Velocity of two identical masses over a frictionless pulley

[Nicki]

Homework Statement

"Consider the following arrangement of masses. Mass 1 is connected to mass 2 by a very light string and moves over a frictionless pulley so that both masses move with the same speed and move the same distances (m2 to the right and m1 down).
Assume m1 = 15 kg, m2 = 15 kg and the coef. of kinetic friction is 0.73. The masses start with an initial velocity of 1.00 m/s. What is their speed after moving 0.0100 m?"

Homework Equations

dE = E1 + E2 = dU1 + dK1 + dU2 + dK2

E1 = dU1 + dK1 (0) = m*g*dH

E2 = dU2 (0) + dK2 = m/2 * (vxf^2 - vxi^2) - friction?

Friction = - 0.73 * mg * d

The Attempt at a Solution

[/B]
This is where I'm getting stuck.
So do I set

m*g*dH = m/2 * (vxf^2 - vxi^2) - friction

and solve for vf through this equation?

or do I use m/2 * (vxf^2 - vxi^2) - m*g*dH = - friction?

I'm just not sure how to set up the final equation in order to solve for velocity.

 

[NTW]

In my humble opinion, the best way of solving this (and many similar) problems is not to throw a number of equations on the table and then trying to make sense with them...

I suggest, first of all, to make a clear sketch, with the masses and forces. Then, and in the case of this problem, you should calculate the acceleration by using f = m * a, plugging in the equation the total mass and the net force. And then you can easily find the solution with one of the equations of uniformly accelerated motion...

 

[Nicki]

Unfortunately this is how we're expected to solve these for the time being

I also realized I was missing KE for one of the masses,so now...

I believe that dE = dE1 + dE2

and E1 = dU = mgHf - mGHi and dK = m/2 (vf^2 - vi^2)

and E2 = dU = 0 and that dK = m/2 (vf^2 - vfi^2)

also friction = -(coef friction)*mgd

 

[BiGyElLoWhAt]

Here's my honest opinion:
I honestly can't make sense of your equations. It looks like you're trying to use energy's, which you should be able to do, but I don't know what your equations mean. What is E1? What is E2? what is dE? is dE differential energy, whereas E1 and E2 are total energy? If so, how do you get dE = E1 + E2?
I'm assuming mghf is final GPE and mghi is initial GPE, which works, but there's a significant lack of organization IMHO.
Also, forces definitely work in this situation as well, as NTW said.

 

[HalfLostAndSearching]

I would approach this problem by first setting up the equations for conservation of energy and solving for the final velocity. The equation for conservation of energy in this case would be:

Ei = Ef
Where Ei is the initial energy and Ef is the final energy.

The initial energy (Ei) would consist of the potential energy of mass 1 (m1), which is m1gh, and the initial kinetic energy of mass 2 (m2), which is 1/2*m2*v^2.

The final energy (Ef) would consist of the final potential energy of mass 1, which is m1gh, and the final kinetic energy of mass 2, which is 1/2*m2*v^2. However, since the masses are connected and move together, their final velocity would be the same and can be represented as v.

Therefore, the equation for conservation of energy becomes:

m1gh + 1/2*m2*v^2 = m1gh + 1/2*m2*v^2

Simplifying this equation, we get:

1/2*m2*v^2 = 0

Solving for v, we get:

v = 0 m/s

This means that after moving 0.0100 m, the masses will come to a stop and their final velocity will be 0 m/s.

It is important to note that in this problem, we are assuming that there is no external force acting on the masses, and the only force that is considered is the frictional force. If we take into account the frictional force, the final velocity would be slightly less than 0 m/s. However, since the masses are moving over a frictionless pulley, we can assume that the frictional force is negligible and can be ignored in this problem.