91影视

Masses of the blocks are${\mathrm{m}}_1=2\mathrm{kg},{\mathrm{m}}_2=5\mathrm{kg},{\mathrm{m}}_3=3\mathrm{kg}\mathrm{and}{\mathrm{m}}_4=10\mathrm{kg}$

The problem is based on Newton鈥檚 second law of motion. The law states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object.

Formula:

${\stackrel{\mathbf{鈬赌}}{\mathbf{F}}}_{net}=\mathbf{m}\stackrel{\mathbf{鈫�}}{\mathbf{a}}$

The total mass accelerated to the right by Force $\stackrel{鈬赌}{\mathrm{F}}$is

${\mathrm{M}}_{\mathrm{s}}={\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4$

$$\begin{gathered} {\mathrm{M}}_{\mathrm{s}}=2\mathrm{kg}+5\mathrm{kg}+3\mathrm{kg}+10\mathrm{kg} \\ =20\mathrm{kg} \end{gathered}$$

The total mass accelerated to the right byCord 3 is

${\mathrm{M}}_{3\mathrm{c}}={\mathrm{m}}_2+{\mathrm{m}}_3+{\mathrm{m}}_4$

$$\begin{gathered} {\mathrm{M}}_{3\mathrm{c}}=5\mathrm{kg}+3\mathrm{kg}+10\mathrm{kg} \\ =18\mathrm{kg} \end{gathered}$$

The total mass accelerated to the right byCord 1 is

$$\begin{gathered} {\mathrm{M}}_{1\mathrm{c}}={\mathrm{M}}_4 \\ {\mathrm{M}}_{1\mathrm{c}}=10\mathrm{kg} \end{gathered}$$

As all the masses lie in one system, the acceleration of all the blocks is the same.

Therefore, the rank of the blocks according to their acceleration, greatest first, is a tie.

To find the tension, we can multiply the acceleration by the mass on the right-hand side.

Since for cord 3, mass is maximum and acceleration is the same, we can conclude that tension is maximum in cord 3. Similarly, the tension in cord 1 would be minimum.

The rank of the cords according to their tension is$\mathrm{cord}3,\mathrm{cord}2,\mathrm{cord}1$