Friction-Newtonian Mechanics 
Here we are having two blocks of mass $2~\rm kg$ and $4~\rm kg$ on an inclined angle with angle of inclination being $30~\circ.$ The block of mass 2 kg has a coefficient of friction $\mu_1=0.2$ and the block of mass 4 kg has a coefficient of friction  $\mu_2=0.3$. 
In a book, it was written that as $\mu_1<\mu_2$,  hence  $a_{\mathrm{2~kg}}>a_{\mathrm{4~kg}}$, where $a_{i}= \text{acceleration of the blocks}$.
I checked for the cases separately and that came out to be true. I just wanted to ask, can i do that (the $\mu_1<\mu_2$ stuff) always while solving a problem or should I always check out the cases separately for whose acceleration would be bigger.
 A: The observation that if $\mu_1 < \mu_2$ then $a_1 > a_2$ applies only when the two blocks are separated and on the same incline.  If the two blocks are in contact, with block 2 lower down the incline (as in the illustration), this is obviously false, because it is impossible for the upper block to pass the lower block.
Neither can we conclude that the two blocks accelerate together at $a_2$ calculated when block 2 is isolated on the incline :
$m_2a_2=m_2g\sin\theta - \mu_2m_2g\cos\theta$
$a_2=g(\sin\theta - \mu_2\cos\theta)$.  
If the two blocks are in contact with $\mu_1 < \mu_2 < \tan\theta$ then they accelerate together at the same rate $a$ given by :
$(m_1+m_2)a=(m_1+m_2)g\sin\theta - (\mu_1m_1+\mu_2m_2)g\cos\theta$
$a=g(\sin\theta - \frac{\mu_1m_1+\mu_2m_2}{m_1+m_2}\cos\theta)$.  
The common acceleration $a$ is not (in general) equal to either $a_2$ or $a_1$.
Conclusion: It is not wise to apply a generalisation without examining if the conditions for it are met.
A: (Forgive me if I am stating the obvious.) Taken individually, any body will accelerate at a rate that is inversely proportional to the coefficient of friction between itself and the surface on which it sits, regardless of its mass. This is easily proven algebraically: the mass affects both forces that run parallel to the plane (the parallel component of weight and the frictional force) equally. Therefore, in the above example, one can immediately surmise that if $μ_1$ < $μ_2$, then $a_{2kg}$ > $a_{4kg}$, and we we will always know which acceleration will be larger.
A: Let's prove it:
$$\sum F=ma\\
f-w_x=ma\\
\mu n-mg\cos(\theta) =ma\\
\mu mg\sin(\theta) -mg\cos(\theta) =ma\\
\mu g\sin(\theta) -g\cos(\theta) =a$$
So, smaller $\mu$ gives smaller (more negative) $a$.
So acceleration is growing downwards along the incline for smaller friction coefficient. 
A: You always need to make a force diagram to be safe, however, for this particular configuration (assuming the blocks are not in contact otherwise will have a common acceleration) you get almost always that response. The reason is that if you calculate the forces along an axis parallel to the plane you get, after simplifying:
$a=\sin(\theta)-\mu \cos(\theta)$, 
Thus, regardless of angle (assuming $\theta \in (0,90))$, the larger $\mu$ the smaller $a$, independently of the individual masses that were cancelled when simplifying the equation.
The above argument assumes that the masses are moving and $a$ is positive (downwards), if $\mu$ makes $a$ to becomes negative then the object will not move and $a$ will be zero.
