Understanding Conservation of Energy in Pulley Systems

[moooocow]

I don't think this is homework but if it is sorry for putting it in the wrong spot.

Ok, I was out of physics class for a few days due to a sinus infection, I REALLY did not want to miss it but sometimes its just not an option. I read the book while I was out to try and keep up but my book seems to do a poor job(at least for me) of explaining how kinetic and potential energy go together as far as conservation goes. I really just don't understand exactly how to go about applying these laws in actual problems. For instance, this is a sample problem I don't understand all the way. "A 50.0 kg block and a 100-kg block are connected by a string as in(it shows a picture of a right triangle with theta being 37°, the 50kg block is being dragged up this angle due to a pulley at the top of the triangle being attached to the 100 kg mass which is going straight down) The pulley is frictionless and of negligible mass. Determine the kinetic energy of the 50kg block as it moves a distance of 20 meters" My teacher went over the answer and this is what we have(that i don't understand)
m1 = 50kg
m2 = 100kg
d = 20m
y= 0 is where m1 starts from
y1 = is where m2 starts
y2 = is where m2 ends

m2Gy1 = 1/2m1v1^2 + 1/2m2v2^2 + m1Gdsin(theta) + m2Gy2

In this part I understand its just setting the initial energy = to the final energy, with the end having both potentials and both kinetics, what I don't understand is why there is only 1 potential energy on the left side. I would think that there would be like m1Gdsin(theta) also because it has potential energy to slide? Even though it would be negated by the tension from m2 why doesn't it have to factor into the equation exactly? This is just an example though, I am looking more for the general method used for these problems because there are quite a few similar ones and I am just really confused on how to go about conservation problems in general. So any pointers or methods would be greatly appreciated.

Also one last thing, when simplified m2Gy2 was subtracted from the right side and made the left side m2Gd, I understand that y1-y2 is d but I am not exactly sure of the algebra behind it(im math retarded sometimes) so if anyone could explain why m2Gy1-m2Gy2 is m2gD It would also be appreciated. Thank you for any type of help, i could really use it for my midterm tomorrow : ( Thank you very much

 

[Ambitwistor]

m2Gy1 = 1/2m1v1^2 + 1/2m2v2^2 + m1Gdsin(theta) + m2Gy2

In this part I understand its just setting the initial energy = to the final energy, with the end having both potentials and both kinetics, what I don't understand is why there is only 1 potential energy on the left side. I would think that there would be like m1Gdsin(theta) also because it has potential energy to slide?

On the left-hand side, there is a potential energy term m1 g y1, but y1=0. (The only potential energy in this problem is gravitational potential energy, given by U=mgy.)

Also one last thing, when simplified m2Gy2 was subtracted from the right side and made the left side m2Gd, I understand that y1-y2 is d but I am not exactly sure of the algebra behind it(im math retarded sometimes) so if anyone could explain why m2Gy1-m2Gy2 is m2gD It would also be appreciated.

If m1 moves 20 meters up the angle on a diagonal, then m2 drops 20 meters straight down, assuming that the rope in between them maintains the same length and doesn't stretch or compress. So the change in height y1-y2 of m2 is d=20 m.

 

[HallsofIvy]

One thing you should remember is that (gravitational) potential energy is always "relative" to some reference position. You can always choose the reference to be the height of one of the objects at one time and make the potential energy of that object equal to 0.

 

[keeganisafish]

Does it look something like this?

http://keegan.aexx.net/problem.gif

(I love physics )