Question #213896

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.


1
Expert's answer
2021-07-05T17:59:31-0400

Given,

Mass of the blocks are MM and mm ,

Side length of the cube =a= a

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

Initial velocity of the bottom wedge =u=u



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

Normal reaction on the upper block =Mg=Mg

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

fsF=0\Rightarrow f_s -F = 0

Now, substituting the values,

μ(m+M)g=F\Rightarrow \mu(m+M)g=F

Let the acceleration of the upper block be A

So, Ffs1=MAF-f_{s1}=MA

Now, substituting the values,

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


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

Hence,

v2=u22asv^2=u^2-2as

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


0=u22(μ(m+M)gμMg)Ms0=u^2-2\frac{(\mu(m+M)g-\mu Mg)}{M} s


Hence, s=u22μ(m+M)gμMgMs=\frac{u^2}{2\frac{\mu(m+M)g-\mu Mg}{M}}


s=u2M2(μ(m+M)gμMg)\Rightarrow s = \frac{u^2 M}{2(\mu (m+M)g-\mu Mg)}


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