If the$\mathbf{1}\mathbf{}\mathbf{kg}$standard body has an acceleration of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$at to the positive direction of an xaxis, (a)what is the xcomponent and (b) what is the ycomponent of the net force acting on the body, and (c)what is the net force in unit-vector notation?

• Mass of the body, $\mathrm{m}=1.0\mathrm{kg}$.
• Acceleration of the body, $\stackrel{\rightharpoonup }{a}=2.0m/s^2$
• Angle of acceleration, $\theta =20°$.

Newton’s second law states that the net force on the body is the vector sum of all the forces acting on the body. The product of mass m and acceleration $\overrightarrow{\mathbf{a}}$ of the body is equal to the net force ${\overrightarrow{\mathit{F}}}_{Net}$ on the body. The relation is given by,

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

Using the formula for Newton’s second law, we can find the x and y components of the net force and net force in unit-vector notation.

Formulae:

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

$$\begin{gathered} {\overrightarrow{F}}_{net}=F_1+F_2 \\ =M\overrightarrow{a} \end{gathered}$$ (i)

From equation (i), we get the x-component of force. Substitute the values of mass, and x component of acceleration in equation (i).

$\begin{array}{l}\overrightarrow{F}=Macos\theta \\ =(1.00\mathrm{kg})(2.00{\mathrm{m}/\mathrm{s}}^2)cos\left({20}^0\right)\\ =1.88\mathrm{N}\end{array}$

Hence, the value of the x-component is $1.88\mathrm{N}$.

From equation (i), we get the y-component of force. Substitute the values of mass, and y component of acceleration in equation (i).

Hence, the value of y-component is $0.684\mathrm{N}$.

The net force is equal to the sum of its x and y components. Therefore, from part (a) and (b), we can write as,

$\begin{array}{lll}\overrightarrow{F}=F_x\hat{i}+F_y\hat{j}& & \\ =(1.88\mathrm{N})\hat{\mathrm{i}}+(0.684\mathrm{N})\hat{\mathrm{j}}& & \end{array}$

Hence, the force is $\left(1.88\mathrm{N}\right)\hat{i}+\left(0.684\mathrm{N}\right)\hat{j}$.