Download Ch 6: Work and Energy

Ch 6: Work and Energy
6.1 Work and Kinetic Energy
‘Member the definition of work, in the physics sense?
Q:
Is a waiter doing any work, in the physics sense, when he carries your tray from
the kitchen to your table? Does he deserve a tip? (I’d still be careful, as this person has
direct opportunity, and possible motivation, to say, provide your meal with “extra
moisture”… )
DEF: Work = F x d
= F (or the component of F in the direction of the motion) x distance
= F, or F component, at the point of application that moves an object
through a distance in the direction of the velocity of the F’s point
of application
= The thing that causes a change in Kinetic Energy of an object (otherwise,
you wouldn’t be doing a true work, right?)
DEF: Kinetic Energy (KE) = the energy of motion
DEF: Potential Energy (PE) = the energy associated with the position of an object, or
the configuration of a system, such as the separation
distance between two objects that attract each other
There are many of types of energy, but this unit focuses primarily on KE and PE (thus
ME) as they pertain to work as well as Newton’s Laws.
Q: If an object moves through a horizontal displacement, ∆xi, then the work done on it is
given by
Q: How would this change for vertical motion?
So work is a scalar that is dependent on the signs of the F and the ∆x
Q: What would be the units for work?
A: W = F∙d, so N∙m = Joules = J. Just so you know, here in the good ol’ U.S. of A., the
common unit is the foot∙pound = ft∙lb where 1 ft∙lb = 1.356 J. For nuclear physics, the
common unit for work is the electron – volt = eV where 1 eV = 1.6 x 10- 19 J.
Just as a kind of coolie-moolie thing to know, the work done to move an e- might be
a few eV, while the work done to move a proton or a neutron would be morr like a MeV,
or a mega eV. Why?
A: The mass ratio difference is 1:1837, and there are strong and weak nuclear forces that
also impact / restrict motion. Hence, the “million” factor!
Q: What is a particle, really?
A: Any object that moves so that all of its parts undergo identical displacement.
When several forces act on the same particle, the displacements of the points of
application of each of these forces are equal.
Wtotal = F1x∆x + F2x∆x + … + Fnx∆x = Fnet x∆x
The Work – KE Theorem
If Fnetx = ma, and ax = 1 (vf2 – vi2) then
2 ∆x
2
Fnetx ∆x = ½ mvf – ½ mvi2, and the left side is equal to the work done, so
Wtotal = ½ mvf2 – ½ mvi2
Do you recognize ½ mv2?
So really, Wtotal = the change in KE, or W causes a change in KE
This is the Work – KE Theorem Wtotal = ∆ KE, or just Wtotal = ∆ K
See ex 6-1, p 154. Assume 2 sig figs.
A truck of mass 3.0 x 103 kg is to be loaded onto a ship by a crane that exerts an upward
force of 31 kN on the truck. This force, which is strong enough to overcome the
gravitational force and get the truck started upward, is applied over a distance of 2.0 m.
Find:
a) the work done by the crane
b) the work done by gravity
c) the upward speed of the truck after the 2.0 m
Since the direction of the motion is only up / down, your text defines the y-axis as the xaxis, and thus, sticks with using cos Ө! 
a) Wcrane = Fcrane, y ∆y = Fcrane cos 00 ∆y = (31 kN)(1)(2m) = 62 kJ
What is the angle between the motion of the truck and the force of gravity?
What is a N / kg going to give you, unit-wise?
b) Wg = Fg ∆y = mg ∆y = mg cos 1800 ∆y = (3000 kg)(9.81 N / kg)(-1)(2m) = - 58.9 = - 59 kJ
c) Wtotal = ∆ K, so
Wcrane + Wg = Kf – Ki = ½ mvf2 – ½ mvi2
This lets you isolate vf2 so you have
vf2 = vi2 + 2(Wcrane + Wg)
m
Don’t forget to convert kJ to J so you can have your final answer in m/s
= 0 + 2(62000 J – 59000 J)
3.0 x 103 kg
= 2.0 m2 / s2
vf = 1.4 m / s
Take the √ of both sides to give you
See ex 6-2, p 154-155
In a TV tube, an e- is accelerated from rest to a KE of 2.5 KeV over a distance of 80.0 cm
where the F that accelerates the e- is an electric F due to the electric field in the tube. Find
the F on the e-, assuming it to be constant and in the direction of motion.
Wtotal = ∆ K
Fx ∆x = Kf – Ki Solve for Fx, and you get
Fx = Kf – Ki
∆x
= 2500 eV – 0 x 1.6 x 10-19 J
0.80 m
1 eV
= 5.0 x 10-16 N
See ex 6-3, p 155
During your winter break, you enter a dogsled race across a frozen lake. To get started
you pull the sled (total mass 80.0 kg) with a force of 180 N at a 20.0 0 angle above the
horizontal. Find:
a) the work you do
b) the final speed of the sled after it moves ∆x = 5.0 m, assuming it starts from rest and
there is no friction
Fig 6-3
Fig 6-4
a) Do a FBD showing initial and final positions, as show in fig 6-4. Define the x axis as the
direction of the displacement, as shown. NOTE: This will also be the W total, as mg and Fn
have no components in the x direction, as shown by fig 6-3.
W = Fx ∆x = F cos Ө ∆x = (180 N)(cos 20.00)(5.0 m) = 846 = 850 J
b) Wtotal = ½ mvf2 – ½ mvi2
vf2 = vi2 + 2Wtotal = 0 + 2(850 J) = 21.25 m2 / s2
m
80.0 kg
Take the √ of both sides, and you get vf = 4.6 m/s
Work Done By A Variable Force
Figure 6-7 below shows the plot of a constant Fx as a function of position. The work done
on a particle whose displacement is ∆x is equal to the area under the F-versus-position
curve. Does that sound familiar to you? 
Fig 6-7
Think about this; even for a variable force, you can approximate it by constructing a
series of constant forces over very small intervals. Obviously, from calc, you know this
constitutes itsy-bitsy rectangles. Guess what the sum of all of those rectangles gives you?
This leads quite nicely into the following:
Q: Do forces ever vary with position? Can you say “PE?”
Q: For Fig 6-5 below, if the man holds the weight at a fixed position, does he do any work
on the weight? Do his muscles do any work within his body?
Fig 6-5
Fig 6-6
See Example 6-4, p 157
A force Fx varies with x as shown in figure 6-9. Find the work done by the force on a
particle as the particle moves from x = 0.0 m to x = 6.0 m
Fig 6-9
The work is the area under the curve. Because the curve consists of straight-line segments,
the easiest approach is to use the geometric formulas for the area.
W = Atotal = A1 + A2
W = (5 N)(4 m) + ½ (5 N)(2 m)
W = 20 J + 5 J
W = 25 J
See Example 6-5, p 158
A 4-kg block on a frictionless table is attached to a horizontal spring that obeys Hooke’s
Law and exerts a force F = - kxi, where k = 400.0 N/m and x is measured from the
equilibrium position of the block. The spring is originally compressed with the block at
x1 = -5 cm (Figure 6-10). Find
(a) the work done by the spring on the block as the block moves from x1 = -5 cm to its
equilibrium position x2 = 0
(b) the speed of the block at x2 = 0
Fig 6-10
Make a graph of Fx versus x. The work done on the block as it moves from x1 to x2 = 0
equals the area under Fx-versus-x curve between these limits (shaded in figure 6-11) which
can be calculated by integrating the force over the distance. The work done equals the
change in kinetic energy, which is simply its final kinetic energy because the initial kinetic
energy is zero. The speed of the block at x = 0 is found from the kinetic energy of the
block.
(a) The work W done by the spring on the block is the integral of Fx dx from x1 = -5 cm to
x2 = 0. NOTE: Though two other forces act on the block, mg and Fn, neither have a
component in the direction of motion, thus neither contribute to work.
(b) Apply the work-kinetic energy theorem and solve for v2.
6-2 The Dot Product
What we’ve learned so far is that work can be calculated by multiplying force, or the
“component of force in the direction of motion” by the distance an object moves. This
works (excuse the pun) for work done by a constant force, or piecewise-smooth linear
forces. We’ve also learned that we can find work by calculating the area under the forceposition graph, or integrating the force-position graph. But what about calculating work
for a particle subject to non-constant force, thus moving non-linearly along a path? We
need a way to calculate how much work is done on a particle, at any given instant along a
curve.
Fig 6-12 a
Fig 6-12 b
Figure 6-12a shows a particle moving along an arbitrary curve while Figure 6-12 b shows
the perpendicular component of the force which changes the direction of the particle, but
not its speed. The tangential component changes its speed but not its direction. Fs = the
mass times the tangential acceleration, dv/dt. It is only this component that does work. So,
we really only concern ourselves with the “piece” of the force vector that is actually doing
work (not ) on the particle.
What this really means is that if a particle is headed upward, but then starts to curve
downward, there must be a force pulling it in and down. If it starts to head back up again,
there must be a force responsible for that, as well. We can calculate how much work is
done at those points (or at any points along the curve, for that matter) by using something
called The Dot Product.
To properly use The Dot Product, we need to know:
 What the notation means, both conceptually and mathematically
 Why we would want to use The Dot Product in the first place
 Some important properties concerning The Dot Product
That component, Fs, is related to the angle Ǿ between the directions of F and ds by
Fs = F cos Ǿ. This means the work done by F for the displacement ds is
W = Fsds = F cos Ǿ ds. This combination of the two vectors and the angle between them is
called the dot product (or scalar product) of the two vectors, where the dot product is
given by
Really, you can think of the dot product as either vector A times the component of vector
B in the direction of A, or vice versa. 


Calculating A B really means multiplying like-component-to-like-component for both
vectors. The un-bolded AB means the magnitudes of each vector, found by using the
Pythagorean Theorem, if necessary. Cos  means the cosine of the angle between the two
vectors.
Q: Why use cos ?
A: It is arbitrarily assumed the particle is always moving along the implicit x-axis, thus,
the piece of the vector doing work along the x-axis will be F cos .
Below is a table detailing properties of dot products for unit vectors.
You can use unit vector notation to write the dot product in terms of rectangular
components where
A∙B = (Axi + Ayj + Azk) ∙ (Bxi + Byj + Bzk)
The dot product of any unit vector with itself = 1, so what is Axi ∙Bxi = AxBx(1) = AxBx
Q: What would the dot product of any vectors utilizing mutually perpendicular vectors
(like i ∙ j) give you? Why?
A: Zero, because they don’t have any components in the other’s respective direction. 
See Example 6-6, p 160-161
Using the dot product, find
(a) the angle between the vectors A = 3 mi + 2mj and B = 4 mi - 3 mj
(b) the component of A in the direction of B.
We find the angle θ from the definition of the dot product.
(a) 1. Write the dot product of A and B in terms of A, B and cos(θ) and solve for cos(θ)
cos(θ) = A∙B
AB
First, find A∙B by taking A∙B = AxBx + AyBy = (3m)(4m) + (2m)(- 3m) = 12m2 – 6m2 = 6m2
The magnitudes of the vectors are obtained from the dot product of the vector with itself.
A∙A = A2 = Ax 2 + Ay 2 = (3m)2 + (2m)2 = 13m2 so A = √13m
B∙B = B2 = Bx 2 + By 2 = (4m)2 + (-3m)2 = 25m2 so B = 5m
Substitute these back into cos(θ) = A∙B = ___6m2____ = 0.333, so θ = 70.60
AB (√13m)(5m)
(b) The component of A in the direction of B is the dot product of A with the unit vector B,
which equals B/ B, so
AB = A∙B
= A∙ B/ B
= 6m2 / 5 m
= 1.2 m
In dot product notation, the work dW done by a force F on a particle over a displacement
ds is
where ds means the magnitude of ds. The work done on the particle as it moves from
point 1 to point 2 is
NOTE: When several forces act on a particle whose displacement is ds, the total work
done is dW = F1∙ds + F2∙ds + …+ Fn∙ds = ∑Fi ∙ ds
See Example 6-7, p 162
You push a box up a ramp using a 100-N horizontal force. For each 5 m of distance along
the ramp, the box gains

3 m of height. Find the work done by F for each 5 m it moves
along the ramp by
a) directly computing the dot product from the components of F and the displacement of s
b) multiplying the product of the magnitudes of F and s with the cos of Ө between them
c) finding Fs (the component of force in the direction of the displacement) and multiplying
it by the magnitude of the displacement
d) finding the component of the displacement in the direction of the force and multiplying
it by the magnitude of the force
a) F = 100 Ni + 0j
s = 4 mi + 3mj
W = F∙s = Fx∆x + Fy∆y
= (100 N)(4 m) + (0)(3 m) = 400 J
b) F∙s = Fs cos Ǿ and F∙s = Fx∆x + Fy∆y
So, cos Ǿ = Fx∆x + Fy∆y = (100 N)(4m) + 0 = 0.8 and
Fs
(100 N)(5m)
W = Fs cos Ǿ = (100 N)(5m)(0.8) = 400 J
c) Fs = F cos Ǿ = (100 N)(0.8) = 80 N
W = Fss = (80 N)(5m) = 400 J
d) sF = s cos Ǿ = (5m)(0.8) = 4m
W = FsF = (100 N)(4m) = 400 J
Q: Can work ever be negative?
A: Yepper! If the component of force in the direction of the displacement is negative, the
work done will be negative, as well. 
Power
DEF: Power = the rate at which force does work
Remember: Work = dsF where ds represents position. Work also has “F” in it. Thus,
dW/dt would be F ds/dt where ds/dt is velocity!
Q: Watt would be the units for power? I already told you, silly! 
A: The units for Power are J/s = Watts = W. Actually, James Watt is most well known
for coining the term horsepower. If we have time, we’ll look at the doc from PHY.
1 hp = 550 ft∙lb / s = 746 W
1 W = 1 J/s
See Ex 6-9, P 163


Show that d(v2)/dt = 2 a  v


v2 = v  v
Using the product rule, we get:


d 2
d
v v
(v ) 
dt
dt

dv 
dv
=  v v 
dt
dt
=2
=
dv 
v
dt
dv 2
(v )
dt
 
d 2
(v ) = 2 a  v
dt
See Ex 6-10, p 164
Save this info for future reference!!
A small motor is used to operate a lift that raises a load of bricks weighing 800 N to a
height of 10 m in 20 s. What is the minimum power the motor must deliver?


P = F  v = F v cos  = F v cos 00
 10m 
= (800 N) 
 (1)
 20s 
= 400 W
Note: This is slightly morr than ½ hp.
See Ex 6-11, p 164
Show that the power delivered to a particle by the net force acting on it equals the rate at
which the KE of the particle is changing.
From Ex 6-9, we know:
We also know if Fnet = ma
Divide both sides by 2 and multiply by m
Rewrite on the LHS
Rewrite LHS and RHS
 
d 2
(v ) = 2 a  v
dt

F
d 2
(v ) = 2 net  v
dt
m


md 2
(v ) = Fnet  v
2dt


d 1 2
 mv  = Fnet  v
dt  2

dK
= Pnet
dt
Note: This equation is useful for implementing the Work-Energy Theorem in 3-D.
Cool stuff time!! Energy of a spring translated to real life physics!
http://www.hockeymonkey.com/easton-v-series-hockey-sticks.html