Work Done by Weight Homework: Find W

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Homework Help Overview

The problem involves a 700 kg crate on a rough incline of 30°, with a horizontal force applied. The discussion centers on calculating the work done by weight, specifically the work done by gravity, as well as exploring the relevance of other given information in the context of the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the meaning of 'work done by weight' and whether it refers to the work done by the applied force or gravity. There are attempts to clarify the relevance of the provided information and the implications for solving the problem.

Discussion Status

The discussion is active, with participants sharing their interpretations and questioning the necessity of certain details. Some have identified the work done by gravity as a key focus, while others are exploring the implications of changes in the problem's parameters for subsequent parts of the question.

Contextual Notes

Participants note that the problem may have additional parts, particularly regarding the work done by friction, and that changes in velocity and time have been introduced in the follow-up question.

Legerity
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Homework Statement



A 700 kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 N is applied horizontally to the crate. The force pushes the crate a distance of 3.0 m up the incline, in a time interval of 7.3 s, and the velocity changes from v1 = 1.4 m/s to v2 = 2.5 m/s. What is the work done by weight?

75512f21-629f-400f-914b-f9f928f9ba49.png


Homework Equations



W = Fd, KE = 0.5mv^2, PE = mgh

The Attempt at a Solution



I drew the free-body diagram, but I am stumped on how to go about the problem. What is the work done by weight, and how can I find it? I don't exactly know what it is.
 
Last edited:
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I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
 
haruspex said:
I don't know what they mean by 'work done by weight'. My guess is they meant the work done by the force P. (Cut and paste error from another question?)
I figured out the problem, and it is the work done by gravity on the box (ΔPE).
 
Legerity said:
I figured out the problem, and it is the work done by gravity on the box (ΔPE).
Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
 
haruspex said:
Really? So most of the information is irrelevant? (Or maybe there are more parts to the question?)
Yes, there is another part of the question I am stuck on. It is now asking me to find the work done by the frictional force, and my attempts at the problem have been futile. The only things changed in the info in the original post is that now V2 = 2.3 m/s, and time = 8.3 s. Could you possibly help?
 
Last edited:
You'll have to use an equation that relates conservation of energy and the energy lost by friction: KE_{i}+PE_{i}+WE_{i}=KE_{f}+PE_{f}+WE_{f}
 
Legerity said:
It is now asking me to find the work done by the frictional force,
You know the distance, so it remains to calculate the frictional force. What equations do you get from your free body diagram?
 
Legerity said:
I figured out the problem, and it is the work done by gravity on the box ( - [/color]ΔPE).
see important correction in red[/color].
 

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