2 blocks sliding down a incline connected by string (check please)

[Seiya]

Two blocks with masses 4.0kg and 8.0 kg are connected by a string and slide down a 30 degree inclined plane. The coefficient of kinetic friction between the 4.0 kg block and the plane is 0.25; that between the 8.0 kg block and the plane is 0.35
a) calculate the acceleration of each block

This part is easy i got it right, don't even need anyone to check it... 4 kg = 2.775 , 8 kg =1.4750

used for both blocks : mgx-ff=ma

b) calculate the tension in the string

I said since the bottom block accelerates faster it will create a tension which means both blocks will be connected by a rigid string with a tension in it... so this means i could solve them as a system ... this is where i need someone to verify i did it right or if not correct me...

i said that

(mgx1+mgx2)-(ff1+ff2) = (m1+m2)a

They are connected so they will have a new a in common for both which comes out to 1.9 .

So now i can find the tension by using either box and the first equations trying to find their acceleration alone... instead now i have

(for the bottom box)

mgx-ff-T=m(1.9)

So i get T = 3.5N

if i use this t in a Newtons second law equation for the top one ill get 1.9 acceleration for it as well...


Can anyone let me know if i did any mistakes please? Thank you :) I really appreciate it

 

[mukundpa]

the equations are
m1 g sin(theeta) - mu1 m1 g cos(theeta) - T = m1 a
and
m2 g sin(theeta) - mu2 m2 g cos(theeta) + T = m2 a
check your calculations
first for a2 and then for common a.
MP

 

[Seiya]

i used this for the second part... if you look at my equation but i combined them together... am i wrong?(mgx1+mgx2)-(ff1+ff2) = (m1+m2)a

if you combine them the T's cancel but i can find the T by setting it in a separate equation? what's wrong with my method?

Thanks

 

[mukundpa]

No the method is not wrong, I told to check the calculations because I am gatting a different answer for acceleration for the second block alone and the commom acceleration.