Physics - W.E.P. Question with Solution | TestHub

For block A: $T-kx=ma$

For block B: $2mg-T=2ma$

$2mg-kx=3ma\Rightarrow a=\frac{2g}{3}-\frac{kx}{3m}$

$a_{\max }=\boxed{\frac{2g}{3}}$

Maximum speed of block A

$2mg-kx=0\Rightarrow x=\frac{2mg}{k}$

$2mgx=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2+\frac{1}{2}(2m)v^2$

Substitute $x=\frac{2mg}{k}$:

$2mg\left(\frac{2mg}{k}\right)=\frac{1}{2}k{\left(\frac{2mg}{k}\right)}^2+\frac{3}{2}m{v}^2$

$\frac{4m^2{g}^2}{k}=\frac{1}{2}k\frac{4m^2{g}^2}{k^2}+\frac{3}{2}m{v}^2$

$\frac{2m^2{g}^2}{k}=\frac{3}{2}m{v}^2$

$v^2=\frac{4m^2{g}^2}{3mk}=\frac{4m{g}^2}{3k}$

$v_{\max }=\boxed{2g\sqrt{\frac{m}{3k}}}$