Finding acceleration in two blocks, one hanging and other on board.

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DrClone
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Homework Statement


Two blocks are connected by a string as in the figure below.

Free-body diagram of a 2.0 kg block on a table with a green velocity arrow labeled v pointing right, connected by a string over a pulley at the table edge to a hanging 1.0 kg block.


What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.10?

Express your answer to two significant figures and include the appropriate units.


Homework Equations



?

The Attempt at a Solution


Attempt: I drew a free-body diagram for two blocks.

m1 = mass one (2.0kg)
m2 = mass two (1.0kg)
∫= friction
T= tension
w= weight

Block 1: Fx=m1(a) = T -∫

Fy= m1(a)=n-m1g

Block 2: Fx=0

Fy= m2(a)=T-m2g

I would find the normal force for Block 1 Fy then plug it into friction. Then...I got lost.
 
Last edited:
The rope and pulley are most likely massless.

The y-component of acceleration for block 1 should be zero. See how that works.
 
This is what I got so far...am I correct?

f+m1a=m2g-m2a?

Then

m1a+m2a=m2g-f

THEN

a(m1+m2)=m2g-f

a=(m2g-f)/(m1+m2)

?
 
I can't really follow your formula, but what did you get for acceleration once you solved?
 
The first box:

x-forces:
T - K = ma
T - (0.1)(N) = 2a

y-forces:
N - w = o
N = w
N = 2g
N = 9.8

The second box:

T - w = m2a
T - m2g = 1a
T - 9.8 = a

Solve that.
 
2.6 m/s^2

it looks like I answered my own question...LOL

Thanks lolcat, I forgot the negative in the second box.
 
Not a problem, but I actually got 7.84 m/s2 lol. Anyone else want to check our answers?