Analysis of Situations Involving
External Forces
The previous part discussed
the relationship between work and energy change. Whenever work is done upon an
object by an external force, there will be a change in the total
mechanical energy of the object. If only internal
forces are doing work (no work done by external forces), there is no change in
total mechanical energy; the total mechanical energy is said to be conserved. Because external forces are capable of changing the total mechanical
energy of an object, they are sometimes referred to as non-conservative forces.
Because internal forces do not change the total mechanical energy of an object,
they are sometimes referred to as conservative forces. In this part of Lesson
2, we will further explore the quantitative relationship between work and
energy.
Relating
Work to Energy
The quantitative relationship between work and mechanical
energy is expressed by the following equation:
TMEi + Wext = TMEf
The equation states that the initial amount of total
mechanical energy (TMEi) plus the work done by
external forces (Wext) is equal to the final amount
of total mechanical energy (TMEf). A few
notes should be made about the above equation. First, the mechanical energy can
be either potential energy (in which case it could be due to springs or gravity) or kinetic energy. Given this fact, the above equation can be
rewritten as
KEi + PEi + Wext = KEf + PEf
The second note that should be made about the above equation
is that the work done by external forces can be a positive
or a negative work term. Whether the work term takes on a positive or
a negative value is dependent upon the angle between the force and the motion.
Recall from Lesson 1 that the work is dependent upon the angle between the force and the
displacement vectors. If the angle is 180 degrees as it occasionally is, then
the work term will be negative. If the angle is 0 degrees, then the work term
will be positive.
The above equation is expresses the quantitative relationship between
work and energy. This equation will be the basis for the rest of this unit. It
will form the basis of the conceptual aspect of our study of work and energy as
well as the guiding force for our
approach to solving mathematical problems. A large
slice of the world of motion can be understood through the use of this
relationship between work and energy.
Raising a Barbell Vertically
To begin our investigation of the work-energy relationship,
we will investigate situations involving work being done by external forces (nonconservative forces). Consider a weightlifter who
applies an upwards force (say 1000 N) to a barbell to displace it upwards a
given distance (say 0.25 meters) at a constant speed. The initial energy plus
the work done by the external force equals the final energy. If the barbell
begins with 1500 Joules of energy (this is just a made up value) and the
weightlifter does 250 Joules of work (F•d•cosine of angle = 1000 N•0.25 m•cosine 0
degrees = 250 J), then the barbell will finish with 1750 Joules of mechanical
energy. The final amount of mechanical energy (1750 J) is equal to the initial
amount of mechanical energy (1500 J) plus the work done by external forces (250
J).
Catching
a Baseball
Now consider a baseball catcher who applies a rightward force
(say 6000 N) to a leftward moving baseball to bring it from a high speed to a
rest position over a given distance (say 0.10 meters). The initial energy plus
the work done by the external force equals the final energy. If the ball begins
with 605 Joules of energy (this is just another made up value), and the catcher
does -600 Joules of work (F•d•cosine of
angle= 6000 N•0.10 m•cosine 180
degrees = -600 J), then the ball will finish with 5 Joules of mechanical
energy. The final energy (5 J) is equal to the initial energy (605 J) plus the
work done by external forces (-600 J).
A
Skidding Car
Now consider a car that is skidding from a high speed to a
lower speed. The force of friction between the tires and the road exerts a
leftward force (say 8000 N) on the rightward moving car over a given distance
(say 30 m). The initial energy plus the work done by the external force equals
the final energy. If the car begins with 320 000 Joules of energy (this is just
another made up value), and the friction force does -240 000 Joules of work (F•d•cosine of angle = 8000
N•30 m•cosine 180 degrees = -240 000 J),
then the car will finish with 80 000 Joules of mechanical energy. The final
energy (80 000 J) is equal to the initial energy (320 000 J) plus the work done
by external forces (-240 000 J).
Pulling a Cart Up an
Incline at Constant Speed
As a final example, consider a cart being pulled up an
inclined plane at constant speed by a student during a Physics lab. The applied
force on the cart (say 18 N) is directed parallel to the incline to cause the
cart to be displaced parallel to the incline for a given displacement (say 0.7
m). The initial energy plus the work done by the external force equals the
final energy. If the cart begins with 0 Joules of energy (this is just another
made up value), and the student does 12.6 Joules of work (F•d•cosine of angle = 18
N•0.7 m•cosine 0 degrees = 12.6 J), then
the cart will finish with 12.6 Joules of mechanical energy. The final energy
(12.6 J) is equal to the initial energy (0 J) plus the work done by external
forces (12.6 J).
In each of these examples, an external force does work upon
an object over a given distance to change the total mechanical energy of the
object. If the external force (or nonconservative force)
does positive work, then the object gains mechanical energy. The
amount of energy gained is equal to the work done on the object. If the
external force (or nonconservative force)
does negative work, then the object loses mechanical energy. The
amount of mechanical energy lost is equal to the work done on the object. In
general, the total mechanical energy of the object in the initial state (prior
to the work being done) plus the work done equals the total mechanical energy
in the final state.
Your Turn to
Try It
The work-energy relationship presented here can be combined
with the expressions for potential and kinetic energy to solve complex
problems. Like all complex problems, they can be made simple if
first analyzed from a conceptual viewpoint
and broken down into parts. In other words, avoid treating work-energy problems
as mere mathematical problems. Rather, engage your mind and utilize your
understanding of physics concepts to approach the problem. Ask "What forms
of energy are present initially and finally?" and "Based on the
equations, how much of each form of energy is present initially and
finally?" and "Is work being done by external forces?" Use this
approach on the following three practice problems. After solving, click the
button to view the answers.
Practice Problem #1
A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car
experiences an 8000 N force of friction. Determine the stopping distance of the
car.
Initially:
PE = 0 J (the
car's height is zero)
KE =
0.5*1000*(25)^2 = 312 500 J
Finally:
PE = 0 J (the
car's height is zero)
KE = 0 J (the
car's speed is zero)
The work done is (8000 N) • (d) • cos 180 = - 8000*d
Using the equation,
TMEi + Wext = TMEf
312 500 J +
(-8000 • d) = 0 J
Using some algebra it can be shown that d=39.1 m
Practice Problem #2
At the end of the Shock Wave roller coaster ride, the 6000-kg train of
cars (includes passengers) is slowed from a speed of 20 m/s to a speed of 5 m/s
over a distance of 20 meters. Determine the braking force required to slow the
train of cars by this amount.
Initially:
PE = 0 J (the
car's height is zero)
KE =
0.5*6000*(20)^2 = 1 200 000 J
Finally:
PE = 0 J (the
car's height is zero)
KE = 0.5*6000*(5)^2 = 75 000 J
The work done is F • 20 • cos 180 = -20•F
Using the equation,
TMEi + Wext = TMEf
1 200 000 J +
(-20*F) = 75 000 J
Using some algebra, it can be shown that 20*F = 1 125 000 and
so F = 56
250 N
Practice Problem #3
A shopping cart full of groceries is sitting at the top of a 2.0-m hill.
The cart begins to roll until it hits a stump at the bottom of the hill. Upon
impact, a 0.25-kg can of peaches flies horizontally out of the shopping cart
and hits a parked car with an average force of 500 N. How deep a dent is made
in the car (i.e., over what distance does the 500 N force act upon the can of
peaches before bringing it to a stop)?
The question pertains to the can of peaches; so focus on the can
(not the cart).
Initially:
PE = 0.25 kg *
9.8 m/s/s * 2 m = 4.9 J
KE = 0 J (the
peach can is at rest)
Finally:
PE = 0 J (the
can's height is zero)
KE = 0 J (the
peach can is at rest)
The work done is 500 N*d*cos 180 = -500*d
Using the equation,
TMEi + Wext = TMEf
4.9 J + (-500*d)
= 0 J
Using some algebra, it can be shown that d = 0.0098 m (9.8 mm)
Stopping Distance
All three of the above problems have one thing in common:
there is a force that does work over a distance in order to remove mechanical
energy from an object. The force acts opposite the object's motion (angle
between force and displacement is 180 degrees) and thus does negative
work. Negative work results in a loss of the object's total amount of
mechanical energy. In each situation, the work is related to the
kinetic energy change. And since the distance (d) over which the force does
work is related to the work and since the velocity squared (v^2) of the object
is related to the kinetic energy, there must also be a direct relation between
the stopping distance and the velocity squared. Observe the derivation below.
TMEi + Wext = TMEf
KEi + Wext = 0 J
0.5•m•vi2 + F•d•cos(Theta)
= 0 J
0.5•m•vi2 = F•d
vi2 
d
The above equation depicts stopping
distance as being dependent upon the square of the velocity. This means that a twofold increase in velocity would result in a
fourfold (two squared) increase in stopping distance. A threefold increase in
velocity would result in a nine-fold (three squared) increase in stopping
distance. And a fourfold increase in velocity would result in a sixteen-fold
(four squared) increase in stopping distance. This is one more example in which
an equation becomes more than a mere algebraic recipe for solving problems.
Equations can also be powerful guides to thinking about how two quantities are
related to each other. In the case of a horizontal force bringing an object to
a stop over some horizontal distance, the stopping distance of the object is
related to the square of the velocity of the object.