When a particle is located a distance x meters from the origin, a force of 2cos(pix/6) newtons acts on it. How much work W is done in moving the particle from x = 1 to x = 5?

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When a particle is located a distance x meters from the origin, a force of 2cos(pix/6) newtons acts on it. How much work W is done in moving the particle from x = 1 to x = 5?

Mathematics
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Is this calculus based physics?
I will assume that this is a calculus based physics question. W = F * D Here the force is variable, so: integral 2cos(pix/6) dx from lower limit 1 to upper limit 5.
right, i got stumped because of the negative force. what does that mean? doesnt force have to be positive

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whats the positive and negative direction here?
A force can always work against a system in which you ascribe a vector to be positive.
Does that make sense?
so what is positive and what is negative
can you make an example
The most simple example would probably be gravity.
Don't let semantics play with your mind, you can designate whichever direction you choose as postive or negative.
so a particle along the x axis , or x meters from the origin experiences a force 2cos (pix/6). what direction is positive and what is negative. it can be up or down?
*cos
but positive and negative are like east /west or north south, right?
yes
graph the function that you have, look at interval x=1 to x=5 and you'll understand how you can have a negative force here.
yes it oscillates
you are finding the area under that curve, and there is more area under the x-axis than above.
right its symmetric and adds up to zero
yes, sorry, you're right it's zero, I was just trying to explain how it could be negative. does it make sense now? sometimes, it's really helpful to look at a graph.
a force is just a vector, it can be negative, because it has magnitude & direction*
ok thanks, can i ask a quick question
i just want to reiterate whats going on
crap my browser froze

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