Confused with inclined plane and work done by gravity

AI Thread Summary
The discussion revolves around the confusion regarding the work done by gravity on an inclined plane. It is clarified that while the angle between the gravitational force and displacement may seem to be zero at first, it actually varies along the path, particularly when considering different points on the incline. The work done by gravity is path-independent when only conservative forces are involved, meaning it depends solely on the initial and final heights, not the specific path taken. The participants emphasize the need to understand the role of the angle in the work formula and the nature of gravitational force as conservative. Ultimately, the work done by gravity remains the same across different paths if the height is constant.
nothingsus
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Homework Statement


4XZMUkA.png


Homework Equations


W_g = mgdcos(phi)

The Attempt at a Solution


So the angle between the force doing work (x component of F_g) and the displacement (down the hill) is 0 degrees. aka phi is 0 degrees. This is true for (a), (b) and (c).

so since cos(0) = 1, the work done by gravity is just mgd. d is greatest for (c), then (b) then (a), so W_g is greatest for (c) then (b) then (a)

However the answers say that work done by gravity is same in all cases.

The textbook says work is generally path dependent, however when only conservative forces (like gravity) are acting, work is not path dependent. So I have a flaw in my understanding somewhere.

potential gravitational energy = mgh, and height is the same in all cases. Can someone explain how this all works?
 
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nothingsus said:
So the angle between the force doing work (x component of F_g) and the displacement (down the hill) is 0 degrees. aka phi is 0 degrees. This is true for (a), (b) and (c).
This is not correct. Consider when the pig is midway along slide (c). In what direction is Fg pointing? In what direction is the displacement d?
 
DrClaude said:
This is not correct. Consider when the pig is midway along slide (c). In what direction is Fg pointing? In what direction is the displacement d?

F_g is straight down (x component is still down the hill) and displacement is up the ramp? So the angle isn't 0 degrees anymore?
 
nothingsus said:
F_g is straight down (x component is still down the hill) and displacement is up the ramp?
Not exactly "up" the ramp, belong along it.

nothingsus said:
So the angle isn't 0 degrees anymore?
Yes. To find the total work done by gravity, you would have to integrate over the entire path, with the angle φ not constant but varying with position depending on the exact shape of the slide.
 
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DrClaude said:
Not exactly "up" the ramp, belong along it.Yes. To find the total work, you would have to integrate over the entire path, with the angle φ not constant but varying with position depending on the exact shape of the slide.
Not the total work was asked but the work of gravity. What do you think about the gravitational force? Is it conservative?
 
nothingsus said:

Homework Statement


4XZMUkA.png


Homework Equations


W_g = mgdcos(phi)

The Attempt at a Solution


So the angle between the force doing work (x component of F_g) and the displacement (down the hill) is 0 degrees.
You should know the meaning of the notations you use. So what is phi in the formula for work?
 
ehild said:
Not the total work was asked but the work of gravity.
o:) Yes, that's what I meant. I'll correct my post.
 
DrClaude said:
o:) Yes, that's what I meant. I'll correct my post.
Is the force of gravity conservative? What do you know about the work done by a conservative force?
 
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