE = Wnc = Wfriction
•
The block starts at rest and ends at rest such that
ΔKE = 0. This leaves us with:
•
ΔPE = ΔPEgrav + ΔPEspring = Wfriction
•
mgΔh + (1/2)k(Δx)2 = –(Ffriction)dBC
Work and Emec
Answer
Looking at the left side of the equation:
mgΔh = (10kg)(9.8N/kg)(0m – 3m) = -294J
(1/2)k(Δx)2 = (1/2)(2,250N/m)(0.3m)2 = 101J
Giving us:
-294J + 101J = -193J = –(Ffriction)dBC
Since the block is not accelerating in the y-direction
while it is crossing the rough surface, we can say:
ΣFy = FN – Fg = 0
•
•
FN = Fg = mg
Ffriction = μk FN = μk(mg)
Work and Emec
Answer
This gives us:
193J = (Ffriction)dBC = (μk)mgdBC
Remember to think about if non-conservative
forces are present before starting energy
problems.
Either way you can try and apply the mechanical
energy equation.
For Next Time (FNT)
Finish the Homework for
Chapter 5.
Start Reading Chapter 6.