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This may be quite off-topic but please help me.

Is there any work done when I run in a rectangle?

I thought that the answer should be no. But my teacher says that we should calculate each side of the rectangle individually and then add it all together. Which one is correct and why?

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Yes you do work.

The net displacement vector is zero (you end where you start), and in such cases the work against a conservative force would be also zero (for example, against gravity). But if we consider running, one mostly does work against a non-conservative force like air-drag. In such instance, the work-done is path dependent and you have to do as your teacher says adding the work in each segment and adding them up.

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  • $\begingroup$ I rather doubt the original problem went to this level of reality. :-) . $\endgroup$ – Carl Witthoft May 22 '14 at 11:50
  • $\begingroup$ @CarlWitthoft, depends on how they incorporate the drag. Do they just look at some simple model which is proportional to the normal force, or do they look at a model which is quadratic in the velocity ? $\endgroup$ – Nick May 22 '14 at 12:31
  • $\begingroup$ In this case, "work" is proportional to displacement, so if you end up at the exact same spot, you haven't done any work at all. $\endgroup$ – Jason Chen May 23 '14 at 4:31
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Undoubtedly, you are right in your answer, and so is your teacher. What he means to tell is that the vector-sum of displacement should be calculated treating all sides of rectangle as individual displacement-vectors, so that resultant displacement-vector is a null-vector.enter image description here

This answer is valid only until you do not consider non-conservative forces like air(fluid)-drag, friction due to ground. Otherwise, you do work, and proceed in the way your teacher suggests.

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In this case, if you integrate the dot product F.dx over the loop, you will see that F has different directions on different sides, in such a way the F.dx is always of same sign, and so net work done is not zero. Whereas in a conservative field (gravity as suggested in the answers), F.dx will change sign for different sides of the rectangle - totaling to zero.

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Generally, if you are within the domain of a vector field $\vec{F}$, the work done along a contour is given by,

$$W = \int_{C} \vec{F} \cdot \mathrm{d}\vec{\ell}$$

where the integral is the line integral over the path $C$. For a rectangle, or any closed curve,

$$W= \oint_{C} \vec{F} \cdot \mathrm{d}\vec{\ell} = \iint_{A} \left( \nabla \times\vec{F} \right) \, \mathrm{d}A$$

by Green's theorem, which follows from Stokes' theorem. If the field is conservative,

$$\vec{F}=\nabla \phi$$

for a scalar field $\phi$. Notice $\nabla \times \nabla\phi=0$, and hence the work done along a closed contour by a conservative field is precisely zero. Alternatively, recall the gradient theorem,

$$\oint_C \vec{F} \cdot \mathrm{d}\vec{\ell} = \phi(\mathbf{x'})-\phi(\mathbf{x}) = 0$$

if the endpoints are the same, i.e. $\mathbf{x'}=\mathbf{x}$, and $\vec{F}$ is conservative. At the Newtonian level, the gravitational field is simply the gradient of a scalar potential function, which is why, by the gradient theorem, the work done is independent of the path chosen; it only depends on the start and end point.

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