View Full Version : Formula
simmer_27
Nov20-04, 06:07 PM
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.
Just divide both sides by umgcosfeta+mgsinfeta
So W/umgcosfeta+mgsinfeta= d
simmer_27
Nov20-04, 07:58 PM
well you dont know "d" so that formula won't work?
Astronuc
Nov20-04, 08:44 PM
'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'.
simmer_27
Nov20-04, 11:48 PM
you don't know "d", you can't have W=something d. u can only have one variable at the end
rayveldkamp
Nov20-04, 11:54 PM
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.
simmer_27
Nov20-04, 11:57 PM
could W=ucosfetaE+E work?
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