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 . . . . . . .

$A$ force $F = 10t$ acts on a block of mass $m = 10 \, kg$ as shown in the figure. The coefficient of friction is $\mu = 0.4$. Find the time $t$ when the block loses contact with the surface.

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

$A$ rope of length $10 \, m$ and linear density $0.5 \, kg/m$ is lying lengthwise on a smooth horizontal floor. It is pulled by a force of $25 \, N$. The tension in the rope at a point $8 \, m$ away from the point of application is ....... $N$

$A$ 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 ......... $N$.

For the given figure,find the value of $m$ (in $kg$) for which the blocks have an acceleration of $\frac{8}{13}g$. The block on the incline has a mass of $100 \, g = 0.1 \, kg$ and the hanging block on the left has a mass of $50 \, g = 0.05 \, kg$. The surface is frictionless $(\mu = 0)$.