Work done on an object traveling down a slope

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SUMMARY

The discussion centers on Feynman's assertion that the work done by the force of constraint on an object traveling down a slope is zero, despite the object having both x and y components of motion. The key takeaway is that the force of constraint is always perpendicular to the displacement along the slope, leading to a zero dot product between the force vector and the displacement vector. This conclusion is supported by the equation W = F·dr, where the dot product results in zero when the vectors are perpendicular. The confusion arises from the presence of x components in both the force and motion, which is clarified by understanding the relationship between the components of the vectors involved.

PREREQUISITES
  • Understanding of gravitational potential energy (U(grav) = mgh)
  • Familiarity with the concept of work (W = F·dr)
  • Knowledge of vector components and dot products
  • Basic principles of physics regarding forces and motion
NEXT STEPS
  • Study the concept of work done by forces in physics, focusing on the dot product of vectors.
  • Explore the implications of forces being perpendicular to displacement in various physical scenarios.
  • Learn about gravitational potential energy and its applications in different contexts.
  • Review Feynman's lectures on physics for deeper insights into classical mechanics.
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Students of physics, educators teaching mechanics, and anyone seeking to deepen their understanding of work and energy principles in classical physics.

Roo2
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Homework Statement



I'm reading Feynman's lectures on physics to brush up on material I haven't looked at in a while, and I got confused by one of his statements. He claims that the change in potential energy of an object traveling down a slope can be calculated solely by the change in gravitational potential between the starting and ending point, irrespective of the force of constraint. While this seems intuitively correct to me, I'm confused by his explanation, in which he states that the work by gravity is nonzero (mgΔh) while the work done by the force of constraint is zero. To illustrate this, he provides the figure below, to which I have added an x-axis for reference:

work_zps315ca862.png


I'm confused because work is f*dr. Clearly, the object travels along the y-axis (I forgot to label, but assume perpendicular to x) and the force of gravity has an (entirely) y component, so f*dy is a nonzero number. However, the resultant force is shown to have an x component, and the direction of motion also has an x component, so f*dx is also nonzero. Why is the work done by the resultant force considered to be zero?

Homework Equations



W = f*dr
U(grav) = -G(m1m2/r^2) =~ mgh

The Attempt at a Solution



Described above.

Not a homework question per se, but I figured that it resembles a homework question more so than a "real" physics discussion in the general physics forum.

Thanks for any help!
 
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Feynman's statement is that the force of constraint does no work. The force of constraint is always perpendicular to the slope, whereas the direction of motion is always parallel to the slope. Hence, the force is always perpendicular to the displacement, and the work (their dot product) is zero.

Makes sense?
 
Yes and no. In the diagram above, the force of constraint has an x component, as does the direction of motion. Therefore, how can the dot product be zero?

edit: it can be zero if Fx * dx = -Fy * dy. But why does that equality hold?
 
Roo2 said:
Yes and no. In the diagram above, the force of constraint has an x component, as does the direction of motion. Therefore, how can the dot product be zero?

edit: it can be zero if Fx * dx = -Fy * dy. But why does that equality hold?

That relation between the components of the vectors F and dr MUST hold, precisely because these two vectors are always perpendicular to each other, and hence their dot product is always zero. The rest comes from the definition of the dot product:

$$\mathbf{F} \cdot d\mathbf{r} \equiv F_xdx + F_ydy = 0 $$

Edit: I use boldface to denote vectors, which is pretty standard.
 
A-ha! I get it. Thanks for showing me how to think about it!
 

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