For the given system,what is the contact force (in $N$) acting on the $4 \, kg$ block?

$A$ string tied to a roof can bear a maximum tension of $50 \,kg$ wt. The minimum acceleration with which a man of $98 \,kg$ can descend is ...... $m/s^2$ [Take $g = 9.8 \,m/s^2$].

$A$ cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$. Work done by the cord on the block is

$A$ body of mass $8\,kg$ is hanging from another body of mass $12\,kg$. The combination is being pulled upward by a string with an acceleration of $2.2\,m/s^2$. The tensions $T_1$ and $T_2$ will be respectively:

Three identical particles of mass $m$ are joined together by a thread as shown in the figure. All three particles are moving in a horizontal plane in a circle about the point $O$. If the velocity of the outermost particle is $v_0$,then the ratio of tensions in the three sections of the string is:

$A$ 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 velocity of $M$ is: