What is the minimum acceleration with which a person can slide down a rope,if the breaking strength of the rope is $\frac{2}{3}$ of the person's weight?

Three blocks of masses $m$,$2m$,and $3m$ are pushed with a force $F$ across a frictionless table as shown in the figure. Let $N_{1}$ be the contact force between the left two blocks ($m$ and $2m$) and $N_{2}$ be the contact force between the right two blocks ($2m$ and $3m$). Then:

Two blocks of masses $5\, kg$ and $4\, kg$ are in contact with each other on a frictionless table. When a horizontal force of $9\, N$ is applied to the block of mass $5\, kg$,then the value of the force of contact between the two blocks is ........ $N$.

$A$ uniform rope of mass $1.0\, kg$ is connected to a box of mass $2.0\, kg$,which is placed on a smooth horizontal surface. The free end of the rope is pulled horizontally by a force of $6\, N$. Find the tension at the midpoint of the rope in $N$.

Two particles each of mass $m$ are moving in horizontal circles with the same angular speed $\omega$. If both strings are of the same length $l$,find the ratio of tension in the strings $\frac{T_1}{T_2}$.

Four blocks are connected as shown in the figure on a horizontal frictionless surface. If $m_1 = m_2 = m_3 = m_4 = m$,then the ratio $T_3 / T_4$ will be: