#### Transcript 141S13-NotesCh5a-May27

```Chapter 5
Work and Energy
Every† branch of science is concerned with energy. That being the
case, you would think that we could come up with a good definition
of exactly what energy is. However, the best we can do for now is
this: energy is the capacity to do work.
In this chapter, we will define work and its relation to force. Then,
we will introduce kinetic energy and potential energy, and relate
them to work. Finally, we will draw a connection between energy
and power.
†except
math. math is weird.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 1
5.1 Work Done by a Constant Force
Work describes what is accomplished by a force when it moves
an object through a particular displacement. Its mathematical
definition is as follows:
The work done by a constant force is equal to the product of
the magnitudes of the displacement and the component of the
force that is parallel to the displacement.
That’s a rather confusing sentence. The situations depicted on
the next two slides will clarify it.
The symbol for work is W (it’s capitalized, so as to not confuse it
with the magnitude of weight, w).
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 2
5.1 Work Done by a Constant Force
In part (a), there is a force F acting on the crate, but it does not
move. Since there is no displacement, no work is done.
In part (b), there is a force F acting on the waterskier, and this
force is in the same direction as his displacement d. In this case,
𝑾 = 𝑭𝒅.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 3
5.1 Work Done by a Constant Force
In part (c), there is a force F and a displacement d, but they are
not in the same direction. Instead, the force acts at an angle 𝜃 to
the displacement. From chapter 3, we know that the component
of F along the direction of d is 𝐹∥ = 𝐹 cos 𝜃 (the subscript “∥”
indicates the “parallel component”). Here, the work is
𝑾 = 𝑭∥ 𝒅 = 𝑭𝒅 𝐜𝐨𝐬 𝜽
Don’t be fooled into thinking that there are
three different equations for work. The
one above works in every case…for the
waterskier, 𝜃 = 0, and (since cos 0 = 1),
the equation reduces to 𝑊 = 𝐹𝑑. Do note
however that this equation is only valid for
constant forces.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 4
5.1 Work Done by a Constant Force
To complete the case in part (c), we note
that the perpendicular component of
force, 𝑭⫠ , does no work, since there is no
displacement in this direction. This fact will
end up being incredibly useful in simplifying
problems later on in this chapter.
Since work is the product of a force and a distance, it should be
clear that it has SI units of N·m (“Newton-meters”). One N·m is
also called a “Joule” (J).
Although force and displacement are both vectors, the formula
for work only involves their magnitudes and the cosine of their
relative directions. Therefore, work is a scalar quantity. It may
have a negative value, however.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 5
5.1 Work Done by a Constant Force
In chapter 2, we learned about graphical methods to calculate
changes in velocity (position) based on the area under a curve of
acceleration (velocity) vs. time.
Work can be analyzed graphically as well. If an object moves a
distance x while it is acted upon by a constant x-directed force F,
a plot of F vs. x is shown below. The area under the curve is Fx,
which is equal to the work done on the object.
This is a trivial example, of course…we’re all
capable of multiplying two numbers
together. However, it will become
important in the next section, when we
study the work done by a non-constant
force.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 6
5.1 Work Done by a Constant Force
When there are several forces acting on an object as it moves
through a displacement d, it is possible for many of them (or
many of their parallel components) to do work on the object.
Their sum is called the net work. When calculating net work,
you have two options:
1. Calculate the net force (the vector sum of each individual force)
and then find the work done by this net force; or
2. Calculate the work done by each individual force, and then
perform a scalar sum of these to find the net work.
I don’t usually use class time to discuss examples that are worked
out in the text. However, Example 5.3 (pages 146-147) is too
good to pass up…
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 7
5.1 Work Done by a Constant Force – Example 5.3
A 0.75-kg block slides with a
uniform velocity down a 20°
incline.
a) How much work is done by
the frictional force?
b) What is the net work done
on the block?
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 8
Problem #1: Catching a Baseball
WBL LP 5.3
A pitcher throws a fastball. When the catcher catches it…
A
…positive work is done
B
…negative work is done
C
…the net work is zero
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 9
Problem #2: Work Done by Gravity
A ball is thrown vertically upward. Unsurprisingly, it eventually reaches a
maximum height, then returns to the point at which it was thrown. What
can we say about the work Wg done by the force of gravity on the ball?
A
Wg is positive
B
Wg is negative
C
Wg = 0
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 10
Problem #3: Sliding Down a Ramp
WBL Ex 5.4/5.5
A 3.00-kg block slides down a frictionless plane inclined 20° to the horizontal. If
the length of the plane’s surface is 1.50 m, how much work is done on the block,
and by what force?
Then, suppose that there is kinetic friction between the block and the plane, with
𝝁𝒌 = 𝟎. 𝟐𝟕𝟓. In this case, what is the work done on the block?
Solution: In class
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 11
Problem #4: Pushing a Desk
WBL Ex 5.15
A man pushes horizontally on a desk that rests on a rough wooden floor. The
coefficient of static friction between the desk and floor is 0.750 and the
coefficient of kinetic friction between the desk and floor is 0.600. The desk’s
mass is 100 kg. The man pushes just hard enough to get the desk moving and
continues pushing with that force for 5.00 s. How much work does he do on the
desk?
Solution: In class
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 12
5.2 Work Done by a Variable Force
Recall that the entire discussion in section 5.1 assumed that the
force was constant. This is a bit of a restrictive condition, since
many forces vary with time and/or position.
One example of a variable force is a spring. If an applied force Fa
stretches or compresses the spring, it is countered by an
oppositely-directed “spring force” Fs.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 13
5.2 Work Done by a Variable Force
For “well-behaved” springs, the spring force is proportional to
the change in length of the spring from its equilibrium position
x0, and opposite in direction. This gives Hooke’s Law,
𝐹𝑠 = −𝑘∆𝑥 = −𝑘 𝑥 − 𝑥0
Often, we set x0 = 0, in which case
𝐹𝑠 = −𝑘𝑥
Here, we see that the spring force is a function of position. That
is, it is variable, rather than
constant.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 14
5.2 Work Done by a Variable Force
The 𝑘 in the previous equations is called the spring constant or
force constant, and it has SI units of N/m. A high value of 𝑘
indicates a stiff spring, while a low 𝑘 indicates that the spring is
very compliant.
Hooke’s law – that the spring force varies linearly with x – is only
valid for ideal springs. Real springs follow this law for reasonably
small values of |x|, but past their elastic limit, the linear relation
no longer applies.
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 15
5.2 Work Done by a Variable Force
Calculating the work done by an arbitrary variable force – that is,
for an arbitrary function F(x) – usually requires techniques of
calculus. However, for a linear force 𝐹 = 𝑘𝑥, it’s actually quite
easy using the graphical method of the last section.
Recalling that the area of a triangle is ½ (base)(height), we see
that
1
1
1 2
𝑊 = 𝐹𝑥 = 𝑘𝑥 𝑥 = 𝑘𝑥 .
2
2
2
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 16
Problem #5: Compressing a Spring
WBL CQ 5.7
If a spring is compressed 2.0 cm from its equilibrium position and then
compressed an additional 2.0 cm, how much more work is done in the second
compression than in the first?
Solution: In class
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 17
Problem #6: Overstretched Spring
WBL Ex 5.27
In stretching a spring in an experiment, a student
inadvertently stretches it past its elastic limit; the force-vsstretch graph is shown below. Basically, after it reaches its
limit, the spring acts as if it were considerably stiffer. How
much work was done on the spring?
Assume that each tick mark on the force axis is 10 N and
every tick mark on the distance axis is 10 cm.
Solution: In class
PC141 Intersession 2013
Day 9 – May 27 – WBL 5.1-5.2
Slide 18
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