Work done on an object traveling down a slope

In summary, Feynman's statement is that the force of constraint does no work because it is always perpendicular to the slope, whereas the direction of motion is always parallel to the slope. This results in a dot product of zero between the force and displacement vectors. This can be explained by the definition of the dot product, which states that the dot product of two perpendicular vectors is always equal to zero.
  • #1
Roo2
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0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
A-ha! I get it. Thanks for showing me how to think about it!
 

Related to Work done on an object traveling down a slope

1. What is meant by "work done" on an object?

Work done on an object refers to the amount of energy transferred to or from the object as a result of a force acting on it. It is measured in joules (J) and can be either positive (work is done on the object) or negative (work is done by the object).

2. How is work done calculated on an object traveling down a slope?

Work done on an object traveling down a slope is calculated by multiplying the force acting on the object by the distance traveled in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

3. Can work done on an object traveling down a slope be negative?

Yes, work done on an object traveling down a slope can be negative if the force acting on the object is in the opposite direction of the object's motion. This means that the object is doing work on its surroundings, and energy is being transferred away from the object.

4. How does the angle of the slope affect the work done on an object?

The angle of the slope affects the work done on an object by changing the amount of the force acting on the object in the direction of its motion. A steeper slope will have a greater component of force in the direction of motion, resulting in more work being done on the object.

5. What is the relationship between work done and the speed of an object traveling down a slope?

The relationship between work done and the speed of an object traveling down a slope can be described by the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Therefore, as the speed of the object increases, the amount of work done on it also increases.

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