Energy, work, 2 masses on incline

[alphysicist]

After I've substituted the numbers I get:
15.68 - 9.408= 2v1^2' + 2.5v2^2'
im not sure what to do with the right side mathematically

How are v1 and v2 related? Can you rewrite it so that you only get one unknown?

 

[FurFur]

well v1 and v2 must be the same.
15.68 - 9.408= 2 (v1^2' + 1.25v2^2')
this still leaves me with 2 unknowns..but both unknowns are equal..hmm

 

[alphysicist]

well v1 and v2 must be the same.
15.68 - 9.408= 2 (v1^2' + 1.25v2^2')
this still leaves me with 2 unknowns..but both unknowns are equal..hmm

If they are both the same (and they are), give them the same name. Call them both v. Then you have only one unknown.


Or, if you want to look at it another way, you know that v1=v2, so replace v1 with v2 in your equation.

 

[FurFur]

oh right! Okay so the equation is:
15.68 - 9.408= 2v^2 + 2.5v^2
6.272= 4.5v^2
6.272/4.5 = v^2
1.39]square root = v
1.18 m/s= v
that seems right.

 

[alphysicist]

oh right! Okay so the equation is:
15.68 - 9.408= 2v^2 + 2.5v^2
6.272= 4.5v^2
6.272/4.5 = v^2
1.39]square root = v
1.18 m/s= v
that seems right.

That looks right to me.

So the interpretation of the energy formula might be, m2 loses potential energy, which is transformed into increasing potential energy for m1 and also increasing kinetic energy for m1 and m2.

 

[FurFur]

right, since m2 is going down and m1 is going up. It makes perfect sense. Well thank you so much for the help:) and for ur time, it took a while (what a tedious question -_-.. maybe because I am reali slow at physics). By the way, how would I mark that the question has been solved?

 

[alphysicist]

right, since m2 is going down and m1 is going up. It makes perfect sense. Well thank you so much for the help:) and for ur time, it took a while (what a tedious question -_-.. maybe because I am reali slow at physics). By the way, how would I mark that the question has been solved?

Sure, glad to help! And the more you practice on these problems, the faster you'll go. It's easy to write down the equations, but it can sometimes be difficult to understand how to apply them until you have enough experience.

I don't think you can mark threads solved right now; that function went away recently after an upgrade, but I think it is planned on being fixed.

 

[FurFur]

oh okay. Yea I'm going to keep practicing since practice makes perfect.