For the given arrangement of blocks on a frictionless surface,the tension $T_2$ is:

$A$ block of mass $48 \ kg$ kept on a smooth horizontal surface is pulled by a rope of length $4 \ m$ by a horizontal force of $25 \ N$ applied to the other end. If the linear density of the rope is $0.5 \ kg \ m^{-1}$,the force acting on the block is (in $N$)

$A$ mass of $2 \, kg$ is suspended with a thread $AB$ as shown in the figure. $A$ thread $CD$ of the same type is attached to the other end of the $2 \, kg$ mass. The lower thread is pulled gradually,harder and harder in the downward direction,so as to apply force on $AB$. Which of the threads will break and why?

$A$ small block of mass $m$ is placed on a wedge of mass $M$ as shown,which is initially at rest. All the surfaces are frictionless. The spring attached to the other end of the wedge has a force constant $k$. If $a'$ is the acceleration of $m$ relative to the wedge as it starts coming down and $A$ is the acceleration acquired by the wedge as the block starts coming down,then the maximum retardation of $M$ is:

$A$ block of mass $18.5 \ kg$ kept on a smooth horizontal surface is pulled by a rope of $3 \ m$ length by a horizontal force of $40 \ N$ applied to the other end of the rope. If the linear density of the rope is $0.5 \ kg \ m^{-1}$ and initially the block is at rest,the time in which the block moves a distance of $9 \ m$ is: (in $s$)

An $80\, kg$ person is parachuting and is experiencing a downward acceleration of $2.8\, m/s^2$. The mass of the parachute is $5\, kg$. The upward force on the open parachute is ........... $N$ (Take $g = 9.8\, m/s^2$)