Question

In the shown figure the magnitude of acceleration of centre of mass of the system is $(g=10m{s}^{-2})$:

• A. $4m{s}^{-2}$
• B. $10m{s}^{-2}$
• C. $2\sqrt{2}m{s}^{-2}$
• D. $5m{s}^{-2}$

Answer 1

The correct option is C $2\sqrt{2}m{s}^{-2}$

Fraction on horizontal moving block

$fr=\mu ms=0.2\times 5\times 10$

$=10N$

$F.B.D$ of vertical block

Net force $=mass\times a$

$\Rightarrow sg-T=sa$

$8sa-T=sa---\left(1\right)$

$F.B.D$ of horizontal block

$T-fr=sa$

$8T-10=sa---\left(2\right)$

adding $\left(1\right)$ & $\left(2\right)$

$a=4m/s^2$

as one mass is moving only horizontal

$\Rightarrow a_x=\frac{ma+0}{m+m}=\frac{a}{2}=2m/s^2$

one mass is moving only in verticaly

so $ay=\frac{mx0+ma}{m+m}=\frac{a}{2}=2m/s^2$

Net $=\sqrt{a{x}^2+9y^2}=2\sqrt{2}m/s^2$