anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Is this calculus based physics?
anonymous
  • anonymous
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.
anonymous
  • anonymous
right, i got stumped because of the negative force. what does that mean? doesnt force have to be positive

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anonymous
  • anonymous
whats the positive and negative direction here?
anonymous
  • anonymous
A force can always work against a system in which you ascribe a vector to be positive.
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
so what is positive and what is negative
anonymous
  • anonymous
can you make an example
anonymous
  • anonymous
The most simple example would probably be gravity.
anonymous
  • anonymous
Don't let semantics play with your mind, you can designate whichever direction you choose as postive or negative.
anonymous
  • anonymous
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?
anonymous
  • anonymous
*cos
anonymous
  • anonymous
but positive and negative are like east /west or north south, right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes it oscillates
anonymous
  • anonymous
you are finding the area under that curve, and there is more area under the x-axis than above.
anonymous
  • anonymous
right its symmetric and adds up to zero
anonymous
  • anonymous
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.
anonymous
  • anonymous
a force is just a vector, it can be negative, because it has magnitude & direction*
anonymous
  • anonymous
ok thanks, can i ask a quick question
anonymous
  • anonymous
i just want to reiterate whats going on
anonymous
  • anonymous
crap my browser froze

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