When calculating the work done over a distance due to a variable force, you're supposed to use $$\int^{b}_{a} F(x)dx$$ Would the force $F(x)$ be an equation representing the resultant force acting on the particle in the direction of travel? Or would the force be one specific force that acts on the particle to find the work done by that force? Or can you interpret it either way?
1
-
1$\begingroup$ If the above is for a 1-dimensional problem then you have only one direction, so direction of force and direction of movement are parallel to each other. $\endgroup$– user3257624Commented May 24, 2018 at 14:54
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Yes, $F(x)$ here is the component in the direction of travel. More generally you can write the formula as
$$W=\int_a^b{\vec F(x)\cdot d\vec{x}}=\int_a^b{F(x)\cos\theta dx}$$ where $\ \cdot\ $ is the dot product between vectors and $\theta$ is the angle between the force and the direction of travel. This formula holds for all forces.
answered May 24, 2018 at 13:01
AccidentalTaylorExpansionAccidentalTaylorExpansion
18.1k3 gold badges33 silver badges65 bronze badges
$\begingroup$
$\endgroup$
2
Work is the integrated force in the direction of travel. Use the vector dot product to automatically generate the component in the direction of travel.
answered May 24, 2018 at 12:51
-
$\begingroup$ so it's a component of the resultant force (that acts in the direction of travel) that you integrate over? $\endgroup$– BigWigCommented May 24, 2018 at 12:54
-
$\begingroup$ @BigWig: Correct - net force in the direction of travel $\endgroup$ Commented May 24, 2018 at 12:59
$\begingroup$
$\endgroup$
First of all you might find this other question useful In the formula for work, $W=\vec{F}\cdot \vec{d}$, is $\vec{F}$ the resultant force?.
Second let us try addressing your question directly. So imagine some mass $m$ subject to different forces, let's say $\vec{F}_1$, $\vec{F}_2$ and $\vec{F}_3$. Notice they are vectors, so they have a direction. So if there is nothing else in this problem one has to write down Newton's equations which will lead you to differential equations of second order which you will have to solve and obtain a trajectory $\vec{r}(t)=(x(t),y(t),z(t))$ for example (you may have constraints too on a generic problem and specific initial conditions). Then you can ask yourself what is the total work done so you use $\vec{F}=\vec{F}_1+\vec{F}_2+\vec{F}_3$ for the computation, or you can ask yourself what is the work done by force 1 and use only $\vec{F}_1$ for the computation. Now, about the trajectory, you can specify two points in time which are interesting to the problem (could be the whole trajectory), in any case the important thing is to understand the vectorial nature of forces $$\vec{F}(x(t))\color{red}{\Large\cdot} d\vec{x}(t)$$ and the scalar nature of work. So work is a result of the component of the given force in the direction of the trajectory. (counter example: if you move something horizontally, then the gravitational force does no work.)
So in the end it can be either. It depends on the question you want to answer, or the question you were asked. If you want to use the Energy-Work theorem then you only include non-conservative forces as for example friction.
