Physics - NLM Question with Solution | TestHub

The free body diagrams are :



${\mathrm{F}}_1$ is force of friction between blocks

${\mathrm{F}}_2$ is force of friction between block and ground

Making FBD of pulley



Since pulley is light

$\mathrm{T}=\frac{\mathrm{F}}{2}=3\mathrm{mg}$

Let blocks move together with common acceleration a

$a=\frac{6mg-F_2}{3m}=\frac{6mg-3mg}{3m}=g$

then for upper block

$T-F_1=ma\Rightarrow 3mg-F_1=mg\Rightarrow F_1=2mg$

but $F_1$ should be less than $f_{\max }=\mu mg$

So, our assumption of no relative motion b/w blocks is incorrect that means there is relative motion

$\Rightarrow F_1$ is kinetic $=\mu mg$

$a_m=\frac{3mg-\mu mg}{m}2g\left(\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{ }\mathrm{s}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\right)$

$a_{2m}=\frac{3mg+\mu mg-3\mu mg}{2m}=\frac{g}{2}\left(\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{ }\mathrm{s}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\right)$

acceleration of pulley = accelerated of hand

$a_p=\frac{a_m+a_{2m}}{2}=\frac{2g+\frac{g}{2}}{2}=\frac{5g}{4}\left(\mathrm{t}\mathrm{o}\mathrm{w}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{ }\mathrm{s}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\right)$