Determine the work done by the force due to gravity

Perseverence

1. The problem statement, all variables and given/known data
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

2. Relevant equations
W=Fd cos (theta )
Force due to gravity perpendicular to incline= mg cos(theta)

3. 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

Related Introductory Physics Homework News on Phys.org

haruspex

Homework Helper
Gold Member
2018 Award
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?

Perseverence

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

Homework Helper
Gold Member
2018 Award
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?

Perseverence

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.

Homework Helper
Gold Member
2018 Award

Perseverence

Cos (theta) = d/h

haruspex

Homework Helper
Gold Member
2018 Award
Cos (theta) = d/h
d>h, so that would make cos(θ)>1.
Draw yourself a diagram, if you have not done so.

Perseverence

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

Perseverence

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

Homework Helper
Gold Member
2018 Award
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?

Perseverence

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

Perseverence

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.

Perseverence

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

Homework Helper
Gold Member
2018 Award
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.

"Determine the work done by the force due to gravity"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving