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.
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?