Particle is located a distance x meters

  • Thread starter Jbreezy
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  • #1

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

The Attempt at a Solution

  • #2
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.
  • #3
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?
  • #4
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 [itex]\int_1^2 f(x)dx[/itex]
  • #5
Yeah so you do no work then with this problem. The integrals are equal above and below

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