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## Homework Statement

A 2.0 kg block rest on a 4.0kg block that is on a frictionless horizontal table. The coefficients of friction are u_s = 0.3 and u_k = 0.2. What is the maximum force F that can be applied to the 4.0 kg block if the 2.0 kg block is not to slip? If F has half this value, find the acceleration of each block and the force of friction acting on each block .If F is twice the value you determined in the first question, find the acceleration of each block.

## Homework Equations

Fx = F = m1a

Fx = ff = m2a

## The Attempt at a Solution

What I need help with is the final part of this problem... part C. Here's what is needed to solve:

I solved for 18 N in part(a), the first part, as the maximum force required to move the object.

Now, I solved for the acceleration of block 2 in the last part; block 2 = ff/m2 = u_k*g = (.2)(9.81) = 2.0 m/s^2

But now I'm trying to solve for the acceleration of the 4.0kg block

F - ff = m1a1 + m2a2

(F - ff - m2a2)/m1 = a1 = 7.019 m/s^2

Not right... the book says it is 7.9 m/s^2. So I tried again ignoring ff

(F - m2a2)/m1 = a1 = 8.019 m/s^2

where am i going wrong here? sorry, no diagram given in problem so I can't scan one.. :X:X

i note that, although they list 18 N as the force, if you leave it unrounded:

17.76 *2 = 35.52

35.52 - m2 = m1a1

35.52 - 4 = 4a

a = 7.88 = 7.9 m/s^2

that can't be it, can it?

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