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**A 2.0 kg block (m1) and a 6.0 kg block (m2) are connected by a massless string. Applied forces, F1 = 12.0 N and F2 = 5.0 N, act on the blocks, as shown in the figure below m1 is on the left side of the system, F1 is pulling on m1 to the left; m2 is to the right, F2 is is pulling m2 to the right (both forces are parallel to the x axis). The coefficient of friction between the blocks and the table is μK = 0.16.**

A) Draw a free body diagram for EACH block. [3 points]

B) Write Newton’s Second Law equations for each block. [3 points] C) Determine the acceleration of the blocks. [2 points]

D) What is the tension on the string connecting the blocks? [2 points]

**Relevant equations:**

ƩF=m*a

F friction=[itex]\mu[/itex]*Fn

F gravity=m*g

**Attempt to solve problem:**

I am assuming the system is moving in the direction of F1, I believe this is where I made a mistake. Anyway, for m1:

ƩFy=m1*g=Fn1

ƩFx=F1-m1*g*μ-T=m1*a

For m2:

ƩFy=m2*g=Fn2

ƩFx=T-m2*g*μ-F2=m2*a

T=m2*a+m2*g*μ+F2

Replacing T for m1:

m1*a=F1-m1*g*μ-(m2*a+m2*g*μ+F2)

∴

a=(F1-m1*g*μ-m2*g*μ-F2)/(m2+m1)=-0.694 m/s^2

**I wasn't provided with a solution to the problem, I am sure I made a mistake. Could use some help**