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### Course: Physics library > Unit 5

Lesson 1: Work and energy- Introduction to work and energy
- Work and energy (part 2)
- Conservation of energy
- What are energy and work?
- What is kinetic energy?
- What is gravitational potential energy?
- What is conservation of energy?
- Work and the work-energy principle
- Work as the transfer of energy
- Work example problems
- Work as area under curve
- Thermal energy from friction
- What is thermal energy?
- Work/energy problem with friction
- Conservative forces
- Power
- What is power?

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# What are energy and work?

Here we learn what work and energy mean in physics and how they are related.

## What does energy and work mean?

**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 ${}^{\circ}$ Celsius.

**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 1This is equal to 4184 joules per Calorie, 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.

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, $W$ , is below:

The work we need to do to burn the energy in the candy bar is $E=280\mathrm{cal}\cdot 4184\mathrm{J}/\mathrm{cal}=1.17\mathrm{MJ}$ .

Therefore, the distance, $\Delta x$ , we need to move the box through is:

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:

**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.,

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.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:

Similarly, the work done for the final 40 m of displacement would be:

## 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, $W=F\cdot \mathrm{\Delta}x$ , 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, ${F}_{||}$ , 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, $F\text{cos}(\theta )$ , 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:

Which is more often written as,

**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?

## 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, $F$ , required to lift a weight with mass $m$ straight up, placing it on a rack which is at a height $h$ above us:

The change in position—previously $\mathrm{\Delta}x$ —is simply the height, so the work, $W$ , that we have done in lifting the weight is then

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 $mgh$ , and the work done by gravity is $-mgh$ . 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

OK, so how many 280 Calorie—i.e., $1.17\times {10}^{6}$ joule—chocolate bars is this? Well, 245.25 J is about $\frac{1}{4770}$ 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 $\frac{1}{1190}$ 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!

## 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.

## Want to join the conversation?

- "However, remember that our bodies are about 25% efficient at transferring stored energy from food into work."

Why is it that it is 4x more efficient, rather than 1/4 times? Doesn't it say 25%? I'm confused.(19 votes)- If you use up 100% of your energy, you are getting only 25% of the work done. That means you have only 25% efficiency. 100 is 4x25. That is why, the percentage of work done is being multiplied by 4 to get the actual amount of energy you have spent doing the work.(15 votes)

- in the numerical for how long do i have to push a heavy box......, further down it is given that work done = 200 * 30 but the distance travelled at that point is 0 if we refer to the X coordinate, so why is it given 200*30??(20 votes)
- To calculate the work done one has to estimate the area under the force vs. position graph. To do so for the initial displacement of 30 m, one has to split up the area into a rectangle and a triangle. So the rectangle is equal to 200N*30m and the rest (0.5((500 N−200 N)⋅30 m)) is equal to the triangle.(22 votes)

- what is chemical potential energy?(19 votes)
- Chemical potential energy is the amount of potential energy tied up in a substance. The term is a little ambiguous because it can refer to the amount of energy you can get from breaking atomic bonds, as in the energy you get out of propane when you burn it, or the amount of energy stored as an ionic potential between molecules, such as in a neuron. Chemical energy can define the amount of work you can do with the proteins in your muscles, or how much energy you can store in a battery.(23 votes)

- In lifting the weight, we do work. What about when we slowly lower the weight? For example, if we lower it at constant velocity, aren't we doing negative work which is equal in magnitude to the work we did lifting the weight? Thus, I guess, in lifting and lowering the weight, our net work is 0. *But does this mean we burn equal amounts of calories lifting the weight as lowering the weight?*(19 votes)
- if you really want to cancel out the work of lifting the weight by "lowering" it, you have to not simply slowly lowering it down but accelerate it downward with the same force of gravity (-9.8m/s^2)

cause

F_up=-mg*d #g=-9.81m/s^2

F_up + F_down = 0

F_down = 0 - F_up

F_down = 0 - (-mg*d)

F_down = mg*d #g=-9.81m/s^2

in other words, when you say (and do) lowering the weight slowly, you are actually meaning (and doing) holding and keeping the weight from falling downward by gravity

in short, you're still holding it up by offsetting gravity, but with "lowering" the height of the point of offset slowly. that's why your net work might be not 0 but bigger than that of lifting it up at last

plus, with some calculation, you might get the proper velocity to "lower" it to the ground, making the force for that and that for lifting same, which results in the same consumption of calories in both exercises(2 votes)

- Doesn't it make more sense to say that because of our body's 25% efficiency the available energy for work is 4 times smaller, instead of saying the work done is 4 times larger?(12 votes)
- same

and you can even say like human body is wasting 75% of its energy while working

which one is fit to your taste is surely up to you!(3 votes)

`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.`

But what if we just let go of that weight? It does have some gravitational potential energy (U=mgh), meaning we had done some work on it.(5 votes)- The key point here is the difference between lifting the weight and holding the weight. While you are lifting the weight from the ground to over your head, it is changing distance, and thus you are performing work. If you hold it over your head, it is not changing distance, and thus you are not performing work. Just because something has potential energy, doesn't mean that work is being done on it - you have gravitational potential energy right now, but no work is being done unless you move through space. In other words, you did work on the weight to get it above your head, but you aren't doing work on the weight right now if you are currently holding it over your head.(7 votes)

- In the example, when we are pulling on the box using a rope with an angle 30 between the rope and the ground.We used this formula W=F*dcos(30) .We just took the horizontal component of the applied force into account . I Knew that that's true about getting the horizontal work done .However , why we neglect the vertical component ,although,it has an influence on our effort to pull the box .(7 votes)
- Since the box is only being displaced in the horizontal direction, the only work being done is in the horizontal direction. Therefore, any forces applied in the vertical direction are not important for calculating the work done to the box.(0 votes)

- I have a question about this 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?"

I get that, because the force applied is diagonal, the work done on the box only comes from the horizontal component of that force. However, the person is still exerting 500N in total. Shouldn't the total force that the person is exerting determine how much of the chocolate bar they should eat?(6 votes) - "W=(50kg)(9.81m/s2)(0.5m)=245.25J" in this equation, should't the acceleration due to gravity be negative since it's accelerating in the opposite direction?!(5 votes)
- kinetic energy(2 votes)

- If we take friction into account, is the force we use in the work equation with or without the friction? (Do we use net force or the total force exerted?)(3 votes)
- Friction is just another force on the object. There's the work done by the moving force, and then there's work done by friction, and you have to net them out to find out how the energy of the object changes.(5 votes)