Answer to Question #213896 in Classical Mechanics for Chintha

Question

Two wooden cubes of side-length 'a' are placed one on another on a horizontal plane. 'M' and 'm' represent the masses of upper cube and lower cube respectively(M>m). The coefficient of dynamic friction between each surface is μ. The bottom wedge is given 'u' initial velocity and it is sufficient for that cube to slip away from the tower. Find the final horizontal displacement of the upper cube.

Answer 1

Given,

Mass of the blocks are $M$ and $m$ ,

Side length of the cube $= a$

Coefficient of dynamic friction between each surface $= \mu$

Initial velocity of the bottom wedge $=u$

Normal reaction on the lower block $=(m+M)g$

Normal reaction on the upper block $=Mg$

So, As the systems are moving with velocity u, it means net force on the blocks are zero.

$\Rightarrow f_s -F = 0$

Now, substituting the values,

$\Rightarrow \mu(m+M)g=F$

Let the acceleration of the upper block be A

So, $F-f_{s1}=MA$

Now, substituting the values,

$\Rightarrow \mu(m+M)g-\mu Mg= MA$

$\Rightarrow A = \frac{\mu(m+M)g-\mu Mg}{M}$

Hence,

$v^2=u^2-2as$

Final velocity of the upper block $(v)=0$

$0=u^2-2\frac{(\mu(m+M)g-\mu Mg)}{M} s$

Hence, $s=\frac{u^2}{2\frac{\mu(m+M)g-\mu Mg}{M}}$

$\Rightarrow s = \frac{u^2 M}{2(\mu (m+M)g-\mu Mg)}$