# What does the 'displacement' refer to in the definition of work?

The definition of work given in books is The work is said to be done by a force on a body, when the body is moved by the force through some 'displacement'. Now let a body of mass $m$ at rest. When a force $F$ is applied on it, it gets accelerated and starts moving. It will keep moving until an another force is applied on the body. If no other force is applied on the body to prevent its state of motion, it will be continuously covering more and more displacement (in the direction of applied force). So, in the above definition what does the word 'displacement' refer to?

• Even though this question is about the definition of displacement, I'm not sure why this detail was added: "... It will keep moving until an another force is applied on the body." It seems as if you might be asking "How can work be done if the object would keep moving on its own anyway?" If that might be what is causing confusion, the answer is that as long as a force is applied, the object isn't just moving it is accelerating. So the work being done (i.e, the result of the energy expended) is to accelerate the object. Jul 19, 2018 at 22:38
• @RandallStewart Even then, the work done on an object doesn't really have much to do with the acceleration. Work is meant to represent the change in energy along a specific path through a force field, and an object's energy can change whether it accelerates along this path or not. If I move a box downward at constant velocity, its gravitational potential energy is still changing, and gravity is still doing work on the box (which is countered exactly by the energy dissipated through the chemical processes in my arm muscles, the energy deposited in the air that was moved aside, etc). Jul 20, 2018 at 2:26

The distance referred to in the definition of work is, specifically, the distance that the object moves while the force is being applied. This is because the actual definition of work is a line integral, where the work $W$ along a path $C$ with tangent vector $d\vec{s}$ is defined as:

$$W=\int_C \vec{F}\cdot d\vec{s}$$

In general, calculating the work done requires that you specify a path along which the object moves first, which is why the notion of "force applied over a distance" makes sense; it's really more like "force applied along a specific path." For a special class of forces (conservative forces), the work done does not depend on the path taken, but only on the starting and ending points. These forces turn out to be exactly the kinds of forces for which you can define a potential energy, since the work done between any two points by that force is just the difference in the potential energy between those points.

It's true that terminology can sometimes be confusing. Let's go to the clear math:

If the body moves in straight line, you have to take the component along the force's direction.

For example, if you push a block during a 3m path in front of you, then the distance is 3m.

However, if the block moves 3m to North-west but the force is north-directed, then you must take only the component along the direction of force, i.e. the northern component. You'll have $3m\cdot \cos(45º)$

In sum, if the trajectory $s$ and the force $F$ make an angle $\alpha$, then

$$W=F\cdot s \cdot \cos(\alpha)$$

This applies only to straight lines. If the movement is curve, you must divide the curve into very small segments of small distance $ds$, and the small amount of work in each segment is:

$$dW= F \cdot ds \cdot \cos(\alpha)$$

In the definition, distance refers to the distance along which the force is applied.

In the example you have given, you must distinguish between an instaneous push and continuous pushing on the mass. If the force is applied continuously, the object will accelerate to arbitrarily close to the speed of light, and the work done approaches infinity.

Imagine pushing a box with your hand in a perfect vacuum, if your hand remains in contact with the box for a distance $d$, then that is the distance used in the definition of work.