# Work when there is more than 1 force

I know that for an object with an applied force, the work done is

$$W = Fd \cos \theta.$$

I was wondering what would happen when there is another force (e.g. friction)? Is it better to say that the work done for a general case is

$$W = F_{net} d \cos\theta.$$

• If there is more than one force, determine the resultant first: $\vec{F_{net}}=\Sigma_{i}^n\vec{F_i}$. – Gert Mar 11 at 0:59
• @Gert is correct. In addition, sometimes there are situations with multiple forces, but we are interested in the work done by only one of them. E.g. the work done by gravity when lifting a book. Then we must specify what we mean. Work done by xxxxx, or Total work or some other phrase. Net force will give total work done; individual forces give work done by those forces. – garyp Mar 11 at 1:17

I think @garyp has answered the question in his comment. That is, we can either discuss about work done on the object by a specific force $$F$$ or total net force $$F_{net}$$. Therefore, how the work is calculated depends on what force you want to discuss with.

First, you must recognize that your formula only works for constant forces and motion in one dimension.

So, if all of those conditions are met, then just add up the work done by each force.

$$W=F_1d\cos\theta_1+F_2d\cos\theta_2$$

If both forces act in the same direction ($$\theta_1=\theta_2=\theta$$), then you can combine the terms

$$W=(F_1+F_2)d\cos\theta$$

In general, work by a force is given by $$W=\int\mathbf F\cdot\text d\mathbf x$$ so for multiple forces we just do

$$W=\sum_i\int\mathbf F_i\cdot\text d\mathbf x=\int\left (\sum_i\mathbf F_i\right)\cdot\text d\mathbf x$$

I am not going to say $$\sum_i\mathbf F_i$$ is the net force, because we can calculate the work done by a subset of the forces. It doesn't just have to be one or all.

To specify a work one must specify a force. E.g. the work of friction depends on the friction force, the net work depends on the net force, and so forth.

$$\let\th=\theta \def\vF{\vec F}$$ Let's begin by specifying the exact meaning of your formula $$W = F\,d\,\cos\th.\tag1$$ Here it's understood:

1) that a force $$\vF$$ constant in magnitude and direction is given

2) that the point of application of $$\vF$$ undergoes a displacement of length $$d$$.

Note that it's not required that the motion of that point is rectilinear; what is requested (if force is constant as we said) is that

3) distance between starting point A and arrival point B is $$d$$

4) angle between $$\overrightarrow{\rm AB}$$ and $$\vF$$ is $$\th$$.

Now if there is another force or more (and if your "object" isn't a single mass point) then each force will have its own point of application. All points will have their own displacements, generally different in magnitude and direction. Then there is only one reply: you have to compute works for each force using (1) and sum works together.

Using net force is only justified for a mass point, or in the special case when displacements are equal (as vectors) even if they apply to different points. This certainly happens if your object is undergoing a translational motion (i.e. with no rotation).

Net force would give a wrong result in all other cases. (Not to mention that if there are several forces, applied to different points undergoing different displacements, you had to determine which one to choose.)

What I've been saying holds if you are interested in total work. This is not always the case, however. But if you want to know work done by one or a subset of forces, nothing changes in the general rule of computing each work one by one and summing.

There are other important points concerning work done on a mechanical system and its relation with variation of kinetic energy. But I guess this is a rather advanced topic for your level of study, so I refrain from entering it. In case you're interested, ask again.