How Far Do the Blocks Slide After Spring Release?

[vande060]

Homework Statement

Two blocks M1 and M2 rest motionless on
a surface of friction coefficient μk. A powerful
spring of force constant k, compressed by
amount x, is in a cavity inside of them. A
hook holds them together. When the hook is
released, they are flung apart. How far apart
will they be after they stop sliding?

http://s861.photobucket.com/albums/ab174/alkaline262/?action=view&current=book2question2.jpg

Homework Equations

W = 1/2m1v1^2 - 1/2m1v1^2

The Attempt at a Solution

mew = u

okay here we go(maybe)

the potential function for a spring -V = 1/2kx^2
potential function for kinetic friction = ukmgx

not sure about the works above, and if i did the 100% right

work energy theorem 1/2m1v1^2 + 1/2m2v2^2 = 1/2kx^2 - ukmgx

i don't suppose i could just set v1 = v2 = 0 and solve the quadratic for x

 

[Xerxes1986]

i don't suppose i could just set v1 = v2 = 0 and solve the quadratic for x

No, because that is a trivial solution.

You are thinking along the right lines, but you don't have to think about the velocities of the blocks at all. You know the P.E. of the spring is the total energy of the system before the hook is released. Then after the blocks come to rest the total energy is 0. What happened to all that energy? (Hint, work done by friction, which is proportional to distance moved ;) )

 

[vande060]

No, because that is a trivial solution.

You are thinking along the right lines, but you don't have to think about the velocities of the blocks at all. You know the P.E. of the spring is the total energy of the system before the hook is released. Then after the blocks come to rest the total energy is 0. What happened to all that energy? (Hint, work done by friction, which is proportional to distance moved ;) )

I don't really understand what you are saying here. could you rephrase it please? I am thinking all of that energy is lost during the displacement, where friction is causing loss of energy by heat and noise

so at the risk of reinventing the wheel: kx^2 - ukmgD = 0

d=displacement
i have never really been sure what x represents in the spring function, i don't know if it is the same as displacement

 

[Xerxes1986]

Think about it a little harder, you have the general equation right but there are some coefficients and simplifications you can make. (Specifically a factor of 1/2 in front of the PE? and D=x+d, and d is what you want.)

 

[vande060]

Think about it a little harder, you have the general equation right but there are some coefficients and simplifications you can make. (Specifically a factor of 1/2 in front of the PE? and D=x+d, and d is what you want.)

okay, i really understand the difference between d and x now. Also, I believe that the coefficient of 1/2 only belongs before the PE of the the spring, correct me if i am wrong but:

1/2kx^2 -ukmg(x+d) = 0

1/2kx^2 -ukmgx - ukmgd = 0

1/2kx^2 -ukmgx = ukmgd

(1/2kx^2 -ukmgx)/ukmg = d

i would have to how much the spring was originally compressed by, along with masses to solve this any further i think