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Quick question on work

  1. Oct 30, 2007 #1
    I know that the work done by a particular force is defined to be:

    [tex] \int \vec{F} \cdot d\vec{r}[/tex]

    and this dot product is defined as:

    [tex] |F||dr| cos(\theta) [/tex]

    I want to show that the work done by the force of gravity on a falling object is [tex] -\Delta U[/tex], using [tex]h_0 [/tex] and [tex] h_f [/tex] as my endpoints. So plugging in the information, I have:

    [tex] \int^{h_f}_{h_0} F*cos(\theta)*dr = mg*cos(0) \int^{h_f}_{h_0} dr = mg(h_f - h_0)[/tex]

    This is only the change in potential energy, rather than the negative change in potential energy. If I place a negative in front of mg to denote the direction of the force, it works out. But given the definition of the dot product where it uses absolute values, I don't see how I can do that. I'm having trouble knowing when to put in the minus sign, and when to leave it out, in other words, I'm not completely certain when it is necessary to take direction into account. If someone could explain my flawed reasoning, I would appreciate it.
  2. jcsd
  3. Oct 30, 2007 #2


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    There's nothing wrong. Note that [itex]h_0>h_f[/itex], and so, if we denote net change of height by h, [itex]h_f-h_0=-h[/itex].
  4. Oct 30, 2007 #3
    I see, I suppose the inequality tripped me up. So when is it necessary to take the direction of the force into account?

    Edit: If [tex] h_f < h_0 [/tex], that means gravity does negative work on the object. This can't be true since its kinetic energy increases.
    Last edited: Oct 30, 2007
  5. Oct 31, 2007 #4

    Doc Al

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    It is always necessary to take direction into account when computing work. The force (mg) points down, thus your distance differential must also point down. Since dy points up, you need -dy:

    [tex]W = \int \vec{F} \cdot d\vec{r} = \int mg (-dy) = -mg\int^{h_f}_{h_0}dy = -mg(h_f - h_0) = -\Delta U[/tex]
  6. Oct 31, 2007 #5
    Thank you for the reply. If the force and the distance differential (I'm assuming that's dr) point down, I take it that we're using the coordinate system where down is positive. What is dy representitive of, and how do we know its direction?
  7. Oct 31, 2007 #6

    Doc Al

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    If you used a coordinate system where down was positive, you wouldn't have had the problem. Even though you called it r, you actually used height as your position variable. That's why you got [itex]h_f - h_0[/itex] with the wrong sign.

    I used a vertical axis (y-axis) to represent position, which is more representative of what you actually did. Thus the y-coordinate represents the height above some origin. I used dy to represent a differential along that axis; since up is positive, we need to use -dy to represent the downward distance differential.
  8. Oct 31, 2007 #7
    So, if you took the direction of dy into account, why didn't you take the direction of the force into account?

    The thing that confuses me is the definition of the dot product. If we take the magnitudes of F and dr, and multiply them, and multiply that by the cosine of 0, we can't get a negative in front of [tex]\Delta U[/tex].
  9. Oct 31, 2007 #8

    Doc Al

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    But I did. The direction of the force is downward, in the direction of -y. (That's why I took -dy as the differential: I want to find the work done by gravity, which acts down.)

    Why cosine of 0? If F and dr are in the same direction, then the angle between them is zero. But if F and dr are in opposite directions, the angle is 180 and cos(180) = -1. (So the angle depends on how you define dr.)

    Since gravity acts down, it does positive work on an object which moves down. I must define the directions accordingly, or I'll get the wrong sign.
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