Work done when dragging a crate 15m at 37° angle

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Homework Help Overview

The problem involves calculating the work done when dragging a crate 15 meters at a 37° angle on a horizontal surface. The discussion centers around the implications of forces acting on the crate, particularly in the context of gravity and friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between work, force, and distance, questioning the role of gravity and whether vertical work is being neglected. There is uncertainty about the necessary force to calculate work and the impact of friction.

Discussion Status

The discussion is ongoing, with participants raising questions about the solvability of the problem due to missing information, particularly regarding external forces and friction. Some guidance has been offered regarding the components of work, but no consensus has been reached.

Contextual Notes

Participants note the absence of specific values for mass and friction, which are critical for solving the problem. There is also a suggestion that the problem may be unsolvable without additional information.

Silverbolt
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A crate is dragged 15m along a horizontal surface by a rope which makes an angle of 37° with the horizontal. How much work is done?
 
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How much work do you think is done?
 
Last edited:
Is there gravity, if so Earths? Are we neglecting vertical work?
 
I THINK we are ignoring the vertical
 
@lewando i am not sure where to start with this problem
 
Assuming you know W = F*d, you have d, you need F. By "dragged" assume no acceleration. What force would resist your dragging force?
 
Okay, just assume there's gravity strong enough that no matter how much work we put into vertical work, it will not move.
Work in this case, is split into two components, vertical and horizontal. Since we are ignoring the vertical component (because we'll assume something like the normal force by the ground + Force applied by you upwards = gravitational force).
So we only have to find the horizontal component.
But we don't even know the force applied yet.
 
@Cbray: also m, also μ.
 
Is this problem unsolvable?
 
  • #10
We can't solve the amount of energy given to the crate without an external force, so information wise, yes it seems unsolvable unless you were given extra information.
 
  • #11
The frictional reaction force is unknowable. So good call.
 

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