Download Document

Work and Energy
Work is defined as a constant force exerted on an object in the direction of motion
times the distance that the object moves. W = Fd
Note that work is something that is done on a system by the external world.
Force is in Newtons and distance is in meters, so work has units of Newton-meters
(N∙m) which is an energy unit called a joule (J).
Energy is defined as the ability of an object to produce a change in itself or the world
around it.
Kinetic Energy (KE) is one half times the mass of an object times the speed of the
object squared: KE = (½)m2
Note that the units for kinetic energy are kg∙m2/s2 but a kg∙m/s2 is a Newton so
kgm2/s2 is really a N∙m which as we have just seen is a joule (J).
The Work-Energy Theorem says that work is equal to the change in kinetic energy:
W = DKE
or
W = (½)m(D)2 or
W = (½)m(f )2 - (½)m(i )2
Work is defined as a constant force exerted on an object in the direction of motion
times the distance that the object moves (displacement). W = Fd
What happens if the direction of the force is perpendicular to the motion of the object?
A force that is applied perpendicular to the motion of an object only changes the
direction of the object (it does NOT change the speed).
Therefore if D = 0, the work will also be zero since W = (½)m(D)2 . An example
of this situation is a planet in orbit. Since the force is always perpendicular to the
direction that the planet moves, the planet’s speed is constant and W = 0.
Imagine the force (arrow) below pushing a box across the floor. Since the box
only moves in the x direction, only the component of the force in the x direction
can do work.
Fx

F
Remember from geometry that
Fx = Fcos()
So work when force is applied at an angle to the
displacement is:
W = Fdcos()
Imagine that a worker pushes a box 4.67 m across a floor with a constant force of
7.18 N. The mass of the box is 3.46 kg. Assume that the worker only pushed the
box in the direction of the motion. How much work did the worker do while
pushing the box?
W = Fd = (7.18 N)(4.67 m) = 33.5306 N∙m
F
Click to
see solution
= 33.5
J
What was the change in kinetic energy of the box?
Since the change in kinetic energy is equal to W, the answer would be the same: 33.5 J
Note the sign convention for work is that if work is done on the system (something
from outside the system is doing the work), then work is positive.
See example and practice problems on page 261
Imagine a boy pulling a wagon. The boy pulls the wagon 22.5 m and is pulling
with a constant force of 49.3 N at an angle of 37.4o to horizontal. How much work
does the boy do while pulling the wagon?
FN
F

Fx
Click to see solution
d
Fg
Note that Fx is in the same direction as the displacement (d).
Note that the normal force (FN) and the force due to gravity (Fg) are both perpendicular
to the displacement so they have no impact on the work done.
W = Fdcos() = (49.3 N)(22.5 m)(cos(37.4o)) = 881.2044 N∙m = 881 J
See example and practice problems on page 262
What happens if the force applied is not constant (i.e. the force changes as the
object moves)?
If the force is not constant, then a graph of force (y-axis) versus displacement (x-axis)
is needed. From the graph, the area under the curve will represent the work done.
If the force increases (or decreases) linearly with respect to displacement, then the
area under the graph of force versus displacement will represent a right triangle.
From geometry, the area of a right triangle will be (½)(base)(height). This means that
the work done will be (½)(displacement)(force).
20
For the triangle the base would be 2.0 m
and the height would be 20.0 N
15
10
Therefore, W = (½)(2.0 m)(20.0 N)
5
= 20. J
0
0
1
2
Power by definition is the work done divided by the time taken to complete the
work.
W
P=
t
The units for power will be J/s which are defined as watts (W).
Imagine that we want to lift 100 bricks up onto a scaffold. The amount of work
required to put pick up 100 individual bricks and put them onto the scaffold will be
the same whether we take 5 minutes or 20 minutes.
What will be different is how tired we get while doing this. If we do it in 5 minutes,
we will need to move 20 bricks each minute (quick pace). If we do it in 20 minutes,
we only need to move 5 bricks each minute (slow pace). This is because doing the
work quickly will require more power than doing is slowly.
A similar example could be riding a bicycle. If we want to ride 10 km, then the
amount of work we do is determined by that distance. However, if we ride the 10 km
in 10 minutes (60 km/hr) compared to 30 minutes (20 km/hr), we know from
experience that going that much faster will require more power.
Since watts are relatively small units, we often use kilowatts (kW) in problems.
1000 W = 1 kW
An electric motor lifts an elevator 9.00 m over 15.0 s by exerting an upward force
of 1.20X104 N. How much power is produced by the motor to accomplish this?
F
d
Click to see solution
Note that F is in the same direction as the displacement (d).
W = Fd = (1.20X104 N)(9.00 m) = 1.08X105 J
P=
W
t
1.08X105 J
=
15.0 s
= 7.20X103 J/s = 7.20X103 W = 7.20 kW
See practice problems on page 264
Power (P) by definition is the work done (W) divided by the time (t) taken to
complete the work.
W
P=
t
but W = Fd so we would write the power equation as: P =
Fd
t
d
However, is speed (). Therefore P = F is also a valid equation.
t
6
1200
5
1000
4
800
3
600
2
400
1
200
0
0
0
100
200
300
400
500
Force (N)
600
700
800
Power (W)
Velocity of muscles (m/s)
A plot of Force applied to the wheels of a bike versus velocity of muscles could give:
So if your legs move very
fast but exert no force on the
wheels, you have no power.
As we change gears, we get
more force going to the
wheels, but the velocity of
our muscles slows down.
Multiplying F times  will
give power delivered.
The idea that the velocity of something used to apply a force is somehow inversely
related to how much force is being applied is not an uncommon idea.
Imagine that we have a manual transmission car, and we have fixed the accelerator
in place about one third of the way to the floor.
If the engine is in neutral, the engine’s rpms (like velocity for our discussion here)
will be high, but nothing happens because the power is zero since there is no force
going to the wheels.
If we shift from neutral into first, the rpm’s will go down as we now get some force
delivered to the wheels. The power will go up.
If we shift from first into second, the rpm’s will go down again and the power
continues to go up.
At some point, we will probably reach a gear where the rpm’s will be so low that
the engine starts to stall, so the power will go down.
Many machines have this type of relationship between force, velocity and power.
There will be some combination of force and velocity that gives the highest power,
but it will not be where velocity and force are both at their highest.
A machine is something that eases the load by changing either the magnitude or
direction of a force to match the force to the capability of the machine or person.
As an example let’s imagine opening a bottle of soda with a bottle opener.
Obviously, trying to open the bottle with just our hands is either very difficult or
impossible for most of us.
When we use a bottle opener (a simple machine), the effort required to open the
bottle becomes much less. Why?
The bottle opener acts as a lever. When we use a lever, the force we have to exert
goes down (or said in reverse: the effect of our efforts are magnified by the lever).
Please note that the work that we do to remove the bottle cap is not reduced. This
is because we have to move the lever farther than the cap will move. Remember
that work is force times distance. We push the bottle opener a long distance
compared to how far the bottle cap moves while at the same time the force we
exert on the opener is less than what the opener exerts on the cap.
Mechanical Advantage is a measure of the amount of force exerted by a machine
compared to the amount of force a person exerts while using the machine.
Mechanical Advantage is defined as the resistance force (force exerted BY the
machine) divided by the effort force (force exerted ON the machine).
Mechanical Advantage in equation form is:
Fr
MA =
Fe
When the Mechanical Advantage is greater than one, the machine increases the
force exerted by a person.
When the Mechanical Advantage is equal to one, the machine likely only changes
the direction that the force needs to be applied. For example, a simple pulley has
a mechanical advantage of 1 (the effort force and the resistance force are equal in
size), but when using a pulley, you may apply the effort force in a different
direction than the resistance force which may make a big difference.
Pulling downward on a rope while using a pulley system to lift an object may be
easier than standing above the object and pulling upward.
Fr
MA =
Fe
Mechanical Advantage in equation form is:
Since work is related to force, we can talk about work performed BY the machine
(output work, Wo) and the work done ON the machine (input work, Wi).
Wo = Frdr
and
Wi = Fede
where dr is the displacement caused by the resistance force and de is the
displacement caused by the effort force.
Note: in an ideal machine Wo would be equal to Wi because a machine can only
increase force not energy.
Therefore:
Wo = Wi
and
So:
Fr
Fe
Frdr = Fede
=
de
dr
Ideal Mechanical Advantage is defined as the displacement of the effort force
divided by the displacement of the resistance force: IMA =
de
dr
However, in real machines the output work (Wo) will be less than the input work (Wi)
because some energy is always lost to things like friction.
The ratio of the output work to the input work is known as the efficiency of the
machine (e).
e =
Wo
Frd r
=
Wi
Fede
Wo
* 100
Wi
therefore we can also write efficiency as: e =
MA
* 100
IMA
The more efficient a machine is the closer the amount of output is to the input work.
Said another way, a less efficient machine will require a greater effort force to obtain a
given resistance force than a more efficient machine would.
There are six types of simple machine: You need to know these. See page 269 for
diagrams illustrating each one.
A lever is a simple machine that magnifies the effort force. Note that a lever
requires some form of pivot point.
A pulley is a simple machine that redirects the effort force.
A wheel and axle is a simple machine that can magnify and redirect the effort force.
Note that the axle acts as a pivot point.
An inclined plane is a simple machine that can magnify and redirect the effort force.
An wedge is a simple machine that can redirect the effort force.
An screw is a simple machine that can redirect and magnify the effort force.
The IMA of each of these six machines is the ratio of the distances moved. For
machines like the lever and the wheel and axle, this ratio can be replaced by the
ratio of the distance between where the force is applied and the pivot point.
Compound machine: a compound machine is a machine consisting of two or more
simple machines such that the resistance force of the first machine becomes the
effort force of the second machine.
An ax is an example of a compound machine.
The handle is a lever that you put your effort force into. This effort force becomes
the resistance force for the end of the handle. Since the handle fits into a wedge, the
resistance force of the handle becomes the effort force for the wedge. The wedge
will then apply this resistance force to whatever the ax hits.
The mechanical advantage of a compound machine is the product of each of the
mechanical advantages of the individual simple machines.
This same concept is true for the ideal mechanical advantage of compound
machines.
A wheel has a diameter of 0.650 m. The wheel is attached to an axle that has a
diameter of 0.0974 m. When a drive belt is wrapped around the axel and is pulled
with a force of 146.4 N, the wheel will advance 0.195 m. A) What is the IMA of
the machine? B) If the efficiency of this wheel and axel machine is 96.7%, what is
the MA? C) What is the resistance force? D) How far was the belt moved?
From the information given, we know: rr = (½)(0.650 m) = 0.325 m
re = (½)(0.0974 m) = 0.0487 m
Fe = 146.4 N
dr = 0.195 m
e = 96.7%
Click to see solution
r
0.0487 m
A) IMA = e =
= 0.1498 = 0.150
rr
0.325 m
B) e =
MA
* 100 so
IMA
C) MA =
D) IMA =
Fr
Fe
so
MA =
e∗IMA (96.7)∗(0.1498)
=
= 0.1449 = 0.145
100
100
Fr = FeMA = (146.4 N)(0.1449) = 21.21 N = 21.2 N
de
so de = drIMA = (0.195 m)(0.1498) = 0.02921 m = 0.0292 m
dr
See example problem on page 271 and practice problems on page 272