# Particle is located a distance x meters

## Homework Statement

When a particle is located a distance x meters from the origin, a force of cos(PIx/3) newtons acts on it. How much work is done in moving the particle from x = 1 to x =2? Interpret your answer by considering the work done from x =1 to x = 1.5 and from x = 1.5 to x = 2

## Homework Equations

So the integral isn't a problem.
If you graph this function I'm sure you know the sign change where cos(pix/3) dips below the x axis at 1.5. I'm just wondering if I should give my answer as the sum of the neg work and positive work. If the work is negative doesn't it mean it is on the opposite direction? But I still say oh this has done. Neg + pos work

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
Since they are talking about "work" I think they expect a more "physical" explanation. Here, you are doing work from x= 1 to x= 3/2 but then works is being done from x= 3/2 to 2. It's like doing work to push an object to the top of a hill, then getting work out by allowing the object to run down the other side.

Ah so rolling a rock up a hill and letting it roll the other side? Do I consider the negative work as postive so I add the area above the graph and the area below for total work?

HallsofIvy
Science Advisor
Homework Helper
No, you do NOT "consider the negative work as positive". The total work done is the "area above the graph minus the area below the graph"- in other words precisely $\int_1^2 f(x)dx$

Yeah so you do no work then with this problem. The integrals are equal above and below