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 $, N$)

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:

At $t = 0$, a body of mass $100 \text{ g}$ starts moving under the influence of a force $(5\hat{i} + 10\hat{j}) \text{ N}$. After $2 \text{ s}$, its position is $(2x\hat{i} + 5y\hat{j}) \text{ m}$. The ratio $x : y$ is . . . . . . .

$A$ smooth right-angled wedge of mass $M$ is kept on a smooth horizontal surface as shown. $A$ mass $m$ is released from the top of the wedge. When $m$ reaches the ground, its speed is $V$. Find the work done by the normal contact force on $m$ while it comes down to the ground.

$A$ rod of length $L$ and mass $M$ is acted on by two unequal forces $F_1$ and $F_2$ $(F_2 < F_1)$ as shown in the figure. The tension in the rod at a distance $y$ from the end $A$ is given by: