Energy is of no use without work.
A spring can be compressed to store potential energy, but it's of little use to us if we can't do something with that stored energy. The spring is a storehouse of energy – almost like a battery. We have to do work to compress it, and at a later time it can unleash its stored energy to do the work of pushing or pulling something.
Work transfers energy from one place or time to another, and possibly from one form to another.
Work is caused by a force. Without any force, there is no work. When a force works over a distance, work has been done, and its value, in units of energy, can be calculated.
Work ($w$) is force ($F$, see Newton's laws) multiplied by the distance ($d$) over which the force was exerted:
$$w = F \cdot d$$
The units of work are the units of energy, Joules (J). Joules are the SI unit of energy. Here's a unit analysis of F·d :
Units of work
$$\left( \frac{Kg \cdot m}{s^2} \right) \cdot m = \frac{Kg\cdot m^2}{s^2} = \bf J$$
Notice also that 1J = 1 N·m, or 1 Newton-meter (N·m). Once in a while you'll see N·m used instead of Joules.
Definition of work
Work is force exerted over a distance.
$$w = F \cdot d$$
The SI unit of work is the unit of energy, the Joule (J). $1 \, J = 1 \, Kg \cdot m^2 \cdot s^{-2}.$
Work can be interchanged with potential energy (PE) and kinetic energy (KE).
Equivalence of work and energy
Work is equivalent to (or can be "traded for" either kinetic energy ($KE$) or potential energy ($PE$). Two examples might help.
Gravitational PE
We know that the potential energy of an object of mass m lifted to a height h is $PE = mgh,$ where $m$ is the mass (in Kg), $g$ is the acceleration of gravity ($g = 9.81 m/s^2$ near the surface of Earth), and $h$ is the height in meters. The work that it takes to lift an object to such a height, giving it that $PE$, is exactly equal to the $PE$ gained. In this case we express the equivalence of work and $PE$ as
$$w = -\Delta (PE)$$
The negative is because the work done to "buy" $PE$ is always in the opposite direction of the motion that would result if the $PE$ were translated into $KE$.
The work of brakes in a car
A car moving in a straight line at a velocity $v$ has kinetic energy, $KE = \frac{1}{2}mv^2.$ If the brakes in that car are applied to cause it to stop, the amount of work done by the brakes in counteracting the $KE$ is exactly equal to the amount of $KE$ that the car initially had. In most vehicles, that energy goes out into the environment in the form of heat, the heating of the brakes, tires and road surface. In electric vehicles, much of that energy can be re-converted into stored electricity by using the motor "backward" as a generator.
There are other kinds of work, such as the pressure-volume ($PV$) work we encounter in chemistry, and we will discuss some others below.
$$w = -\Delta (KE)$$
The sign of the work in this case is dependent on what we get an object to do by doing work on it. We can, for example, speed up or slow down a moving object by doing work — exerting a force over a distance. Each results on a change in $KE$ equal in size to the work done.
Work against gravity
The work done in lifting an object a vertical distance is equivalent to the potential energy gained in doing so. The gravitational $PE$ is
$$PE = mgh$$
where $m$ is the mass, $g$ is the acceleration of gravity near the surface of Earth, and $h$ is the height.
A nice example of this is a roller coaster. Consider the roller coaster diagram below:
The first step in any roller coaster's journey is to be lifted to the top of the tallest hill (A). The amount of work done is $w = mgh$, which is exactly equal to the amount of potential energy gained: $PE = mgh$.
As the coaster begins to descend hill A, it loses potential energy (PE) and gains kinetic energy (KE), the energy of motion. In a sense, the coaster is "trading in" PE for KE.
At point B, which is a little higher than the origin of the coaster, most of the PE has been converted to KE, though some remains.
As the coaster passes B and begins to climb hill C, it loses KE and gains PE. All of this is due to the original work done in lifting the coaster to the top of hill A.
Upon descending hill C, the coaster returns to its initial height, thus converting all of its PE (therefore all of the work that was done) to KE.
A roller coaster could keep going like that forever if it wasn't for friction (between the wheels and the track and because of air resistance). The friction force does work on the coaster, much as the brakes on a car do work to stop it, reducing its KE. Because of friction, such a coaster could never climb another hill as high as hill A after the initial lift.
The pendulum
A pendulum is another great example of how work, potential and kinetic energy can be interconverted. The only thing that makes a pendulum swing back-and-forth is gravity.
We get a pendulum started by doing some work on it to swing the bob to one side, which not only move it to the side, but elevates it as well. The swing of a pendulum is then an ongoing exchange between kinetic and potential energy of the bob (the weight). Here's the basic idea.
In order to swing the pendulum to the left of right, the bob must be raised up against the force of gravity by doing work equal to the PE gained, $w = mgh.$
This work is equivalent to the amount of PE stored in the pendulum at this point in its arc. When the pendulum is let go, the force of gravity converts that potential energy into kinetic energy,
$$KE = \frac{1}{2}mv^2,$$
where $v$ is the velocity of the weight. The velocity at each of the turning points is zero, thus $KE = 0$ there, and $KE$ has its maximum value at the equilibrium (lowest) point of the pendulum.
The figure below illustrates how the motion of the pendulum interconverts between kinetic and potential energy.
The graph below the pendulum diagram, a plot of energy vs. position, aligns with points in its arc.
At any point in the path of the bob, the sum of PE and KE is constant, and equal to the work done in rasing it:
$$mgh + \frac{1}{2}mv^2 = k,$$
where k is a constant. At the equilibrium position, the KE is equal to the PE at the turning points, so we have
$$\frac{1}{2} mv^2 = mgh$$
We can solve this equation for the maximum velocity, $v_{max}:$
$$v_{max} = \sqrt{2 gh}$$
Example 1
A 50 Kg box is moved 15 m across a floor with a constant force of 27 N. How much work was done?
$$ \begin{align} w &= F d \\[5pt] &= 27 \frac{Kg m}{s^2} \cdot 15 \, m \\[5pt] &= 405 \, J \end{align}$$
Notice that we didn't use the 50 Kg mass of the box. Don't be distracted by red herrings like this. Know what you're looking for and calculate is. In this case, work depends on force and distance, so those are the only things we need.
I've expanded all of the units here, but once you're confident that $1 \, N \cdot m = 1 \, J,$ there's no need to.
Example 2
A SpaceX Dragon rocket can lift 6000 Kg of payload into space. The stratosphere of Earth's atmosphere begins at about 10 Km where the Dragon is lanched. How much work does the Dragon do in lifting a 6000 Kg payload to the edge of the stratosphere?
$$ \begin{align} w &= mgh \\[5pt] &= (6000 \, Kg)\left( 9.8 \, \frac{m}{s^2} \right)(1 \times 10^4 \, m) \\[5pt] &= 5.89 \times 10^8 \, J \\[5pt] &= 589 \, MJ \end{align}$$
Notice that I've converted 10 Km to meters (1 Km = 1,000 m).
In reporting numbers like this, it's nice to convert to a larger unit like mega-Joules (MJ), which makes the base number a little easier to comprehend and remember. Most humans have a better sense of what 100 is compared to 1 × 108. That's my theory, anyway.
Example 3
120 mJ of energy is used to push an object a distance of 1 cm. How much force was required to move the object?
$$w = F \, d \phantom{000} \color{#E90F89}{\longrightarrow} \phantom{000} F = \frac{w}{d}$$
So the force is
$$ \begin{align} F &= \frac{w}{d} = \frac{120 \times 10^{-3} \, J}{1 \times 10^{-2} \, m} \\[5pt] &= 12 \, N \end{align}$$
Don't forget your basic algebraic rearrangements of these simple formulas. Algebra is everywhere!
$$ \begin{align} \require{cancel} w &= F \, d \\[5pt] \frac{w}{d} &= \frac{F \cancel{d}}{\cancel{d}} \: \color{#E90F89}{\leftarrow \: \text{divide by d}} \\[5pt] \frac{w}{d} &= F \end{align}$$
Work in chemistry: PV work
The kind of work we encounter most often in chemistry is pressure-volume (PV) work.
The product of pressure and volume has units of energy. Here's how it works: The units of pressure are force divided by area, which we can write as Newtons ($N$) divided by $m^2$,
$$P = \frac{N}{m^2}$$
Multiplying by matching units of volume, $m^3$, gives
$$PV = \frac{n}{m^2} m^3 = N\cdot m = J$$
Recall that 1 N = 1 Kg·m·s-2, so multiplying by meters gives 1 J = 1 Kg·m2·s-2.
PV work is apparent when we heat a container of gas under constant pressure. This can be done using a cylinder fitted with a movable piston, atop of which sits a weight. The weight produces a constant force on the piston (which has a fixed area), and thus a constant pressure.
If we then heat the container, the volume must increase (recall from the ideal gas law that $V = nRT/P$). Play the animation to see how it works.
When the volume expands, we say (by convention) that the surroundings have done work on the system, and we call this positive work ($+w$). When the volume decreases, we say that the surroundings have done work on the system, and we call that negative work ($-w$).
Play the animation (arrow) a few times to get the idea.
The signs of heat and work (+ / -)
When the surroundings do work on the system (e.g. to compress the system), positive work has been done from the point of view of the system.
When the system does work on the surroundings (e.g. to expand its volume), negative work has been done.
Think of the sign conventions for heat and work as system-centric. Heat added to or work done on a system is positive. Otherwise, both are negative.
Equivalence of heat and work
Doing work on a chemical system, such as a liquid, is completely equivalent to adding an equal amount of heat (remember, the units are the same). The total energy input into a system is the heat it receives plus the work it does; it's part of the principle of conservation of energy.
Prescott Joule, one of the early developers of thermodynamics, performed an experiment that showed that mechanical work done on a container of water produces a temperature rise in the water equivalent to the amount of heat needed to cause the temperature change.
In Joule's experiment, a weight of known mass was dropped a known distance, thus producing a known amount of kinetic energy which produced mechanical stirring of a quantity of water. The friction of the stirring process produced a temperature rise equivalent to that kinetic energy.
Joule's experiment
Heat and work are equivalent in chemical systems.
Example
Suppose that the weight in Joules experiment is 1 Kg and drops a distance of 10 m (which could be accomplished by ten 1m drops if the weight was cranked back to the top of its 1m range slowly), and suppose our jar contains 1 L of water (1 Kg because the density of water is 1 g/mL).
Now $PE = mgh$ = (1 Kg)(9.8 m·s-2)(10 m) = 98 J. So that 98 J is "released" into the water through friction between the turning paddle blades and the water. The KE of motion of the paddles is translated to the motion of the water molecules, which is what heat is.
Now to calculate the temperature rise, we need a formula with which you might not be familiar (it's here): The heat ($q$) required to change the temperature of $m$ grams of a substance by $\Delta T$ ˚C is
$$q = mC\Delta T$$
where $C$ is a property (which Joule measured) called the specific heat capacity. The specific heat capacity of water is C = 4.184 J/g˚C, so the temperature rise would be
$$ \begin{align} \Delta T &= \frac{q}{mC} \\[5pt] &= \frac{98 \, J}{(1000 \, g)(4.184 \, J/g˚C)} \\[5pt] &= 0.023 \, ˚C \end{align}$$
which is a small, but measurable temperature rise.
Practice problems
-
How much work (in Joules) is required to lift a 330 Kg piano to a window located 9.5 m from the ground?
Solution
The work of lifting against gravity is $w = mgh,$ where m is mass, h is height, and g is the acceleration of gravity on Earth.
$$ \begin{align} w &= mgh \\[5pt] &= (330 \, Kg)\left( 9.8 \frac{m}{s^2} \right)(9.5 m) \\[5pt] &= 30,273 \, J = 30.3 \, KJ \end{align}$$
-
A force of 90 N is applied to a crate, moving it 20 m along the direction of the applied force. How much work is done on the crate?
Solution
Work = force × distance, so we have
$$ \begin{align} w &= F \cdot d \\[5pt] &= 90 N \cdot 20 m \\[5pt] &= 1800 \, N\cdot m \\[5pt] &= 1800 \, J \\[5pt] &= 1.8 \, KJ \end{align}$$
-
A box rests on a horizontal surface (and we'll ignore friction). One person pushes on the box with a force of 15N to the right, and another pushes with a force of 12N to the left. The box moves 3.0m to the right. Calculate the work done by (a) the first person, (b) the second person, and (c) the net force.
Solution
It's always helpful to sketch a diagram. Let's let force in the rightward direction be a positive force, then the 12 N force will be a negative force.
$$w_{15} = F \cdot d = 15 \, N \cdot 3.0 \, m = 45 \, J$$
$$w_{12} = F \cdot d = -12 \, N \cdot 3.0 \, m = -36 \, J$$
Now the net work done is the difference, + 9.0 J. We could also first find the net force, $F_{net} = 15 - 12 = +3 \, N.$ Then the work done by that net force would be
$$w = F_{net} \cdot d = 3 \, N \cdot 3.0 \, m = 9.0 \, J.$$
-
A climber climbs 15.2 m up a wall, expending 6250 J of energy (work) to do so. Calculate the mass of the climber.
Solution
Work = force × distance, so the work of lifting is $w = mgh,$ where the force is $F = mg$ and the distance is the height. Rearranging to solve for mass gives
$$w = mgh \: \color{#E90F89}{\longrightarrow} \: m = \frac{w}{gh}$$
$$ \require{cancel} \begin{align} m &= \frac{6250 \, J}{9.8 \frac{m}{s^2} \cdot 15.2 \, m} \\[5pt] &= \frac{6260 \frac{Kg \cancel{m^2}}{\cancel{s^2}}}{9.8 \frac{\cancel{m}}{\cancel{s^2}} \cdot 15.2 \cancel{m}} \\[5pt] &= 42 \, Kg \end{align}$$
In the second step above, the units were expanded so you can see that the result has the units we want, units of mass (Kg).
SI units
SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units
| length | meter | (m) |
| mass | Kilogram | (Kg) |
| time | second | (s) |
| force | Newton | N |
| energy | Joule | J |
Red herring
Red herring is a metaphor for a kind of logical fallacy. A school of herring can contain millions of silvery 20-30 cm fish swimming in a synchronized way. The idea is that if you notice that one is red, it will capture your attention, and you might miss the main point: millions of silver fish swimming as one. Sometimes people also say "Don't miss the forest for (looking at) the trees."
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