Transcript 141S13-NotesCh5b-May28

5.3 The Work-Energy Theorem: Kinetic Energy
We’ll start with the form of energy that is most closely related to
work – kinetic energy (KE).
Consider the figure below, in which an object of mass m has an
initial speed v0. A constant force F acts on the object (in the
same direction as its velocity), increasing the speed to v. The
displacement of the object is x.
From the last lecture, we
know that the work done on
the object is 𝑊 = 𝐹𝑥. Then,
from chapter 2, we have
𝑣 2 = 𝑣02 + 2𝑎𝑥, which
implies that the object has
acceleration 𝑎 =
PC141 Intersession 2013
𝑣 2 −𝑣02
.
2𝑥
Day 10 – May 28 – WBL 5.3-5.4
Slide 1
5.3 Work-Kinetic Energy Theorem: Kinetic Energy
Now, we apply Newton’s second law:
𝑣 2 − 𝑣02
𝐹 = 𝑚𝑎 = 𝑚
2𝑥
Finally, the work done is
𝑊 = 𝐹𝑥 = 𝑚
𝑣 2 −𝑣02
2𝑥
𝑥=
1
𝑚𝑣 2
2
−
1
𝑚𝑣02 .
2
We now define the kinetic energy (K) of the object:
1
𝐾 = 𝑚𝑣 2
2
By this definition, the work done on the object is simply equal to
the change in the object’s kinetic energy:
𝑊 = 𝐾 − 𝐾0 = ∆𝐾
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 2
5.3 Work-Kinetic Energy Theorem: Kinetic Energy
Kinetic energy can be thought of as energy possessed by mass
that has motion. It has SI units of Joules, and can not have a
negative value (since both m and v2 are either positive or zero).
The last equation on the previous slide represents the workkinetic energy theorem:
the net work done on a body by all the
forces acting on it is equal to the change
in kinetic energy of the body
When positive net work is done on an object, its kinetic energy
increases (which means its speed increases, assuming that mass
is constant). When negative net work is done, the object’s
kinetic energy and speed must decrease. If net work is zero, the
object’s KE and speed do not change.
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 3
5.3 Work-Kinetic Energy Theorem: Kinetic Energy
Think back to Problem #2 from the last lecture. We found that
when an object is thrown vertically upward, the force of gravity
does negative work on the object on the way up (since force and
displacement are in opposite directions) and positive work on
the way down (since they are in the same direction). Based on
the previous paragraph, we can conclude that the object must
slow down (lose KE) on the way up and speed up (gain KE) on the
way down, which of course is exactly what happens.
Then, recall the example on slide 8 from last lecture…we found
that the net work done on the sliding block was zero…but we
also knew that the block was sliding at constant velocity. Thus,
its KE wasn’t changing, which agrees with the result that no work
was done. On the other hand, in problems 3 and 4 from last
lecture, the objects in question were accelerating, and we found
that positive net work was being done in both cases.
PC141 Intersession 2013
Slide 4
Day 10 – May 28 – WBL 5.3-5.4
Problem #1: Footrace
A father racing his son has half the kinetic energy of the son, who has half the
mass of the father. The father speeds up by 1.0 m/s and then has the same
kinetic energy as the son. What are the initial speeds of both the father and the
son?
Solution: In class
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 5
Problem #2: Accelerating from Rest
WBL Ex 5.31
A constant net force of 75 N acts on an object initially at rest as it moves through
a parallel distance of 0.60 m.
a) What is the final kinetic energy of the object?
b) If the object has a mass of 0.20 kg, what is its final speed?
Solution: In class
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 6
Problem #3: Out-of-Control Truck
WBL Ex 5.35
An out-of-control truck with a mass of 5000 kg is traveling at 35.0 m/s when it
starts descending a steep 15° incline with locked brakes. The incline is icy, so the
coefficient of friction is only 0.30. Use the work-KE theorem to determine how
far the truck will skid before it comes to rest.
Solution: In class
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 7
Problem #4: Stopped by a Spring
A container of mass m = 0.40 kg slides across a horizontal, frictionless counter
with speed v = 0.50 m/s. It runs into and compresses a spring of spring constant
k = 750 N/m. When the container is momentarily stopped by the spring, by what
distance d is the spring compressed?
Solution: In class
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 8
5.4 Potential Energy
A second type of energy is potential energy (U), named as such
because it represents the potential to do work. This potential
arises due to the configuration of a system of objects; it can be
thought of as “stored work”.
This work is done when the position or configuration is altered.
Again, consider Problem #2 from the last lecture. As we
discussed earlier in this lecture, the object loses KE on the way
up and gains KE on the way down. What we didn’t mention is
what happens to this KE. As it turns out, the gravitational force
converts KE to PE on the way up and then converts PE to KE on
the way down. The “system of objects” in this example is the
thrown object and the Earth, which interact via the gravitational
force. We’ll return to gravity in a bit…
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 9
5.4 Potential Energy
For a spring, recall that the work done in stretching or
compressing the spring by a distance 𝑥 from an equilibrium
1
position 𝑥0 is 𝑊 = 𝑘𝑥 2 (this is the work done by the applied
2
force in causing the stretch, it is not the work done by the
oppositely-directed spring force).
This work leads to a change in the spring’s potential energy:
1 2 1 2
𝑊 = ∆𝑈 = 𝑈 − 𝑈0 = 𝑘𝑥 − 𝑘𝑥0
2
2
For reasons that will soon become clear, we are permitted to
define a reference position or configuration where U = 0. Here,
we will define this position to be 𝑥0 . This tells us that the
potential energy of a spring that has been stretched or
𝟏
compressed by a distance x is 𝑼 = 𝒌𝒙𝟐
𝟐
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 10
5.4 Potential Energy
Gravitational Potential Energy
Consider the situation shown in the figure. A can of mass m is
initially at position y0 and then raised to a position y by an
applied force F. We assume that the can has a constant velocity
during this trip. Therefore, a = 0 and (by Newton’s 2nd law), F =
mg.
The work done by the applied force in moving
the can through a displacement Δy = y - y0 is
𝑊 = 𝐹∆𝑦 = 𝑚𝑔 𝑦 − 𝑦0 = 𝑚𝑔𝑦 − 𝑚𝑔𝑦0
But we know that 𝑊 = ∆𝑈 = 𝑈 − 𝑈0 .
Therefore, if we set U = 0 at y0, we have the
gravitational potential energy
U=mgΔy = mgy if y0 = 0
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 11
5.4 Potential Energy
The Zero Reference Point
It is important to note that potential energy can only be
described relative to a “reference point” or “reference
configuration”, where we define U0 = 0. Just as we are free to
choose the origin of our axes, we are also free to choose any
reference point that we wish; however, some are more
convenient than others. For example, in the gravitational
example from the last slide, the tabletop is a convenient choice.
For a spring, the equilibrium position (where it is neither
stretched nor compressed) is convenient.
The reason that we have this freedom is that absolute values of
U are of no consequence – only changes in U (ΔU) have any
physical meaning.
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 12
5.4 Potential Energy
In the figure below, a mass moves from the bottom to the top of
a short stairwell. On the left, we’ve defined the reference point
to be the middle step (here, U0 = 0, and U1 = -mgy, U2 = mgy).
During the move, ΔU = U2 - U1 = 2mgy.
On the right, the lower step is the middle step. Then, U0 = 0, and
U1 = mgy, U2 = 2mgy). During the move, ΔU = U2 - U0 = 2mgy. As
we see, the choice of reference position makes no difference in
calculating ΔU.
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 13
5.4 Potential Energy
Unlike K, U can have a negative value. There’s nothing inherently
strange here, it only means that the reference level was defined
at a relatively higher potential. This was the case on the previous
slide (on the left stairwell).
Path-Independence
It is also important to point out that changes in potential energy
are path-independent. This means that ΔU is only dependent
upon the starting and ending positions, and not upon the
precise path taken between them. This facilitates the solution
of many problems where the exact path taken between two
points is complicated or possibly unknown.
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 14
5.4 Potential Energy
Chemical Potential Energy
Many students in this class are chemistry, biology, or kinesiology
majors, and have undoubtedly encountered the concept of
chemical potential energy. As any physicist will tell you, there is
no difference between CPE and the PE that we have been
describing here. CPE is energy stored in the electromagnetic
attractions (bonds) between atoms…that is, it is stored in the
“configuration of the system of atoms”. During a chemical
reaction, the system of atoms is rearranged in some way, and this
can result in an increase or decrease in the CPE – perhaps
manifested outwardly as a decrease or increase in the system’s
temperature.
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 15
Problem #5: Loop-the-Loop
In the figure, a small block of mass m = 0.032 kg can slide
along the frictionless loop-the-loop, with radius R = 12 cm.
The block is released from rest at point P, which is a height h
= 5.0R above the bottom of the loop.
a) How much work does the gravitational force do on the
block as the block travels from point P to point Q, and
from point P to the top of the loop?
b) If the gravitational potential energy of the block-earth
system is taken to be zero at the bottom of the loop, what
is that potential energy when the block is at the point P,
the point Q, and the top of the loop?
Solution: In class
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 16
Problem #6: Change in Potential Energy
WBL LP 5.11
A change in gravitational potential energy…
A
…is always positive
B
…depends on the reference point
C
…depends on the path taken by the object
D
…depends only on the initial and final positions of the object
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 17
Problem #7: Stacking Books
WBL Ex 5.39
Six identical books, 4.0 cm thick and each with a mass of 0.80 kg, lie individually
on a flat table. How much work would be needed to stack the books one on top
of the other?
Solution: In class
PC141 Intersession 2013
Day 10 – May 28 – WBL 5.3-5.4
Slide 18