Question

While two forces act on it, a particle is to move at the constant velocity

$\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{4}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{)}\hat{\mathbf{j}}$.One of the forces is${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{-}\mathbf{6}\mathbf{}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}$. What is the other force?

Short answer

The other force is $(-2\mathrm{N})\stackrel{⏜}{i}+(6\mathrm{N})\hat{j}$.

Step-by-step solution

The given data

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

Understanding the concept of Newton’s law

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 ${\overrightarrow{\mathbf{F}}}_{\mathbf{Net}}$ is given by the product of mass m and acceleration $\overrightarrow{\mathbf{a}}$ as,

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

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,

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

Calculation of other force

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

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

role="math" localid="1657019045099" $$\begin{gathered} \overrightarrow{F_1}+\overrightarrow{F_2}=M\overrightarrow{a}(\because \overrightarrow{a}=0\mathrm{m}/{\mathrm{s}}^2) \\ \overrightarrow{F_1}+\overrightarrow{F_2}=0 \\ \overrightarrow{F_2}=-\overrightarrow{F_1} \\ =(-2\mathrm{N})\hat{i}+(6\hspace{0.17em}\mathrm{N})\hat{j} \end{gathered}$$

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