The mass of a lift is $2000 \, kg$. When the tension in the supporting cable is $28000 \, N$,then its acceleration is:

$A$ 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 placed on a horizontal surface and it is tied with an inextensible string to a block of mass $m$,as shown in the figure. $A$ block of mass $m_0$ is also placed on $M$. If $\mu < \mu_{min}$ (the minimum friction required to keep the block $m$ stationary),then the downward acceleration of $m$ is:

$A$ rope of mass $5 \,kg$ is moving vertically in a vertical position with an upward force of $100 \,N$ acting at the upper end and a downward force of $70 \,N$ acting at the lower end. The tension at the midpoint of the rope is ......... $N$.

$A$ mass of $1 \ kg$ is suspended by means of a thread. The system is
$(i)$ lifted up with an acceleration of $4.9 \ ms^{-2}$
(ii) lowered with an acceleration of $4.9 \ ms^{-2}$.
The ratio of tension in the first and second case is

$A$ 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