The mass of the lift is $200 \ kg$. When it ascends with an acceleration of $4 \ m/s^2$,the tension in the cable supporting the lift will be [Given: Acceleration due to gravity $g = 10 \ m/s^2$]. (in $N$)

What is the minimum acceleration with which a person can slide down a rope,if the breaking strength of the rope is $\frac{2}{3}$ of the person's weight?

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 $g=10 \,m/s^2$)

$A$ rope of length $L$ and mass $M$ is pulled by a constant force $F$ on a frictionless surface. What is the tension in the rope at a distance $x$ from the end where the force is applied?

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

Three blocks of masses $m_1, m_2$ and $m_3$ are connected by massless strings as shown on a frictionless table. They are pulled with a force $F = 40 \, N$. If $m_1 = 10 \, kg, m_2 = 6 \, kg$ and $m_3 = 4 \, kg$,the tension $T_2$ (between $m_1$ and $m_2$) will be ........ $N$.