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## Homework Statement

Find the acceleration reached by each of the two objects shown in Figure P4.49 if the coefficient of kinetic friction between the 7.00kg object and the plane is 0.250

Here is a picture of the problem: http://a496.ac-images.myspacecdn.com/images01/90/l_1488ef92fb0cdcd21ba1febcabca499f.jpg

## Homework Equations

F=ma

## The Attempt at a Solution

I'm calling the 7.00 kg object M1 and the 12.0kg object M2.

<b>For M1</b>

(Fnet)x=ma

T+mgsin37degrees-fk=m1a

(Fnet)y=ma (a=o)

n1-mgcos37degrees=0

n1=mgcos37degrees

<b>For M2</b>

(Fnet)y=m2a

T-m2g=m2a

T=m2a+m2g

I take this equation and put it in the (Fnet)x equation I got for M1:

T+mgsin37degrees-fk=m1a

m2a+m2g+mgsin37degrees-fk=m1a

m2a+m2g+mgsin37degrees-(0.250)mgcos37degrees=m1a

m2g+mgsin37degrees-(0.250)mgcos37degrees=m1a-m2a

a=m2g+mgsin37degrees-(0.250)mgcos37degrees / (m1-m2)

a=7.64 m/s^s

Now, my question is: It says "Find the acceleration reached by <b>each</b> of the two objects. Now, I know with an ideal pulley the tensions on both side of the equation are equal, but I'm not sure if the acceleration is. I would assume so cause it's one rope and one rope can't move at two different speeds at once, but I could be wrong. So, is that the answer to the acceleration of both blocks or merely the acceleration of the 7.00kg object and there is further work to be done to find the acceleration of the 12.0kg object?