Two blocks, a fixed pulley and friction

[SakuRERE]

Homework Statement

a mass m1 is attached to a second mass m2 by an acme (massless, unstretchable) string. m1 sits on a table with which it has coefficients of static and dynamic friction μs and μk respectively. m2 is hanging over the ends of a table, suspended by the taut string from an acme pulley. at time t=0 both masses are released.
1. what is the minimum mass m2,_min such that the two masses begin to move?
2. If m2= 2m2,_min, determine how fast the two blocks are moving when mass m2 has fallen a height H (assuming the m1 hasn't yet hit the pulley)?

Homework Equations

Σf=ma
fsmax=μs*fn
vf²=vi²+2aΔx[/B]

The Attempt at a Solution

for a:
I am not sure about how I understand the question, but what comes to my mind when reading ( Begin to move) is that still the system acceleration is zero and that they are asking for the F_smax (the maximum static friction) just before the system starts to move and the friction becomes Kinetic friction of Fk.
so:
for block m1
Σƒ= m*a=0
f_smax=T
μs*Fn= T
μs*m1g=T
now for block m2:
Σƒ=m*a=0
m2g=T

and by this m2g=μs*m1g so m2=(μs*m1g)/g

section b)
I guess since the mass of m2 is doubled, that here we will deal with the kinetic friction rather than the static friction.
since both are linked by a rope with the same pulley than they must have the same acceleration, and we should find it.
and after finding it then we will use this equation:
vf²=vi²+2aΔx
where vi= 0
so vf=√(2aΔx).
am I right??

 

[haruspex]

so m2=(μs*m1g)/g

Yes, but you can simplify a little.

section b)
I guess since the mass of m2 is doubled, that here we will deal with the kinetic friction rather than the static friction.
since both are linked by a rope with the same pulley than they must have the same acceleration, and we should find it.
and after finding it then we will use this equation:
vf2=vi2+2aΔx
where vi= 0
so vf=√(2aΔx).
am I right??

Yes.

 

[SakuRERE]

but what is a

yes, of course, I will be substituting a in term of the given variables but I just didn't because I wasn't sure from my solution for section a at the first place. but now after your response, I will continue the solution. and regarding the H and Δx, I will be substituting Δx with H simply, but I wanted to write the equation in its original form so that everyone easily identifies it.
I will be happy if you grant me another help and answered this question:
for a moment I wondered, will the final velocity be the same for both blocks, I know they have the same acceleration and H will be the same as well but doesn't the friction affect the velocity of that block in the incline?

 

[haruspex]

I will be substituting a in term of the given variables

Yes, sorry, I realized that after rereading your original post, so I edited my reply.

will the final velocity be the same for both blocks, I know they have the same acceleration

Same acceleration over the same period of time, so...

 

[SakuRERE]

Same acceleration over the same period of time, so...

Thanks! I just wanted to make sure and to move away any doubts that are used to appear out of nowhere :)