91Ó°ÊÓ

1. Velocity$\stackrel{→}{v}$of particle,$\stackrel{→}{v}=(3m/s)\hat{i}-(4\mathrm{m}/\mathrm{s})\hat{j}$.
2. One force, role="math" localid="1657017870844" ${\stackrel{→}{F}}_1=(2N)\hat{i}+(-6\mathrm{N})\hat{j}$.

Newton’s second law states that the net force acting on the body is equal to the mass of the body multiplied by the acceleration with which the body is moving. The net force ${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{Net}}$ is given by the product of mass m and acceleration $\stackrel{\mathbf{→}}{\mathbf{a}}$ as,

${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{Net}}\mathbf{=}\mathit{m}\stackrel{\mathbf{→}}{\mathbf{a}}$.

Modifying the formula for Newton’s second law, we can find the other force if mass and acceleration of the body and one force acting on it is known.

Formulae:

The net force on a particle according to Newton’s second law,

$$\begin{gathered} {\stackrel{→}{F}}_{net}={\stackrel{→}{F}}_1+{\stackrel{→}{F}}_2 \\ =M\stackrel{→}{a} \end{gathered}$$

As velocity$\stackrel{→}{v}=cons\tan t$so,$\stackrel{→}{a}=0m/s^2$,

Substitute the values of acceleration and mass in equation (i).

role="math" localid="1657019045099" $$\begin{gathered} \stackrel{→}{F_1}+\stackrel{→}{F_2}=M\stackrel{→}{a}(âˆ\mu \stackrel{→}{a}=0\mathrm{m}/{\mathrm{s}}^2) \\ \stackrel{→}{F_1}+\stackrel{→}{F_2}=0 \\ \stackrel{→}{F_2}=-\stackrel{→}{F_1} \\ =(-2\mathrm{N})\hat{i}+(6 \mathrm{N})\hat{j} \end{gathered}$$

Hence, the other force is$(-2\mathrm{N})\stackrel{Áåœ}{i}+(6\mathrm{N})\hat{j}$.