# Is there an intuitive explanation of the work formula?

Upon learning calculus, I decided it was time to derive all of classical mechanics to give myself a good understanding of physics. What I found was that, while trying to do so, I would need some definitions for the more abstract concepts, such as work and energy. What I found was that the definition of work, W = Fs, has weird implications and does not seem intuitively correct as it is based on two variables that affect each other as the system continues.

Work is the least psychologically pleasing (I chose my words carefully) topic in all of physics for me. It implies that it takes more work to move objects that already have a high velocity than objects with velocity 0, but don't different inertial reference frames contradict this definition? i.e. a baseball moving through space at v=a from the reference of a bystander at v=0 would be moving through a lot more space while being accelerated than from the reference of, say, a train that's going at v=a and sees the baseball at v=0.

I want an answer to that question (on the contradiction of the energy formula with different inertial frames), and I'd also like an intuitive explanation of work.

• Have a look here: physics.stackexchange.com/q/51220 if you have any doubts about how the kinetic energy changes in different inertial frames. – Ellie Aug 23 '15 at 19:42
• "It implies that it takes more work to move objects that already have a high velocity than objects with velocity 0." This may point to some incorrect intuition you have. It takes 0 work to move an object with a high velocity and no forces acting on it; you just wait since it's already moving. It takes more work to accelerate an object with high velocity (since $W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$) the same $\Delta v$, but this is intuitive to me since $v_f$ is greater. Worth noting is that it takes less work to reach the same $v_f$ when $v_i$ is greater. – jpmc26 Aug 23 '15 at 22:54

As you point out, work done is a function of the frame of reference. More specifically, if you apply a force on an object, that force typically connects two different objects, and it's the relative velocity of these two objects that really concerns us.

Example: you are walking in a train, and pull a suitcase behind you. The friction between the suitcase and the ground is 5 N. When you cover the 20 m length of the rail car, you have done 100 J of work.

Now look from the perspective of someone outside the train. If they were trying to apply a force of 5 N on that suitcase while running along next to the train, they might have to cover a significantly greater distance than 20 m in order to move the suitcase. If it took 20 seconds in each case to move the length of the rail car, and the train was moving at 10 m/s (just to keep numbers simple), the person running alongside the train would have covered 220 m, all the while applying a force of 5 N. Total work done 1100 J.

Why is the result different although the suitcase must have ended up with the same energy? We need to look at where else forces are applied. When you were walking in the train, your feet were pushing back on the floor of the train with 5 N in order to pull the suitcase; and the suitcase applied 5 N in the opposite direction, in response to the friction. There was no net force on the train. But when the force came from the person walking next to the train, there was no "feet on the floor" - so there was actually a force on the train for the 20 seconds, not just on the suitcase.

Not sure if that makes it clearer for you - but these are the kinds of things you need to consider as you figure out "how work works".

With regards to intuition, it might help to think about situations of mechanical advantage.

For example, consider a simple pulley system.

modified from "Pulley1a". Licensed under Public Domain via Commons - https://commons.wikimedia.org/wiki/File:Pulley1a.svg#/media/File:Pulley1a.svg

You can work out using force that the weight $W/2$ balances the weight $W$. Since they are balanced, we could grab the weight $W/2$ and pull downwards on it with essentially zero effort, lifting the weight $W$ up.

In doing so, we put effectively zero energy into the system. However, work is being done on weight $W$ since it is rising. That energy must come from weight $W/2$. The force of gravity on weight $W/2$ is half as much as on weight $W$, but weight $W/2$ falls twice as far. Thus, the total energy change (work done by gravity) is the same magnitude. This is what allows the pulley to balance, and it shows that the work formula should have force*distance in it.

As others have written there is no contradiction with reference frames or relativity; one simply must keep in mind that kinetic energy is quadratic in velocity. See Ron's answer for a thought experiment about that and the other answers on this page for more intuition and math regarding reference frames.

• I was a bit tired when I first read this, so I didn't give you due credit, but looking back, this is a great representation of joules in action! Many thanks, Mark. Once I earn >=15 reputation, I'll give you the well-deserved upvote. – Striker Aug 25 '15 at 2:42

Work does depend on frame of reference, but so does change in kinetic energy.

Work done and changes in kinetic energy should either both bother you or neither bother you.

To know how much the kinetic energy changes from one location to another you need to know the force (if constant) and how far apart the locations are:

$$\frac{1}{2}mv_b^2-\frac{1}{2}mv_a^2=\int_a^b\left[\frac{d}{dt}\left(\frac{1}{2}mv^2\right)\right]dt=\int_a^b[mav]dt=\int_a^bFvdt=F\int_a^bvdt=F(x_b-x_a).$$