# Homework Help: Finding acceleration in two blocks, one hanging and other on board.

1. Jan 22, 2012

### DrClone

1. The problem statement, all variables and given/known data
Two blocks are connected by a string as in the figure below.

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.

2. Relevant equations

?

3. 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: Jan 22, 2012
2. Jan 22, 2012

### SHISHKABOB

are the rope and pulley massless?

3. Jan 22, 2012

### lolcat

The rope and pulley are most likely massless.

The y-component of acceleration for block 1 should be zero. See how that works.

4. Jan 22, 2012

### DrClone

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)

?????

5. Jan 22, 2012

### lolcat

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

6. Jan 22, 2012

### lolcat

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.

7. Jan 22, 2012

### DrClone

2.6 m/s^2

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

Thanks lolcat, I forgot the negative in the second box.

8. Jan 22, 2012

### lolcat

Not a problem, but I actually got 7.84 m/s2 lol. Anyone else want to check our answers?