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Homework Help: I dont get it-someone please help

  1. Mar 22, 2005 #1
    i dont get it------someone please help

    Hi, i think i might have gotten the first aspect of this question right, but the other parts im not too sure about...ie i have no idea what its going on about.

    Question: An object is attracted toward the origin with a force given by F_x = -k/(x^2)

    part a: Calculate the work done by a force F_x when the object moves in the x-diection from x_1 to x_2.

    for this i found the integral of the force with limits x_1 to x_2 , i ended up with something like

    work = k( (1/x_2) - (1/x_1) )

    this might be totally wrong though

    partb: if x_2 is greater than x_1, is the work done by F_x positive or negative?

    i got negative, because it turns out negative for the integral answer.

    partc: The only other force acting on the object is a force that you exert with your hand to move the obect slowly from x_1 to x_2. How much work do you do?

    i have absolutely no idea whats going on here, they dont tell u what this force is???

    partd: if x_2 is greater than x_1 is the work you do positive or negative? (in relation to part c)

    please help
    Last edited: Mar 22, 2005
  2. jcsd
  3. Mar 22, 2005 #2


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    a) is correct.So is b).For point c),what sense has the force you exert wrt the sense of the force in the points b) & a)...?

    If you figure that out & count the fact that the problem does not mention other force,nor the kinematics of the body,u can solve c) & d) easily;

  4. Mar 22, 2005 #3

    thanks for that verification, i'm glad i got something right.

    arent u doing like negative work, since your taking energy away....or is it positive? If so, i still dont get it, would it be like the same force, but u integrate it positively or something? Bloody hell :grumpy:
  5. Mar 22, 2005 #4
    Work is the integral of F with respect to distance. Here your distance is x_2-x_1, and the force F is -k/x^2. Integrate F with respect to x and evaluate at the points x_2 and x_1. Going WITH a force increases potential, going AGAINST a force decreases potential.
  6. Mar 22, 2005 #5


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    There's no convention here,like in thermodynamics.You just gave to apply calculus & the definition.

    And yes,it is the same force in absolute value,but it has a negative sign (compared to the I-st),therefore the work done would aquire the same "-".

  7. Mar 23, 2005 #6

    what the?!?!? but how can it be the same magnitude of force, i mean the question dosent specify anything about the force in terms of its magnitude, how can we deduce that it is infact the same magnitude with a negative?

    and isnt that force going in the same direction as the other one?

    omg im getting even more confused.

    so what youre saying is that its the same force with magnitude Fx=k/(x^2)
    and so i do the same integration thing.

    i end up with k(1/x_1 - 1/x_2)

    and the next part is positive work
  8. Mar 23, 2005 #7
    Hi dark_angel,

    The fact that the object moves ``slowly'' from [itex]x_1[/itex] to [itex]x_2[/itex] suggests that it is not accelerating, and therefore the force from your hand balances the attractive force. The work done by your hand is thus
    [itex]\int_{x_1}^{x_2}\frac{k}{x^2}\thinspace\mathrm{d}x = k\left(\frac{1}{x_1} -\frac{1}{x_2}\right)[/itex]
    as you said.
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