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Calculate the work done

  1. Mar 23, 2005 #1
    Hey everyone, I'm stuck on one of my homework questions. It reads:

    An object is attracted toward the origin with a force given by Fx = -k/(x^2). (Gravitational and electrical forces have this distance dependence.).

    I have to calculate the work done by the force when the object moves in the x direction from x1 to x2. I tried to do so and got the answer W = (-k/(x^2))*(x2-x1) but apparantly the answer doesn't depend on the variable x so I'm really stuck.

    The 2nd part of the question reads "The only other force acting on the object is a force that you exert with your hand to move the object slowly from x1 to x2. How much work do you do?". I'm also really stuck on this, I don't even know how to approach it.

    I tried to ask some of my friends but they were at a loss as to what to do so I though i'd give this a shot :)
    Last edited: Mar 23, 2005
  2. jcsd
  3. Mar 23, 2005 #2
    Do you know about integrals? Maybe you have seen this formula:

    [tex]W = \int_{x_1}^{x_2}Fdx[/tex],

    knowing that F = F(x).

    Because the object is moved slowly (constant velocity), the force you apply is the opposite of the other force F (equal magnitudes).
  4. Mar 23, 2005 #3
    Oh geez I feel like such an idiot... i completely did not even realise i couldn't use that equation because it wasn't a constant force. Thank you so much. I used integration and came up with the answer: W = K/x2 - K/x1, however i have no idea if that is right or not.

    So that would make the answer to the 2nd part W = -K/x2 + K/x1 ?
    Last edited: Mar 23, 2005
  5. Mar 23, 2005 #4
    That looks totally wrong. Remember as a rule during integration that if the denominator has the variable squared, then the integrated result will have the variable to the power of one. (The power is -2; add 1 during integration and it becomes -1). :smile:
  6. Mar 23, 2005 #5
    Haha, yeh sorry I realised my mistake after i posted it and edited it. Does it look alright now?
  7. Mar 23, 2005 #6
    It is OK now!
  8. Mar 23, 2005 #7
    :) Fantastic! Thankyou so much for your help and for bearing with me. I certainly have to do a lot more study in physics (and maths!)
  9. Mar 24, 2005 #8
    Do you go to the University of Sydney?

    Also, just out of curiosity, how did you get get the answer to the second part? I know the first part, but I just want to know how you got the second part. Any help would be appreciated.
  10. Mar 24, 2005 #9
    It is not likely that janiexo will answer immediately. But what you're asking is pretty simple.

    The force you apply by hand F' is the opposite of the other force in order for the object to move slowly at constant velocity. So,

    [tex]F' = -F = \frac{k}{x^2}[/tex]

    Since you found the work done by F, then

    [tex]W' = \int_{x_1}^{x_2} -Fdx = -\int_{x_1}^{x_2} Fdx = -W = \frac{k}{x_1} - \frac{k}{x_2}[/tex]
  11. Mar 24, 2005 #10
    Okay...then how come the first answer isn't considered to be in the opposite direction? It is attracted to the centre with a negative force, but you integrate this negative force to get the work done by the object's force in the opposite direction. Sorry for being a pain...I understand what you're stating, but how come the force in the first question isnt considered opposite to -k/x^2?
  12. Mar 24, 2005 #11
    The force is [tex] F = \frac{-k}{x^2} [/tex]

    The negative means the force is attractive, the force pulls things to the origin. When you go along a force, the work done is positive. When you move the object back to its original position, you are doing work AGAINST the force, therefore the work done is negative. You are applying the force with the same magnitude as F, but in the opposite direction. The force applied by your hand is

    [tex] F = \frac{-k}{x^2} [/tex] If you find the work done by this force you will find it is positive.
  13. Mar 24, 2005 #12
    OK, we are treating forces as scalars, signed scalars, not as vectors. So the force is negative when its direction is opposite to the direction of movement (therefore doing negative work), and positive when its direction is in the direction of movement (therefore doing positive work).
    In our example, the textbook (correctly) has given force in terms of x with a minus, because the movement is toward the origin while the force pushing it out of the origin. Now we take the opposite force (pushing the object toward the origin) as positive, so it will do positive work.

    Optionally you could not bother with signs, and just take the magnitude of the forces. When the direction is opposite to the motion you could use:

    [tex]W = -\int_{x_1}^{x_2} Fdx[/tex]

    and when it is in the direction of motion you could use:

    [tex]W = +\int_{x_1}^{x_2} Fdx[/tex]

    Either way you will arrive at the same result!
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