Formula for mass up an inclined plane

AI Thread Summary
To derive the work needed to push a mass up an inclined plane, start with the formula W=Fd, considering gravitational potential energy, friction, and the angle of inclination. The initial equation provided, W=(umgcos(theta)+mgsin(theta))(d), needs adjustment since 'd' is the independent variable. It's suggested to express 'd' in terms of height (h) using the relationship d=h/sin(theta) for further calculations. Ultimately, the expression for work can remain in terms of 'd' if the focus is solely on the work done. The discussion emphasizes the importance of keeping variables clear and correctly defining relationships in the formula.
simmer_27
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I was just wondering if anyone could help me out with this homework. You have to derive a formula to figure out the work needed to push a mass up an inclined plane. All you know is the gravitational potential energy, the coeffieciant of friction, and the angle. Your supposed to start out with W=Fd.
All I have so far is W=(umgcosfeta+mgsinfeta)(d). I'm not sure how to get "d", if anyone could help me out I'd really appreciate it. I'm guessing you have to make the masses cancel out somehow too.
 
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Just divide both sides by umgcosfeta+mgsinfeta

So W/umgcosfeta+mgsinfeta= d
 
stupid

well you don't know "d" so that formula won't work?
 
'd' is the independent variable, and 'W' is the dependent variable.

You solved the problem, and you have a formula for 'W' as a function of 'd'.
 
you don't know "d", you can't have W=something d. u can only have one variable at the end
 
Leave W as a function of the angle and d...if u have the change in the gravitational potential energy, then u can work out the final height. So in the triangle d is the hypotenuse, h is the opposite side and hence d can be expressed as

d=h/sin(theta)
Substitute that into your work equation, however from what you wrote, the question only wants an expression for the work done, so it can be left in terms of d.
 
could W=ucosfetaE+E work?
 
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