If the 1 kg standard body is accelerated by only${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{j}}$and ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{j}}$, then what is${\overrightarrow{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$ (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive x-direction? What are the (d) magnitude and (e) angle of $\overrightarrow{\mathbf{a}}$?

1)${\overrightarrow{F}}_1=3\stackrel{⏜}{i}+4\stackrel{⏜}{j}$

2)$\overrightarrow{F_2}=-2\stackrel{⏜}{i}-6\stackrel{⏜}{j}$

3)$m=1\mathrm{kg}$

Newton’s law states the force acting on the body is the product of mass of the body and the acceleration with which body is moving.

Two forces are given in vector notations. We can use vector algebra to add these two forces and find the resultant and magnitude of them. Using Newton’s second law of motion, we can find the magnitude and acceleration of the object.

Formula

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{a}$

Here,${\mathbf{F}}_{\mathbf{net}}$ is the force,m is the mass of the object, and a is the acceleration of the object.

Add both the forces in unit vector notation as below:

$$\begin{gathered} {\overrightarrow{F}}_{net}=F_1+F_2 \\ =3\stackrel{⏜}{i}+4\stackrel{⏜}{j}-2\stackrel{⏜}{i}-6\stackrel{⏜}{j} \\ =1\stackrel{⏜}{i}-2\stackrel{⏜}{j} \end{gathered}$$

Therefore, the net force is$1\stackrel{⏜}{i}-2\stackrel{⏜}{j}$

From part (a), the magnitude of force is calculated as,

$$\begin{gathered} \left|F_{act}\right|=\sqrt{1^2+{\left(-2\right)}^2} \\ =2.24\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the force is 2.24 N.

Using the trigonometry, we can find the angle of force as

$$\begin{gathered} \tan \theta =\frac{y}{x} \\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \\ =-\frac{2}{1} \\ =-63.45° \end{gathered}$$

Therefore, the angle is $-63.45°$.

Use Newtons second law to calculate the acceleration from the net force.

$$\begin{gathered} F_{net}=m\times a \\ a=\frac{F_{net}}{m} \\ =\frac{2.24\mathrm{N}}{1\mathrm{kg}} \\ =2.24\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration is $2.24\mathrm{m}/{\mathrm{s}}^2$.

Direction of acceleration is same as direction of net force that is$-63.45°$