Question

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?

Answer 1

Is this calculus based physics?

Answer 2

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.

Answer 3

right, i got stumped because of the negative force. what does that mean? doesnt force have to be positive

Answer 4

whats the positive and negative direction here?

Answer 5

A force can always work against a system in which you ascribe a vector to be positive.

Answer 6

Does that make sense?

Answer 7

so what is positive and what is negative

Answer 8

can you make an example

Answer 9

The most simple example would probably be gravity.

Answer 10

Don't let semantics play with your mind, you can designate whichever direction you choose as postive or negative.

Answer 11

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?

Answer 12

*cos

Answer 13

but positive and negative are like east /west or north south, right?

Answer 14

yes

Answer 15

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.

Answer 16

yes it oscillates

Answer 17

you are finding the area under that curve, and there is more area under the x-axis than above.

Answer 18

right its symmetric and adds up to zero

Answer 19

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.

Answer 20

a force is just a vector, it can be negative, because it has magnitude & direction*

Answer 21

ok thanks, can i ask a quick question

Answer 22

i just want to reiterate whats going on

Answer 23

crap my browser froze