Friction — Physics STD 11 Science — Question

$\text{R}_1+\text{ma}-\text{mg}=0$

$\Rightarrow\text{R}_1=\text{m(g}-\text{a) = mg}-\text{ma} \ ...(\text{i})$

$\text{T}-\mu\text{R}_1=0\Rightarrow\text{T = m(mg}-\text{ma}) \ ...(\text{ii})$

Again, $\text{F}-\text{T}-\mu\text{R}_1=0$

$\Rightarrow\text{F}-\{\mu(\text{mg}-\text{ma})\}-\text{u}(\text{mg}-\text{ma})=0$

$\Rightarrow\text{F}-\mu\text{mg}+\mu\text{ma}-\mu\text{mg}+\mu\text{ma}=0$

$\Rightarrow\text{F}=2\mu\text{mg}-2\mu\text{ma}\Rightarrow\text{F}=2\mu\text{m(g}-\text{a})$

2. Acceleration of the block be a₁

$\text{R}_1=\text{mg}-\text{ma} \ ...(\text{i})$

$2\text{F}-\text{T}-\mu\text{R}_1-\text{ma}_1=0$

$\Rightarrow2\text{F}-\text{t}-\mu\text{mg}+\mu\text{a}-\text{ma}_1=0 \ ...(\text{ii})$

$\text{T}-\mu\text{R}_1-\text{Ma}_1=0$

$\Rightarrow\text{T}=\mu\text{R}_1+\text{Ma}_1$

$\Rightarrow\text{T}=\mu(\text{mg}-\text{ma})+\text{Ma}_1$

$\Rightarrow\text{T}=\mu\text{mg}-\mu\text{ma + Ma}_1$

Subtracting values of F & T, we get

$2(2\mu\text{m(g}-\text{a}))-2(\mu\text{mg}-\mu\text{ma}+\text{Ma}_1)\\-\mu\text{mg}+\mu\text{ma}-\mu\text{a}_1=0$

$\Rightarrow4\mu\text{mg}-4\mu\text{ma}-2\mu\text{mg}+2\mu\text{ma = ma}_1+\text{Ma}_1$

$\Rightarrow\text{a}_1=\frac{2\mu\text{m(g}-\text{a})}{\text{M + m}}$

Both blocks move with this acceleration but in opposite directions.