Determine the work done by the force due to gravity

In summary: Force due to gravity perpendicular to incline= mg cos(theta)In summary, the homework statement discusses the work done by the force due to gravity as the crate of mass m slides down an incline. The equation relating d and h is provided, as well as the reasoning behind why cosine of theta is not used in the original equation.
  • #1
Perseverence
88
7

Homework Statement


If the height of a frictionless incline is h. Determine the work done by the force due to gravity F as the crate of the mass m slides down the incline

Homework Equations


W=Fd cos (theta )
Force due to gravity perpendicular to incline= mg cos(theta)

The Attempt at a Solution


The solution doesn't have any mention of cosine theta or any consideration of the angle of the incline. Please explain why this would be.

The solution simply states that
W=FD and F=-mg therefore W=mgh
 
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  • #2
By height up the incline it does not mean distance up the slope. It means vertical displacement.
What is the relationship between those two distances?
 
  • #3
But it asks for work done by of gravity which does require force of gravity on the incline which has a cosine theta component.
 
  • #4
Perseverence said:
But it asks for work done by of gravity which does require force of gravity on the incline which has a cosine theta component.
In the formula Fd cos(θ), F and d have directions and θ is the angle between those directions.
With F=mg, that's vertical. If you are taking d as the displacement along the slope then θ is the angle of the slope to the vertical. But you are not given d, you are given h, the vertical component of displacement. What is the relationship between d and h?
 
  • #5
haruspex said:
In the formula Fd cos(θ), F and d have directions and θ is the angle between those directions.
With F=mg, that's vertical. If you are taking d as the displacement along the slope then θ is the angle of the slope to the vertical. But you are not given d, you are given h, the vertical component of displacement. What is the relationship between d and h?
?Cos(theta)? H and d are definitely not perpendicular because d is along a slope.
 
  • #6
Perseverence said:
Cos(theta)?
Write the equation.
 
  • #7
Cos (theta) = d/h
 
  • #8
Perseverence said:
Cos (theta) = d/h
d>h, so that would make cos(θ)>1.
Draw yourself a diagram, if you have not done so.
 
  • #9
I'm sorry, but this isn't helping me understand
 
  • #10
I feel like this question is getting so spread out and granular that I'm getting lost from what I'm trying to understand in the first place
 
  • #11
Perseverence said:
I'm sorry, but this isn't helping me understand
It will. Just try to get the equation relating d and h right. This is very basic stuff.
Draw the slope, with a horizontal line from the base and a vertical ine from the top, meeting.
You have a right angled triangle, length d up the slope and height h on the vertical.
If the angle to the horizontal is θ, what is the equation relating d, h and θ?
 
  • #12
haruspex said:
It will. Just try to get the equation relating d and h right. This is very basic stuff.
Draw the slope, with a horizontal line from the base and a vertical ine from the top, meeting.
You have a right angled triangle, length d up the slope and height h on the vertical.
If the angle to the horizontal is θ, what is the equation relating d, h and θ?
Okay. I was wrong before cosine of theta is h / d. I still don't understand why this would explain why cosine is not used in the original equation
 
  • #13
Perseverence said:
Okay. I was wrong before cosine of theta is h / d. I still don't understand why this would explain why cosine is not used in the original equation
Work of gravity can only be determined by knowing the force of gravity, and the force of gravity could only be determined by mg cosine Theta.
 
  • #14
haruspex said:
It will. Just try to get the equation relating d and h right. This is very basic stuff.
Draw the slope, with a horizontal line from the base and a vertical ine from the top, meeting.
You have a right angled triangle, length d up the slope and height h on the vertical.
If the angle to the horizontal is θ, what is the equation relating d, h and θ?
Ok. I finally understand!. It all cancels out! Wow, that was painful. Thank you so much for sticking with me and helping me understand.
 
  • #15
Perseverence said:
Ok. I finally understand!. It all cancels out! Wow, that was painful. Thank you so much for sticking with me and helping me understand.
Well, I'm not sure that you do, so I'll go ahead with the response I had started to write anyway.

Perseverence said:
cosine of theta is h / d.
Still wrong, unless θ is the angle to the vertical.
Perseverence said:
why cosine is not used in the original equation
You have not given a clear reason why it should be.
You mention that
Perseverence said:
Force due to gravity perpendicular to incline= mg cos(theta)
but why should the gravitational force perpendicular to the incline be interesting? The force up the slope has to balance the gravitational force parallel to the incline, mg sin(θ).
Perseverence said:
W=Fd cos (theta )
That is where θ is the angle between the force and the displacement. If the slope is θ, the angle between g and d is π/2-θ, so W=mg sin(θ) d. Since h= d sin(θ), yes it cancels out to produce mgh.
 
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FAQ: Determine the work done by the force due to gravity

1. What is the formula for determining the work done by the force due to gravity?

The formula for determining the work done by the force due to gravity is W=mgΔh, where W is work, m is the mass of the object, g is the acceleration due to gravity, and Δh is the change in height of the object.

2. How is the work done by the force due to gravity related to the potential energy of an object?

The work done by the force due to gravity is directly related to the potential energy of an object. As an object moves against the force of gravity, its potential energy decreases and the work done by the force of gravity increases.

3. How does the work done by the force due to gravity change when the angle of elevation changes?

The work done by the force due to gravity remains the same regardless of the angle of elevation. This is because the change in height, Δh, remains constant regardless of the angle.

4. Does the work done by the force due to gravity depend on the mass of the object?

Yes, the work done by the force due to gravity is directly proportional to the mass of the object. This means that the greater the mass of the object, the greater the work done by the force of gravity.

5. Can the work done by the force due to gravity be negative?

Yes, the work done by the force due to gravity can be negative if the object is moving in the opposite direction of the force of gravity. This means that the object is gaining potential energy and the work done by the force of gravity is decreasing.

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