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Energy is a word which tends to be used a lot in everyday life. Though it is often used quite loosely, it does have a very specific physical meaning.
Energy is a measurement of the ability of something to do work. It is not a material substance. Energy can be stored and measured in many forms.
Although we often hear people talking about energy consumption, energy is never really destroyed. It is just transferred from one form to another, doing work in the process. Some forms of energy are less useful to us than others—for example, low level heat energy. It is better to talk about the consumption or extraction of energy resources, for example coal, oil, or wind, than consumption of energy itself.
  • A speeding bullet has a measurable amount of energy associated with it; this is known as kinetic energy. The bullet gained this energy because work was done on it by a charge of gunpowder which lost some chemical potential energy in the process.
  • A hot cup of coffee has a measurable amount of thermal energy which it acquired via work done by a microwave oven, which in turn took electrical energy from the electrical grid.
In practice, whenever work is done to move energy from one form to another, there is always some loss to other forms of energy such as heat and sound. For example, a traditional light bulb is only about 3% efficient at converting electrical energy to visible light, while a human being is about 25% efficient at converting chemical energy from food into work.

How do we measure energy and work?

The standard unit used to measure energy and work done in physics is the joule, which has the symbol J. In mechanics, 1 joule is the energy transferred when a force of 1 Newton is applied to an object and moves it through a distance of 1 meter.
Another unit of energy you may have come across is the Calorie. The amount of energy in an item of food is often written in Calories on the back of the packet. A typical 60 gram chocolate bar for example contains about 280 Calories of energy. One Calorie is the amount of energy required to raise 1 kg of water by 1^{\circ}degree Celsius.

^{\circ}degree
^{\circ}degree
This is equal to 4184 joules, so one chocolate bar has 1.17 million joules or 1.17 MJ of stored energy. That’s a lot of joules!

How long do I have to push a heavy box around to burn off one chocolate bar?

Suppose we’re feeling guilty about eating a chocolate bar; we want to find how much exercise we need to do to offset those extra 280 Calories. Let’s consider a simple form of exercise: pushing a heavy box around a room, see Figure 1 below.
Figure 1: A person pushes a box to the right.
Using a bathroom scale between ourselves and the box, we find that we can push with a force of 500 N. Meanwhile, we use a stopwatch and measuring tape to measure our speed. This comes out to be 0.25 meters per second.
So how much work do we need to do to the box to burn off the candy bar? The definition of work, WWW, is below:
\Large W = F\cdot \Delta xW=FΔxW, equals, F, dot, delta, x
The work we need to do to burn the energy in the candy bar is E=280 \mathrm{cal} \cdot 4184 \mathrm{J/cal}=1.17 \mathrm{MJ}E=280cal4184J/cal=1.17MJ.
Therefore, the distance, Δx, we need to move the box through is:
\begin{aligned} W &= F\cdot \Delta x \\ 1.17 \text{ MJ} &= (500 \text{ N})\cdot \Delta x \\ \dfrac{1.17\times 10^6 \text{ J}}{500 \text{ N}} &= \Delta x \\ 2,340 \text{ m}&= \Delta x \end{aligned}W1.17 MJ500 N1.17×106 J2,340 m=FΔx=(500 N)Δx=Δx=Δx
Remember, however, that our bodies are about 25% efficient at transferring stored energy from food into work. The actual energy we will offset is then four times as high as the work done to the box. So, we only need to push the box through a distance of 585 m, which is still over five football fields long. Given the known speed of 0.25 m/s that will take us:
\frac{585 \mathrm{m}}{0.25 \mathrm{m/s}}=2340 \mathrm{s}0.25m/s585m=2340s
Exercise: Suppose that the force that we apply to the box, see Figure 1 above, is initially reduced but increases to a constant value as we warm up. For instance, in the graph below we see that as the box is displaced further—i.e., xxxgets larger—the force, FFF, increases for the first 30 m, see Figure 2 below. How could we find the work done during the period where the force is changing?
\small{10}10\small{20}20\small{30}30\small{40}40\small{50}50\small{60}60\small{70}70\small{100}100\small{200}200\small{300}300\small{400}400\small{500}500\small{600}600F (N)F(N)x (m)x(m)
Figure 2: A varying force on the box.
If the force is not constant, one way to determine the work done is to divide the problem up into small sections over which the change is negligible and sum up the work done in each section. Just as we have learned when looking at velocity time graphs, this can be done by calculating the area under the curve using geometry.
\small{10}10\small{20}20\small{30}30\small{40}40\small{50}50\small{60}60\small{70}70\small{100}100\small{200}200\small{300}300\small{400}400\small{500}500\small{600}600F (N)F(N)x (m)x(m)
The work done by a force is equal to the area under a force vs. position graph. In the case of Figure 2, it would be:
(200\text{ N} \cdot 30 \text{ m})+ \frac{1}{2}\left((500\text{ N}-200\text{ N})\cdot 30 \text{ m}\right)=10500 \mathrm{J}(200 N30 m)+21((500 N200 N)30 m)=10500J for the initial 30 \text{ m}30 m30, space, m of displacement.
Similarly, the work done for the final 40 m of displacement would be:
500\text{ N} \cdot 40 \text{ m}=20,000 \mathrm{J}500 N40 m=20,000J

What if we aren’t pushing straight on?

There is one thing we need to watch out for when doing these problems. The previous equation, W = F \cdot \Delta xW=FΔxW, equals, F, dot, delta, x, doesn’t take into account situations where the force we are applying is not in the same direction as the motion.
For instance, imagine we use a rope to pull on the box. In that case there will be an angle between the rope and the ground. To untangle this situation, we begin by drawing a triangle to separate out the horizontal and vertical components of the applied force.
The key point here is that it is only the component of the force, F_{||}FF, start subscript, vertical bar, vertical bar, end subscript, that lies parallel to the displacement that does work on an object. In the case of the box shown above, only the horizontal component of the applied force, F \text{cos}(\theta)Fcos(θ)F, c, o, s, left parenthesis, theta, right parenthesis, is doing work on the box since the box is being displaced horizontally. This means that a more general equation for the work done on the box by a force at an angle θ could be written as:
W=F_{||}\cdot\Delta xW=FΔxW, equals, F, start subscript, vertical bar, vertical bar, end subscript, dot, delta, x
W=(F \cos{\theta})\cdot\Delta xW=(Fcosθ)ΔxW, equals, left parenthesis, F, cosine, theta, right parenthesis, dot, delta, x
Which is more often written as,
\Large W=F \Delta x\cos{\theta}W=FΔxcosθW, equals, F, delta, x, cosine, theta
Exercise: Suppose we use a rope to pull the box, and the angle between the rope and the ground is 30º. This time we pull along the rope with a force of 500 N. How much of a chocolate bar can we eat this time if we pull the box through the same 585 m?
W_{\text{on box}}=F\Delta x\text{ cos}\theta=(500\text{ N})(585 \text{ m})\text{cos}30^{\circ}W, start subscript, o, n, space, b, o, x, end subscript, equals, F, delta, x, space, c, o, s, theta, equals, left parenthesis, 500, space, N, right parenthesis, left parenthesis, 585, space, m, right parenthesis, c, o, s, 30, degree
W_{\text{on box}}=253,312 \text{ J}W, start subscript, o, n, space, b, o, x, end subscript, equals, 253, comma, 312, space, J
W_{\text{by body}}=4 \times W_{\text{on box}}=4\times 253,312 \text{ J}W, start subscript, b, y, space, b, o, d, y, end subscript, equals, 4, times, W, start subscript, o, n, space, b, o, x, end subscript, equals, 4, times, 253, comma, 312, space, J
W_{\text{by body}}=1,013,248 \text{ J}W, start subscript, b, y, space, b, o, d, y, end subscript, equals, 1, comma, 013, comma, 248, space, J
1.17 \times 10^61, point, 17, times, 10, start superscript, 6, end superscript
\text{ fraction of candy bar}=\dfrac{W_{\text{by body}}}{E_{\text{candy bar}}}=\dfrac{1,013,248 \text{ J}}{1.17 \times 10^6 \text{ J}}=0.87space, f, r, a, c, t, i, o, n, space, o, f, space, c, a, n, d, y, space, b, a, r, equals, start fraction, W, start subscript, b, y, space, b, o, d, y, end subscript, divided by, E, start subscript, c, a, n, d, y, space, b, a, r, end subscript, end fraction, equals, start fraction, 1, comma, 013, comma, 248, space, J, divided by, 1, point, 17, times, 10, start superscript, 6, end superscript, space, J, end fraction, equals, 0, point, 87

What about lifting weights instead?

In the previous example, we were doing work on a box which we were pushing around a floor. In doing so, we were working against a frictional force.
Another common form of exercise is lifting weights. In this case we are working against the force of gravity rather than friction. Using Newton’s laws we can find the force, FFF, required to lift a weight with mass mmm straight up, placing it on a rack which is at a height hhh above us:
F=mgF=mgF, equals, m, g
The change in position—previously \Delta xΔxdelta, x—is simply the height, so the work, WWW, that we have done in lifting the weight is then
W=m g hW=mghW, equals, m, g, h
The exercise we have done in lifting the weight has resulted in energy being stored in the form of gravitational potential energy. It is called potential energy because it has the potential to be released at any moment with a crash as the weight falls back to the ground.
We did positive work on the weight since we exerted our force in the same direction as the displacement of the weight, i.e., upward. The work done by gravity on the weight while it was lifted was negative since the force of gravity is directed in the opposite direction to the displacement. Also, since the weight is stationary after the lift, we know that the work that we have done is exactly canceled out by the work done by gravity. The work done by us is mghmghm, g, h, and the work done by gravity is -mghmghminus, m, g, h. We will talk more about this when we look into kinetic energy.
OK, let’s put some numbers in and find how much of that chocolate bar we would offset by lifting a weight of 50 kg up to a height of 0.5 m. The work done to the weight is
W=(50\mathrm{kg})(9.81\mathrm{m/s^2})(0.5\mathrm{m}) = 245.25 \mathrm{J}W=(50kg)(9.81m/s2)(0.5m)=245.25J
OK, so how many 280 Calorie—i.e., 1.17 \times 10^61.17×1061, point, 17, times, 10, start superscript, 6, end superscript joule—chocolate bars is this? Well, 245.25 J is about \dfrac{1}{4770}47701start fraction, 1, divided by, 4770, end fraction of a chocolate bar. But remember, our bodies are only about 25% efficient, so the work done by the person is actually four times larger, about 981.8 J, which is \dfrac{1}{1190}11901start fraction, 1, divided by, 1190, end fraction chocolate bars. So, if we can lift this weight once every 2 seconds, it will take us about 2380 seconds or 40 minutes of hard work to burn off this chocolate bar!
F=maF, equals, m, a\text{kg} \dfrac{\text{m}}{\text{s}^2}k, g, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction
1 \text{ Joule}=1 \text{ N}\cdot\text{}m1, space, J, o, u, l, e, equals, 1, space, N, dot, m
1 \text{ Joule}=1 \text{ kg} \dfrac{\text{m}}{\text{s}^2}\cdot m=1 \text{ kg} \dfrac{\text{m}^2}{\text{s}^2}1, space, J, o, u, l, e, equals, 1, space, k, g, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction, dot, m, equals, 1, space, k, g, start fraction, m, start superscript, 2, end superscript, divided by, s, start superscript, 2, end superscript, end fraction
\text{ kg} \dfrac{\text{m}^2}{\text{s}^2}space, k, g, start fraction, m, start superscript, 2, end superscript, divided by, s, start superscript, 2, end superscript, end fraction

What about simply holding a weight stationary?

One frequent source of confusion people have with the concept of work comes about when thinking about holding a heavy weight stationary above our heads, against the force of gravity. We are not moving the weight through any distance, so no work is being done to the weight. We could also achieve this by placing the weight on a table; it is clear that the table is not doing any work to keep the weight in position. Yet, we know from our experience that we get tired when doing the same job. So what is going on here?
It turns out that what is actually happening here is that our bodies are doing work on our muscles to maintain the necessary tension to hold the weight up. The body does this by sending a cascade of nerve impulses to each muscle. Each impulse causes the muscle to momentarily contract and release. This all happens so fast that we might only notice a slight twitching at first. Eventually though, not enough chemical energy is available in the muscle and it can no longer keep up. We then begin to shake and eventually must rest for a while. So work is being done, it is just not being done on the weight.

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