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elsternj
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Homework Statement
A block of weight sits on a plane inclined at an angle [tex]\theta[/tex] as shown. (Intro 1 figure) The coefficient of kinetic friction between the plane and the block is [tex]\mu[/tex].
What is the work done by the applied force of magnitude ?
Express your answer in terms of some or all of the following: [tex]\mu[/tex],w,[tex]\theta[/tex], L
Homework Equations
The Attempt at a Solution
now the answer to this is W = w(sin([tex]\theta[/tex])+[tex]\mu[/tex]cos([tex]\theta[/tex]))L
now I somewhat understand this because the force has to be enough to overcome friction and the weight pulling it down. but what I am concerned with is this:
This would be the answer if the sum of the forces were equal to 0 right?
because if:
[tex]\sum[/tex]Fx=F-wsin[tex]\theta[/tex]-[tex]\mu[/tex]wcos[tex]\theta[/tex]=0
when you solve for F you get the answer originally posted and then just multiply it by the length. How is one to know, given the above information, to set that equal to 0 and solve? It doesn't mention anything about moving at a constant speed. Just trying to get some clarification because I always set the sum of my forces equal to ma and I had trouble answering this question because the answer had to only include the variables [tex]\mu[/tex],w,[tex]\theta[/tex], L ... so the only way I could see to solve for F and keep only those variables would be if acceleration happened to be 0.
Homework Statement
Homework Equations
The Attempt at a Solution
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