Two masses connected by a pulley with a frictionless table

[AHinkle]

Homework Statement

Homework Equations

m1
\SigmaF_y=N-m₁g = 0
\SigmaF_x=T=m₁a
(Because there's no friction i see no opposing force to T)

m2
\SigmaF_y=m₂g-T=m₂a
\SigmaF_x=0

The Attempt at a Solution

m₂g-T=m₂a
T=m₁a

(m₁+m₂)a=m₂g-T+T
I notice that the T's cancel when i add the equations together
so it becomes

(m₁+m₂)a=m₂g
so...
a=(m₂g)/(m₁+m₂)

so...
T=m₁a
T=(m₁) (m₂g)/(m₁+m₂)

m₁=6.03kg
m₂=4.68kg

T=(6.03Kg)((4.68Kg)(9.8)/(6.03Kg+4.68Kg))
so...
T=25.8225N

T=m₁a
25.8225N = (6.03Kg)a

a=4.2823 m/s²
did i do this right?

 

[collinsmark]

m₂g-T=m₂a
T=m₁a

(m₁+m₂)a=m₂g-T+T
I notice that the T's cancel when i add the equations together
so it becomes

(m₁+m₂)a=m₂g
so...
a=(m₂g)/(m₁+m₂)

You know, you could have just plugged the numbers in here and saved yourself some work.

a=4.2823 m/s²

'Looks right to me.