Question

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.16?

Answer 1

(Missing figure is here: https://www.physicsforums.com/attachments/ch05-p070-jpg.149243/ )

Let's call $m_1 = 1.0 kg$ and $m_2=2.0 kg$ the masses of the two blocks. We can write Newton's second law for both blocks (sum of all forces acting on the block = ma). On block 1, we have two forces: the weight $m_1 g$ pointing downwards and the tension of the string T poiting upwards. On block 2, we have the tension of the string going right and the friction $\mu m_2 g$ going left. Therefore

$m_1 g - T = m_1 a$
$T-\mu m_2 g = m_2 a$

Summing the two equations, we find
$m_1 g - \mu m_2 g = (m_1 + m_2) a$
and then using $\mu=0.16$ we can find the acceleration:
$a= \frac{m_1 g - \mu m_2 g}{m_1 + m_2}=2.22 m/s^2$