# Intuition behind Work

I have a doubt in understanding the intuition behind the concept of work. First of all, I think this isn't duplicate, I've searched on the site, and the closest thing I've found was this post which is different than what I'm struggling here to understand.

Well, let me show an example of what I mean: momentum. For me, defining the "quantity of motion" to be $p = mv$ is pretty obvious, the amount of motion should be proportional to how fast the thing is moving and how much thing that's making the movement. Position, velocity, acceleration, those all have obvious interpretations too. But what about work ?

I've already heard that "work is the amount of force that was used to make the movement", however, if that's the case, shouldn't we define work simply as the component of the force in the direction of movement ? Why do we multiply by distance ? In other words, how do we get a "feeling" of what's work intuitively so that the equation defining it simply states mathematically our intuition ?

Thanks very much in advance.

• This post is a pretty good explanation of what work means.
– Dan
Mar 26, 2013 at 23:14
• Well, intuitively, the longer you apply a force the more work you have done: Lifting an object higher means you've done more work. Mar 26, 2013 at 23:35
• The statement: "work is the amount of force that was used to create the movement" is not really precise and possibly the source of the confusion. Work is the amount of energy used to create the movement. But you can think of calculating the work as like taking the amount of force applied at each point, and then summing them up for every point along the path. As Lagerbaer points out, you would expect a greater change in energy if you apply the force over a larger distance Mar 26, 2013 at 23:40
• Yes. You multiply the force times the distance. The usual definition of work is "force times distance". Lift a 1-lb weight 1 foot off the floor, and you have done 1 ft-lb of work. Lift 100 lb 10 feet - 1000 ft-lb. Twice as much weight, twice as much work. Twice as much height, twice as much work. Mar 27, 2013 at 0:14

If you want a lengthy-but-most-thorough answer, the link that Dan posted with his comment will help out a lot. However, since you're appealing to intuition (and I believe that training one's intuition is very important in learning physics) I'll try to give a simpler answer. I won't do quite as well, but here we go:

First, like it's been said, it's not correct to say that "work is the amount of force that was used to make the movement." Work is the amount of energy used to make the movement. To understand this intuitively, it sometimes helps to think about how the motion could move the object up—up a hill, for example. If something moves up against earth's gravity, it gains potential energy. As it moves down a hill, it can get that energy back. If you let a ball or a car roll down a hill, and then down another hill twice as high, it won't be going twice as fast at the bottom, only about 40% faster ($K_{\text{E}}=\frac12mv^2$; double the energy doesn't mean double the velocity). This isn't really a great example, but tying energy to gravitational potential energy could help your intuition swallow this.

Now why do we multiply by distance? Maybe you've had to push a car sometime in your life. I don't think that anything makes me more tired than running behind a car trying to push it along. Again, how tired you get pushing something is not a scientific quantity, but it might help you think about it another way. If you push a car that's stuck, it doesn't move. You've gone no distance, and even though there was a lot of force involved, you did no work, and the car doesn't move. Its momentum doesn't increase either, for that matter. Once you get it moving, you follow it along to get it up to speed, and it just kills your legs. But you're applying a force to an object as it moves over a certain distance. That's how we measure work.

Well, I'm sure that's not the best answer, but everyone has to train their intuition to understand this in their own way. I don't know if this is the way I understand it, but it might be a stepping-stone to it. Anyway, kudos to you for asking! I hope this helps.

I think of work as encoding a human value judgement about how useful a certain application of force is for a certain displacement.

This is based on the formula $$W=\vec F \cdot \vec d = \|\vec F \|\|\vec d \| \cos \theta$$, which I read as $$W= (\|\vec F \| \cos \theta) \|\vec d \|$$.

I.e. $$\cos \theta$$ ranges from -1 to 1 and shows how "useful" $$\vec F$$ would be for a displacement $$\vec d$$.

For example, if I want to move my kitchen table, $$\cos \theta$$ encodes the idea that even if I use all my strength to push it towards the floor, it will be totally useless for moving my table since in this case $$W= (\|\vec F \| \cos 90°) \|\vec d \| = (\|\vec F \| 0) \|\vec d \|=0$$

That is, if I put in a ton of force but get no work done, I should reevaluate my methods.

The best use of my strength would be to push the table in exactly the same direction as I want it to go, because then $$\cos 0° = 1$$, and none of my effort is wasted. $$\cos 180° = -1$$ so I'd waste no effort in getting the opposite of what I want.

Just to expand on this intuition, we can rearrange the formula:

$$\frac{W}{\|\vec F \|\|\vec d \|}=\frac{\vec F \cdot \vec d }{\|\vec F \|\|\vec d \|}=\cos \theta$$

If you consider a triangle in the unit sphere with hypotenuse length one, if you take "one step" from the origin to the tip of the hypotenuse $$\cos \theta$$ tells you how much you moved in the $$x$$-direction.

The same is true for any two vectors. If you have $$\vec{v}$$ and $$\vec{w}$$, if you walk "one step" along $$\vec{v}$$, the cosine of the angle between them tells you how much you've walked in the $$\vec{w}$$ direction.

The same idea is why $$\cos \theta$$ of two vectors is the correlation between them.

So to me, work captures the idea of "how much displacement am I getting from what I'm putting in?". It will rarely be 100%.