Work, Energy and Power
Introduction
Energy gives us one more tool to use to analyze physical situations. When forces
and accelerations are used, you usually freeze the action at a particular instant in
time, draw a free-body diagram, set up force equations, figure out accelerations,
etc. With energy the approach is usually a little different. Often you can look at the
starting conditions (initial speed and height, for instance) and the final conditions
(final speed and height), and not have to worry about what happens in between.
The initial and final information can often tell you all you need to know.
Work and energy
Whenever a force is applied to an object, causing the object to move, work is done
by the force. If a force is applied but the object doesn't move, no work is done; if a
force is applied and the object moves a distance d in a direction other than the
direction of the force, less work is done than if the object moves a distance d in the
direction of the applied force.
The physics definition of "work" is:
The unit of work is the unit of energy, the joule (J). 1 J = 1 N m.
Work can be either positive or negative: if the force has a component in the same
direction as the displacement of the object, the force is doing positive work. If the
force has a component in the direction opposite to the displacement, the force does
negative work.
If you pick a book off the floor and put it on a table, for example, you're doing
positive work on the book, because you supplied an upward force and the book
went up. If you pick the book up and place it gently back on the floor again,
though, you're doing negative work, because the book is going down but you're
exerting an upward force, acting against gravity. If you move the book at constant
speed horizontally, you don't do any work on it, despite the fact that you have to
exert an upward force to counter-act gravity.
Kinetic energy
An object has kinetic energy if it has mass and if it is moving. It is energy
associated with a moving object, in other words. For an object traveling at a speed
v and with a mass m, the kinetic energy is given by:
The work-energy principle
There is a strong connection between work and energy, in a sense that when there
is a net force doing work on an object, the object's kinetic energy will change by an
amount equal to the work done:
Note that the work in this equation is the work done by the net force, rather than
the work done by an individual force.
Gravitational potential energy
Let's say you're dropping a ball from a certain height, and you'd like to know how
fast it's traveling the instant it hits the ground. You could apply the projectile
motion equations, or you could think of the situation in terms of energy (actually,
one of the projectile motion equations is really an energy equation in disguise).
If you drop an object it falls down, picking up speed along the way. This means
there must be a net force on the object, doing work. This force is the force of
gravity, with a magnitude equal to mg, the weight of the object. The work done by
the force of gravity is the force multiplied by the distance, so if the object drops a
distance h, gravity does work on the object equal to the force multiplied by the
height lost, which is:
work done by gravity = W = mgh (h = height lost by the object)
An alternate way of looking at this is to call this the gravitational potential energy.
An object with potential energy has the potential to do work. In the case of
gravitational potential energy, the object has the potential to do work because of
where it is, at a certain height above the ground, or at least above something.
Spring potential energy
Energy can also be stored in a stretched or compressed spring. An ideal spring is
one in which the amount the spring stretches or compresses is proportional to the
applied force. This linear relationship between the force and the displacement is
known as Hooke's law. For a spring this can be written:
F = kx, where k is known as the spring constant.
k is a measure of how difficult it is to stretch a spring. The larger k is, the stiffer
the spring is and the harder the spring is to stretch.
If an object applies a force to a spring, the spring applies an equal and opposite
force to the object. Therefore:
force applied by a spring : F = - kx
where x is the amount the spring is stretched. This is a restoring force, because
when the spring is stretched, the force exerted by by the spring is opposite to the
direction it is stretched. This accounts for the oscillating motion of a mass on a
spring. If a mass hanging down from a spring is pulled down and let go, the spring
exerts an upward force on the mass, moving it back to the equilibrium position,
and then beyond. This compresses the spring, so the spring exerts a downward
force on the mass, stopping it, and then moving it back to the equilibrium and
beyond, at which point the cycle repeats. This kind of motion is known as simple
harmonic motion, which we'll come back to later in the course.
The potential energy stored in a spring is given by:
where x is the difference between the spring's length and its unstrained length.
In a perfect spring, no energy is lost; the energy is simply transferred back and
forth between the kinetic energy of the mass on the spring and the potential energy
of the spring (gravitational PE might be involved, too).
Conservation of energy
We'll take all of the different kinds of energy we know about, and even all the
other ones we don't, and relate them through one of the fundamental laws of the
universe.
The law of conservation of energy states that energy can not be created or
destroyed, it can merely be changed from one form of energy to another. Energy
often ends up as heat, which is thermal energy (kinetic energy, really) of atoms and
molecules. Kinetic friction, for example, generally turns energy into heat, and
although we associate kinetic friction with energy loss, it really is just a way of
transforming kinetic energy into thermal energy.
The law of conservation of energy applies always, everywhere, in any situation.
There is another conservation idea associated with energy which does not apply as
generally, and is therefore called a principle rather than a law. This is the principle
of the conservation of mechanical energy.
Conservation of energy
The conservation of mechanical energy
Mechanical energy is the sum of the potential and kinetic energies in a system. The
principle of the conservation of mechanical energy states that the total mechanical
energy in a system (i.e., the sum of the potential plus kinetic energies) remains
constant as long as the only forces acting are conservative forces. We could use a
circular definition and say that a conservative force as a force which doesn't
change the total mechanical energy, which is true, but might shed much light on
what it means.
A good way to think of conservative forces is to consider what happens on a round
trip. If the kinetic energy is the same after a round trip, the force is a conservative
force, or at least is acting as a conservative force. Consider gravity; you throw a
ball straight up, and it leaves your hand with a certain amount of kinetic energy. At
the top of its path, it has no kinetic energy, but it has a potential energy equal to the
kinetic energy it had when it left your hand. When you catch it again it will have
the same kinetic energy as it had when it left your hand. All along the path, the
sum of the kinetic and potential energy is a constant, and the kinetic energy at the
end, when the ball is back at its starting point, is the same as the kinetic energy at
the start, so gravity is a conservative force.
Kinetic friction, on the other hand, is a non-conservative force, because it acts to
reduce the mechanical energy in a system. Note that non-conservative forces do
not always reduce the mechanical energy; a non-conservative force changes the
mechanical energy, so a force that increases the total mechanical energy, like the
force provided by a motor or engine, is also a non-conservative force.
An example
Consider a person on a sled sliding down a 100 m long hill on a 30° incline. The
mass is 20 kg, and the person has a velocity of 2 m/s down the hill when they're at
the top. How fast is the person traveling at the bottom of the hill? All we have to
worry about is the kinetic energy and the gravitational potential energy; when we
add these up at the top and bottom they should be the same, because mechanical
energy is being conserved.
At the top: PE = mgh = (20) (9.8) (100sin30°) = 9800 J
KE = 1/2 mv2 = 1/2 (20) (2)2 = 40 J
Total mechanical energy at the top = 9800 + 40 = 9840 J
At the bottom: PE = 0 KE = 1/2 mv2
Total mechanical energy at the bottom = 1/2 mv2
If we conserve mechanical energy, then the mechanical energy at the top must
equal what we have at the bottom. This gives:
1/2 mv2 = 9840, so v = 31.3 m/s.
Modifying the example
Now let's worry about friction in this problem. Let's say, because of friction, the
velocity at the bottom of the hill is 10 m/s. How much work is done by friction,
and what is the coefficient of friction?
The sled has less mechanical energy at the bottom of the slope than at the top
because some energy is lost to friction (the energy is transformed into heat, in other
words). Now, the energy at the top plus the work done by friction equals the
energy at the bottom.
Energy at the top = 9840 J
Energy at the bottom = 1/2 mv2 = 1000 J
Therefore, 9840 + work done by friction = 1000, so friction has done -8840 J
worth of work on the sled. The negative sign makes sense because the frictional
force is directed opposite to the way the sled is moving.
How large is the frictional force? The work in this case is the negative of the force
multiplied by the distance traveled down the slope, which is 100 m. The frictional
force must be 88.4 N.
To calculate the coefficient of friction, a free-body diagram is required.
In the y-direction, there is no acceleration, so:
The coefficient of kinetic friction is the frictional force divided by the normal
force, so it's equal to 88.4 / 169.7 = 0.52.
Applying energy concepts
With energy, we've got another tool in our physics toolbox to use to attack
problems. Let's try some more examples to see how these energy concepts are
applied.
Air resistance
Air resistance can be a tricky thing to account for. Generally, the force applied by
air to a moving object is proportional to the square of the speed of the object. We
know how to handle constant forces, but a force that depends on speed is a
different story. If we apply energy to the problem, however, we can make some
headway.
Take, for instance, a simple coffee filter. It's very light, and has a relatively large
surface area. If you let a coffee filter fall, it falls at roughly constant speed; this is
because the force of gravity is balanced by the force of air resistance. Note that the
constant speed is known as the terminal velocity - the same kind of thing applies to
skydivers. Let's say we drop a coffee filter (from rest) from a height of 1.5 m. If
there was no air resistance, the filter would take about 0.55 seconds to hit the
ground, and be traveling at a final speed of 5.4 m/s.
The actual time for a coffee filter to fall 1.5 m is more like 1.5 seconds (try timing
it yourself). Doing some estimating, making the assumption that the speed is
constant over this time interval gives a speed of 1.0 m/s. How much work is done
by air resistance? Approximately what is the force of air resistance on the filter?
Applying conservation of energy makes it easy to determine the work done. This is
as long as the mass is known - weighing a stack of filters on a balance, I found that
one filter weighs approximately 1 gram. This gives the coffee filter an initial
potential energy, relative to the floor, of 0.001 x 1.5 x 9.8 which is approximately
0.015 J. When it hits the ground the kinetic energy is 0.0005 J. Clearly, energy has
been lost - in fact, almost all the energy has been lost! The work done by friction
equals the change in energy, so it's almost -0.015 J. It's negative because it opposes
the motion: the displacement is down and the force is up.
The work is the force times the displacement. The average force in this case is
therefore about 0.01 N. This is the same answer you get if you set the gravitational
force equal to the force of air resistance, so that's a good sign that we're doing
things correctly.
An example, using the PE of a spring
A 0.123 kg block sits on a plane inclined at 20°. The block is pushed back against
a spring (k = 23.4 N / m), compressing the spring by 0.345 m. When the block is
let go, it is accelerated up the incline by the spring. The coefficient of kinetic
friction between the block and the incline is 0.220.
How far up the incline does the block go?
Attack this problem using work and energy. The initial energy (stored in the
spring) is equal to the final energy (gravitational PE) plus whatever gets lost to
friction. Writing this as an equation gives:
Energy before = energy after + energy lost to friction
A free-body diagram tells us that:
Substituting this in to
transforms the energy equation to:
Solving this for d, the distance the block travels up the slope, gives:
Plugging in all the numbers gives:
Power
Introduction
Being able to do work is not just what's important; how fast you can do work is
also an important factor. Power is the measure of how fast work is done.
Computers have more calculating power than we do; a sports car generally has a
more powerful engine than an economy car. Power is the rate at which work is
done and the rate at which energy is used. The unit for power is the watt (W).
An interesting calculation is the average power output of a human being. This can
be determined from the amount of energy we consume in a day in the way of food.
Most of us take in something like 2500 "calories" in a day, although what we call
calories is really a kilocalorie; assuming we use up all this energy in a day (a
reasonable assumption considering we'll have to eat tomorrow, too) we can use this
as our energy output per day.
First, take the 2.5 x 106 cal and convert to Joules, using the conversion factor 4.18
J / cal. This gives roughly 1 x 107 J. Figuring out our average power output, we
simply divide the energy by the number of seconds in a day, 86400, which gives a
bit more than 100 W. In other words, on the average, we're just a little brighter
than your average light bulb.
Calculating power from speed
Power is work over time, and work is force multiplied by distance. Power can be
written as:
Power : P = F s / t (F is the force in the direction of s, the displacement)
Displacement over time is velocity, so power can also be written in this form:
Power : P = F v (F is the force in the direction of the velocity)
Here's an example of when you might use this. Let's say you're riding your bicycle
on a level road at a constant speed of 10 m/s. You're riding into a headwind, and
you're burning up energy at the rate of 500 J/s. If you assume that 80% of this
energy is going to overcome air resistance, how much force is the air exerting on
you?
The power used to overcome air resistance is 80% of 500 W, which is 400 W.
Assuming there aren't any other forces acting against you, then dividing this by
your speed should give you the force the air exerts on you. This works out to 40 N.
Example - A car climbing a hill
A car with a mass of 900 kg climbs a 20° incline at a steady speed of 60 km/hr. If
the total resistance forces acting on the car add to 500 N, what is the power output
of the car in watts? In horsepower?
Note that the gravitational force is the only force which needs to be split into
components. mg sin20° acts down the slope; mg cos20° acts into the slope.
Fr represents the resistance forces.
A good place to start here is with the free-body diagram. The power output by the
car's engine goes into the force directed up the slope. This force is actually static
friction exerted on the drive wheels by the road - the road exerts this force because
the engine causes the drive wheels to rotate.
The velocity is constant, so the forces must balance. Applying Newton's second
law in the x-direction gives:
F - Fr - mg sin20° = 0
The force up the slope is then F =Fr + mg sin20° = 500 + 3017 = 3517 N
Converting the car's speed to m/s gives 16.67 m/s. The power output can then be
found from
P = Fv = (3517) (16.67) = 58620 W.
This can be converted to horsepower, using the conversion 746 W = 1 hp. This
gives a power output of 78.6 hp.
Most cars have engines with power outputs of about 100 hp, so this is a reasonable
value (and there's nothing in the question to say that this has to be the maximum
power output of the car).
Momentum
Introduction
There are two kinds of momentum, linear and angular. A spinning object has
angular momentum; an object traveling with a velocity has linear momentum. For
now, and throughout chapter 7, we'll deal with linear momentum, and just refer to
it as momentum, without the linear.
There are 4 really important things to know about momentum. The first is how
momentum is defined, as the product of mass times velocity:
momentum : p = mv
The second note is built into this equation; momentum is a vector, and the
momentum has the same direction as the velocity.
The third point is the relationship between momentum and force. We've talked a
lot about forces in the last few weeks, and there is a strong connection between
force and momentum. In fact, Newton's second law was first written (by Newton
himself, of course) in terms of momentum, rather than acceleration. A force acting
for a certain time (this is known as an impulse) produces a change in momentum.
Again, this is a vector equation, so the change in momentum is in the same
direction as the force.
The fourth really important point about momentum is that momentum is
conserved; the total momentum of an isolated system is constant. Note that
"isolated" means that no external force acts on the system, which is a set of
interacting objects. If a system does have a net force acting, then the momentum
changes according to the impulse equation.
Momentum conservation applies to a single object, but it's a lot more interesting to
look at a situation with at least two interacting objects. If two objects (a car and a
truck, for example) collide, momentum will always be conserved. There are three
different kinds of collisions, however, elastic, inelastic, and completely inelastic.
Just to restate, momentum is conserved in all three kinds of collisions. What
distinguishes the collisions is what happens to the kinetic energy.
Types of collisions: (momentum is conserved in each case)
•
•
•
elastic - kinetic energy is conserved
inelastic - kinetic energy is not conserved
completely inelastic - kinetic energy is not conserved, and the colliding
objects stick together after the collision.
The total energy is always conserved, but the kinetic energy does not have to be;
kinetic energy is often transformed to heat or sound during a collision.
A1-D collision example
A car of mass 1000 kg travels east at 30 m/s, and collides with a 3000 kg truck
traveling west at 20 m/s.
(a) If the collision is completely inelastic, how fast are the car and truck going, and
in what direction, after the collision? What percentage of the kinetic energy is lost
in the collision?
(b) What happens if the collision is elastic?
(a) Car crashes are often completely inelastic, with much of the kinetic energy
going into deforming the cars. Momentum is always conserved, though, so, using c
for car and t for truck, (and f for final) the conservation of momentum equation is:
If we take east as the positive direction, then the truck's velocity goes into the
equation with a negative sign, so: vf = [ (1000) (30) + (3000) (-20) ] / (1000 +
3000) = -7.5 m/s, which is 7.5 m/s west
The change in kinetic energy can be found by adding up the kinetic energy before
and after the collision:
KE lost = 1050000 - 112500 = 937500 J
Percentage of KE lost = 100% x 937500 / 1050000 = 89.3%
So, a great deal of the kinetic energy is lost in the collision.
(b) What would happen if the car and truck were both made out of rubber and the
collision was elastic, with no loss of kinetic energy. In this case the calculations are
a lot more complicated, because we have to combine the energy conservation
equation with the momentum conservation equation:
In this case, after some nice algebraic manipulation (which is worth trying on your
own), the final velocities of the car and truck work out to:
Note that if you were driving the car, you would experience a much greater force
in the case of an elastic collision than in a completely inelastic collision, in which
much of the energy is absorbed by the deformation of the car. Let's say you have a
mass of 50 kg, and that the collision lasts for 0.1 seconds. In the case of the
completely inelastic collision, your momentum would change from 50 kg x 30 m/s
east = 1500 kg m/s east to 50 kg x 7.5 m/s west = -375 kg m/s east, which is a net
change of 1875 kg m/s. This change in momentum is produced by an average force
acting for the 0.1 s of the collision, so the force works out to 18750 N.
In the elastic collision, your momentum would change from 1500 kg m/s east to 50
kg x 45 m/s west = -2250 kg m/s east, for a net change of 3750 kg m/s, exactly
twice that in the completely inelastic case. The force you would experience would
therefore also be doubled.
Back to impulse
Before doing an example of a collision in 2 dimensions, let's look at a short
example of how the impulse equation is applied. Recall that impulse is a force
acting for a particular time, producing a change in momentum:
Consider a hose spraying water directly at a wall. If 3 kg of water emerge from the
hose every second, and the speed of the water is 10 m/s, how much force is exerted
on the wall by the water?
The first step in coming to an answer is making an assumption, that the water does
not bounce back from the wall, but is simply stopped by the wall. In this case, the
change in momentum for one second's worth of water is -30 kg m/s. To produce
this change in momentum, the wall must exert a force on the water of -30 N, which
is 30 N in the direction opposite to the direction the water travels from the hose.
The water exerts an equal and opposite force on the wall, 30 N in the direction the
hose points.
Note that is the water bounced off the wall and came back with a momentum of 30
kg m/s towards the hose, that would represent a net change in momentum of 60 kg
m/s towards the hose, because momentum is a vector. In that case the force exerted
by the water on the wall would be twice as high, 60 N.
Collisions
A 2-D collision
Because momentum is a vector, whenever we analyze a collision in two or three
dimensions the momentum has to be split up into components. Consider the
following example to see how this works. A 1000 kg car traveling at 30 m/s, 30°
south of east, collides with a 3000 kg truck heading northeast at 20 m/s. The
collision is completely inelastic, so the two vehicles stick together after the
collision. How fast, and in what direction, are the car and truck traveling after the
collision?
To solve this problem, simply set up two conservation of momentum equations,
one for the y-direction (positive y being north) and another for the x-direction
(positive x being east). Setting up a vector diagram for the momentum is a good
idea, too, like this:
To set up the two momentum conservation equations, simply write down the
equation for the momentum before the collision in the y-direction and set it equal
to the momentum after the collision in the y-direction, and then do the same thing
in the x-direction.
The y equation can be rearranged to solve for the y component of the final
velocity:
Similarly, in the x-direction:
It would be easy to figure out the final velocity using the Pythagorean theorem, but
let's find the angle first instead, by dividing the y equation by the x equation:
Now let's go back to get the final velocity from the Pythagorean theorem :
This gives a final velocity of 18.4 m/s at an angle of 21.8° north of east.
We could figure out how much energy is lost during the collision if we wanted to;
because energy is a scalar rather than a vector, this is done the same way in 2-D
(and 3-D) as it is in 1-D.