# Is the $d$ in $W=F*d$ displacement or distance?

My textbooks say that work=force times displacement but when I was considering conservative and non-conservative forces I got a bit confused. I know that the work done by non-conservative forces onto an object depends on the path taken. But if I consider the work done by frictional force onto an object and use displacement, won't the amount of work done be the same even if the object had taken different paths? So, in this case, I wondered why is it displacement and not distance?

But if I consider the work done by frictional force onto an object and use displacement, won't the amount of work done be the same even if the object had taken different paths? So, in this case, I wondered why is it displacement and not distance?

No. The work will definitely not be the same. The complete definition of work over some path $$C:{\bf \vec r}(t)$$ is $$W=\int_C{\bf \vec F}\cdot \mathrm d {\bf \vec r}$$

You're integrating over the object's trajectory.

Essentially, you "chop" up the path the object takes into very tiny displacements $$\delta {\bf \vec r}$$,* and then the "tiny" amount of work done over that tiny displacement is $$\delta W= {\bf \vec F} \cdot \delta{\bf \vec r}$$. After that, you use the integral to sum up all these tiny little bits of work to end up with the work $$W$$.

*If you were to sum up all the little $$\delta {\bf \vec r}$$'s, you'd end up with the displacement $${\bf \vec r}$$, whereas if you were to sum up $$\left|\delta{\bf \vec r}\right|$$, you'd end up with the distance.

This is a dot product: $$dW = \mathbf{F}\cdot \mathbf{dr},$$ where $$\mathbf{dr}$$ is the displacement.

Is the $$d$$ in $$W=F*d$$ displacement or distance?

Displacement is a vector quantity. Distance is a scalar quantity.

In the equation

$$W=F*d$$

$$d$$ is both displacement and distance because the equation only applies to a force of constant magnitude and direction acting over a distance $$d$$. But this is a special case of the general definition of work where work is the dot product of two vectors, force $$\vec F$$, and differential displacement $$d\vec r$$, or

$$W=\int\vec F \! \cdot d\vec r$$

But if I consider the work done by frictional force onto an object and use displacement, won't the amount of work done be the same even if the object had taken different paths? So, in this case, I wondered why is it displacement and not distance?

The work will not be the same because if you integrate

$$W=\int\vec F \cdot d\vec r$$

over different paths you will get different values for friction work. This is due to the fact that the direction of a non conservative force such as friction will vary over different paths.

That is not the case for a conservative force, such as gravity. In the case of gravity, the work is the same between the same two points because the direction of the force of gravity does not vary between the two points. Only displacements in the direction of the force contribute to work.

Hope this helps.