An elevator of mass $500 \,kg$ is ascending upwards with a constant acceleration $a=2 \,m/s^2$. What is the work done by the tension in the elevator cable during its climb by $12 \,m$ (in $\,kJ$)? (Take $g=10 \,m/s^2$)

$A$ block of mass $1 \, kg$ is suspended by a string of mass $1 \, kg$ and length $1 \, m$ as shown in the figure. Taking $g = 10 \, m/s^2$,calculate the force exerted by the support on the string. (in $, N$)

$A$ block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $m$. If a force $P$ is applied at the free end of the rope,the force exerted by the rope on the block will be

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

Three blocks of different masses connected with inextensible strings are pulled by a force $F$ on a frictionless surface as shown in the figure. The ratio of tensions $T_1$ to $T_2$ is

Two particles of equal mass $m$ are connected to a rope $AB$ of negligible mass such that one is at end $A$ and the other divides the length of the rope in the ratio $1:2$ from $B$. The rope is rotated about end $B$ in a horizontal plane. The ratio of the tension in the smaller part (between $B$ and the middle particle) to the tension in the larger part (between the two particles) is (ignore the effect of gravity).