Work on an Incline: Calculating Force and Work

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Homework Help Overview

The discussion revolves around calculating work done on a crate being moved up an inclined ramp and comparing it to lifting the crate vertically. The subject area includes concepts of work, force, and motion on an incline.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the formula for work, questioning whether the distance should be the length of the ramp or the vertical height. There is also discussion about the forces involved in both scenarios of moving the crate up the ramp versus lifting it straight up.

Discussion Status

Participants are actively engaging with the problem, clarifying the application of the work formula and the forces at play. Some guidance has been provided regarding the direction of forces and distances relevant to the calculations, but no consensus has been reached on the final calculations.

Contextual Notes

Participants are considering the implications of using different distances for the work calculations and the specific forces acting on the crate in each scenario. There is an acknowledgment of the need for proper units in the final answers.

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A 60-kg crate is slid up an inclined ramp 2.0-m long onto a platform 1.0m above the floor level. A 400-N forcem, parallel to the ramp, is needed to slide the crate up the ramp at a constant speed.
a) How much work is done in sliding the crate up the ramp?
b) How much work would be done if the crate were simply lifted straight up from the floor to the platform?

I believe I use W = F*D, but is the distance equal to the hypotenues of the ramp?

And how come you can't just do W = (400)*(1.0) for the second question?

Help please :confused:
 
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Format said:
I believe I use W = F*D, but is the distance equal to the hypotenues of the ramp?
Right. The applied force is parallel to the ramp and the distance the crate moves in that direction is given as 2.0 m. The work done by a force equals Force x Distance in the direction of the force.

And how come you can't just do W = (400)*(1.0) for the second question?
In the second problem the 400 N force doesn't exist. The force needed to lift the crate equals the weight of the crate and points upward. The upward distance the crate is moved is 1.0 m.
 
So its 400*2.0?

And for the second part is it m*g*1.0m?
 
Right! Be sure to give the proper units when you present your answers.
 
Awsome, thanks!
 

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