Clarification on this specific problem (work done on incline plane)

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 10K views
elsternj
Messages
42
Reaction score
0

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

2994_a.jpg

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

 
Last edited:
Physics news on Phys.org
hi elsternj! :smile:
elsternj said:
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

It doesn't mention anything about moving at a constant speed.
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.

yes, you're right, it's a rubbish question :redface: