# Determine the work done by the force due to gravity

## 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

## Answers and Replies

haruspex
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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?

But it asks for work done by of gravity which does require force of gravity on the incline which has a cosine theta component.

haruspex
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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?

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.

haruspex
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Cos(theta)?
Write the equation.

Cos (theta) = d/h

haruspex
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Cos (theta) = d/h
d>h, so that would make cos(θ)>1.
Draw yourself a diagram, if you have not done so.

I'm sorry, but this isn't helping me understand

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

haruspex
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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?

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

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.

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.

haruspex
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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.

cosine of theta is h / d.
Still wrong, unless θ is the angle to the vertical.
why cosine is not used in the original equation
You have not given a clear reason why it should be.
You mention that
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).
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.

• Nojins