Download The Work-Energy Relationship

ENERGY LESSON 4
The Work-Energy Relationship
Internal vs. External Forces
We will learn that there are certain types of forces, which when present and when
involved in doing work on objects will change the total mechanical energy of the object.
And there are other types of forces which can never change the total mechanical energy
of an object, but rather can only transform the energy of an object from potential energy
to kinetic energy (or vice versa).
The two categories of forces are referred to as internal forces and external forces.
Internal Forces
External Forces
Fgrav
Fapp
Fspring
Ffrict
Fair
Ftens
Fnorm
EXTERNAL FORCE
When net work is done upon an object by an external force, the total mechanical energy
(KE + PE) of that object is changed.
If the work is positive work, then the object will gain energy.
If the work is negative work, then the object will lose energy.
The gain or loss in energy can be in the form of potential energy, kinetic energy, or both.
Under such circumstances, the work which is done will be equal to the change in
mechanical energy of the object.
INTERNAL FORCE
When the only type of force doing net work upon an object is an internal force (for
example, gravitational and spring forces), the total mechanical energy (KE + PE) of that
object remains constant.
In such cases, the object's energy changes form. For example, as an object is
"forced" from a high elevation to a lower elevation by gravity, some of the
potential energy of that object is transformed into kinetic energy.
Yet, the sum of the kinetic and potential energies remain constant.
When the only forces doing work are internal forces, energy changes forms - from
kinetic to potential (or vice versa); yet the total amount of mechanical is
conserved. This is called energy conservation.
EXAMPLES: In the following descriptions, the only forces doing work upon the objects
are internal forces - gravitational and spring forces. Thus, energy is transformed from KE
to PE (or vice versa) while the total amount of mechanical energy is conserved.
Read each description and indicate whether energy is transformed from KE to PE or from
PE to KE.
KE to PE or PE to KE?
Description of Motion
1.
2.
A ball falls from a height of 2
meters in the absence of air
resistance.
Explain.
PE to KE
The ball is losing height (falling) and gaining
speed. Thus, the internal or conservative force
(gravity) transforms the energy from PE (height)
to KE (speed).
A skier glides from location A to PE to KE
location B across a friction free ice.
The skier is losing height (the final location is
lower than the starting location) and gaining
speed (the skier is faster at B than at A). Thus,
the internal force or conservative (gravity)
transforms the energy from PE (height) to KE
(speed).
3.
A baseball is traveling upward
towards a man in the bleachers.
KE to PE
The ball is gaining height (rising) and losing
speed (slowing down). Thus, the internal or
conservative force (gravity) transforms the
energy from KE (speed) to PE (height).
4.
A bungee cord begins to exert an KE to PE
upward force upon a falling bungee
jumper.
The jumper is losing speed (slowing down) and
the bunjee cord is stretching. Thus, the internal
or conservative force (spring) transforms the
energy from KE (speed) to PE (a stretched
"spring"). One might also argue that the
gravitational PE is decreasing due to the loss of
height.
5.
The spring of a dart gun exerts a PE to KE
force on a dart as it is launched from
an initial rest position.
The spring changes from a compressed state to a
relaxed state and the dart starts moving. Thus,
the internal or conservative force (spring)
transforms the energy from PE (a compressed
spring) to KE (speed).
When work is done by external forces, the total mechanical energy of the object is
altered.
The work that is done can be positive work or negative work depending on
whether the force doing the work is directed opposite the object's motion or in the
same direction as the object's motion.
If the force and the displacement are in the same direction, then positive work is
done on the object. If positive work is done on an object by an external force, then
the object gains mechanical energy.
If the force and the displacement are in the opposite direction, then negative
work is done on the object; the object subsequently loses mechanical energy.
The following descriptions involve external forces (friction, applied, normal, air
resistance and tension forces) doing work upon an object.
Read the description and indicate whether the object gained energy (positive work) or
lost energy (negative work). Then, indicate whether the gain or loss of energy resulted in
a change in the object's kinetic energy, potential energy, or both.
Change PE or
Description
+ or - Work?
Megan drops the ball and
hits an awesome forehand.
The racket is moving
horizontally as the strings
apply a horizontal force
while in contact with the
ball.
The work is +.
KE or Both?
The force is to the
Kinetic energy.
right and the
displacement is to the
The applied force of the racket
right. F and d are in causes the ball to gain speed. Thus,
the same direction.
the external or nonconservative
Thus, positive work is force alters the kinetic energy of
done.
the ball.
A tee ball player hits a long
Both.
The work is +.
ball off the tee. During the
contact time between ball
The applied force of the bat causes
The force is up and to
and bat, the bat is moving at
the ball to gain both height and
the right and the
a 10 degree angle to the
speed. Since the force has an up
displacement is up and
horizontal.
component, it contributes to a
to the right. >F and d
height change. Thus, the external
are in the same
or nonconservative force alters the
direction. Thus,
both the kinetic energy and the
positive work is done.
potential energy of the ball.
Rusty Nales pounds a nail
The work is +.
Kinetic energy.
into a block of wood. The
hammer head is moving
The force is to the left The applied force of the hammer
horizontally when it applies and the displacement causes the nail to gain speed. Thus,
force to the nail.
is to the left. F and d
the external or nonconservative
are in the same
force alters the kinetic energy of
direction. Thus,
the nail.
positive work is done.
The frictional force between
The work is highway and tires pushes
Kinetic energy
backwards on the tires of a The force is to the left
skidding car.
and the displacement The friction force on the car causes
is to the right. F and d
the car to lose speed. Thus, the
are in the opposite
external or nonconservative force
direction. Thus,
alters the kinetic energy of the car.
negative work is done.
A diver experiences a
horizontal reaction force
exerted by the blocks upon
her feet at start of the race.
Kinetic energy.
The work is +.
The applied force of the starting
The force is to the
blocks causes the diver to gain
right and the
speed. Thus, the external force
displacement is to the
alters the kinetic energy of the
right. F and d are in
diver. (NOTE: there is another
the same direction. force - gravity - which changes the
Thus, positive work is diver's height; but the blocks are
done.
not responsible for this height
change.)
A weightlifter applies a force
The work is +
Potential energy
to lift a barbell above his
head at constant speed.
The force is up and the The applied force of the causes the
displacement is up. F
barbell to gain height. Thus, the
and d are in the same external or nonconservative force
direction. Thus,
alters the potential energy of the
positive work is done.
barbell.
Note that in the five situations described above, a horizontal force can never change the
potential energy of an object.
Potential energy changes are the result of height changes and only a force with a vertical
component can cause a height change.
The Work-Energy Relationship
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.
The quantitative relationship between work and mechanical energy is expressed by the
following equation:
TMEi + Wext = TMEf
NOTE: the mechanical energy can be either potential energy (in which case
it could be due to springs or gravity) or kinetic energy.
KEi + PEi + Wext = KEf + PEf
NOTE: the work done by external forces can be a positive or a negative work term.
To begin our investigation of the work-energy relationship, we will investigate situations
involving work being done by external forces.
EXAMPLE: 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. If
the barbell begins with 1500 Joules of energy, what is the final mechanical energy of the
barbell?
EXAMPLE: 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). If the ball begins with 605 Joules of energy, what is the final mechanical
energy of the ball?
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)?
EXAMPLE:
Answer: 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)
The work done is (500 N) • (d) • cos 180 = - 500*d
Finally:
PE = 0 J (the can's height is zero)
KE = 0 J (the peach can is at rest)
TMEi + Wext = TMEf
0 = PEi + Wext
0 = 0.25 kg * 9.8 m/s/s * 2 m + -500*d
d = 9.8 x 10-3 m
Go through ANIMATIONS – see directions for computer and overhead
Animations found at
http://www.glenbrook.k12.il.us/gbssci/Phys/mmedia/index.html#work
ANIMATION “WHICH PATH REQUIRES THE MOST ENERGY”
ANIMATION “HOW FAR WILL IT SKID”
Regarding the equation discussed in the ANIMATION “HOW FAR WILL IT SKID”,
Stopping distance is dependent upon the square of the velocity, in the case of a horizontal
force bringing an object to a stop over some horizontal distance
TMEi + Wext = TMEf
KEi + Wext = 0 J
0.5•m•vi2 + F•d•cos(Theta) = 0 J
0.5•m•vi2 = F•d
v i2 d
This means that a twofold increase in velocity would result in a fourfold (two squared)
increase in stopping distance.
A fourfold increase in velocity would result in a sixteen-fold (four squared) increase in
stopping distance.
ENERGY LESSON 4 HOMEWORK
1. A car which 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). What is the final mechanical energy of the
car, if the car begins with 320 000 Joules of energy?
2. 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).
What is the final mechanical energy of the cart?
3. Test your understanding by predicting the stopping distance values in the table below
for a car with a horizontal force acting on it to bring it to a stop.
Velocity (m/s)
Stopping Distance (m)
0 m/s
0
5 m/s
4m
10 m/s
15 m/s
20 m/s
25 m/s
4. 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.
5. 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.
HOMEWORK KEY
1. –80000 J
2. 12.6 J
3. 16 m, 36 m, 64 m, 100 m
4. 39.1 m
5. 56 250 N
ENERGY LESSON 4 HOMEWORK
1. A car which 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). What is the final mechanical energy of the
car, if the car begins with 320 000 Joules of energy?
2. 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).
What is the final mechanical energy of the cart?
3. Test your understanding by predicting the stopping distance values in the table below
for a car with a horizontal force acting on it to bring it to a stop.
Velocity (m/s)
Stopping Distance (m)
0 m/s
0
5 m/s
4m
10 m/s
15 m/s
20 m/s
25 m/s
Answer: (a) d = 16 m
From 5 m/s to 10 m/s is a two-fold increase (10 / 5 = 2) in velocity. Thus there should be
a four-fold increase in stopping distance. Multiply 4 m by 4.
(b) d = 36 m
From 5 m/s to 115 m/s is a three-fold increase (15 / 5 = 3) in velocity. Thus there should
be a nine-fold increase in stopping distance. Multiply 4 m by 9.
(c) d = 64 m
From 5 m/s to 20 m/s is a four-fold increase (20 / 5 = 4) in velocity. Thus there should be
a 16-fold increase in stopping distance. Multiply 4 m by 16.
(d) d = 100 m
From 5 m/s to 25 m/s is a five-fold increase (25 / 5 = 5) in velocity. Thus, there should be
a 25-fold increase in stopping distance. Multiply 4 m by 25.
4. 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.
Answer: Initially:
PE = 0 J (the car's height is zero)
KE = 0.5*1000*(25)^2 = 312 5000 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
5. 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.
Answer:
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