What is the Relationship Between Work and Friction?

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The discussion focuses on calculating work done against friction, specifically in the context of a tractor moving a certain distance. Participants highlight the importance of clearly stating the entire question to avoid confusion. The formula for work done (WD = F x D) is reiterated, with a specific example given where force is 11,000 N and distance is 6.0 meters, resulting in 66,000 J of work done. There is a mention of inclined planes, but the main calculation pertains to horizontal movement. Clarity in the problem statement is emphasized to facilitate accurate calculations.
ellieee
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Homework Statement
how to calculate work done against friction? I saw alot of people suggesting to use the formula involving cosine if I'm not wrong but it's not in my syallbus, so is there another way to calculate it?
Relevant Equations
WD = F x D
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Do you consider inclined plane ?
 
ellieee said:
Homework Statement:: how to calculate work done against friction? I saw a lot of people suggesting to use the formula involving cosine if I'm not wrong but it's not in my syallbus, so is there another way to calculate it?
Relevant Equations:: WD = F x D

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You need to state the whole question or we're just guessing.
 
haruspex said:
You need to state the whole question or we're just guessing.
qn 3c
 

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anuttarasammyak said:
Do you consider inclined plane ?
refer to qn 3c pls:) not inclined plane
 

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As you referred WD=F*D
F=11,000 N
D is a distance meter the tractor goes in 5 seconds.
 
anuttarasammyak said:
As you referred WD=F*D
F=11,000 N
D is a distance meter the tractor goes in 5 seconds.
oh but the answer is (11 000)(6.0) = 66 000 J though
 
ellieee said:
oh but the answer is (11 000)(6.0) = 66 000 J though
So, what is D?
 
onatirec said:
So, what is D?
oh I was careless and didn't see properly , sorry
 

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