Physics - Center of Mass Question with Solution | TestHub

When a length $x$ of the chain has fallen, its velocity is $v=\sqrt{2gx}$. The mass of the chain segment that has fallen and accumulated on the table is $m_x=(M/L)x$. The force exerted by the chain on the table has two components: the weight of the accumulated chain, $F_w=m_{x}g=(M/L)xg$, and the force due to the momentum change of the falling chain segment. The rate at which mass hits the table is $dm/dt=(M/L)v$. The force due to this impact is $F_m=v(dm/dt)=v(M/L)v=(M/L)v^2$. Substituting $v^2=2gx$, we get $F_m=(M/L)(2gx)=2Mgx/L$. The total force on the table is $F=F_w+F_m=(M/L)xg+2Mgx/L=3Mgx/L$.