Only two horizontal forces act on a 3.00 kgbody that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting $\mathbf{62}\mathbf{°}$north of west. What is the magnitude of the body’s acceleration?

1. The mass of the body,m=3.0kg
2. Two forces: 9.0N acting due east and 8.0 N acting $62°$north of west

The net force on the body is the sum of all the forces acting on the body. The net force ${\overrightarrow{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}$on a body with a mass m related to the body’s acceleration $\overrightarrow{\mathbf{a}}$by,

${\overrightarrow{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{a}}$

Using the formula for force, we can find the magnitude of the acceleration of the body.

Formula:

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

role="math" localid="1657011060953" $$\begin{gathered} {\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2 \\ =\mathrm{M}\overrightarrow{\mathrm{a}} \end{gathered}$$ (i)

For force${\overrightarrow{F}}_1=9N\mathrm{and}\mathrm{\theta }=0°\left(\mathrm{considering}\mathrm{the}\mathrm{angle}\mathrm{of}\mathrm{force}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{east}\right)$

For x-direction,

$$\begin{gathered} {\overrightarrow{F}}_{1x}=9\mathrm{N}.\cos \left(0\right) \\ =9\mathrm{N} \end{gathered}$$

For y-direction,

$$\begin{gathered} {\overrightarrow{F}}_{1y}=9N.\sin \left(0\right) \\ =0 \end{gathered}$$

For force${\overrightarrow{F}}_2=9\mathrm{N}$

The angle can be calculated as,

$$\begin{gathered} \theta =180°-62° \\ =118°\left(\mathrm{considering}\mathrm{the}\mathrm{angle}\mathrm{of}\mathrm{force}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{east}\right) \end{gathered}$$

For x-direction,

$$\begin{gathered} {\overrightarrow{F}}_{2_x}=8\mathrm{Ncos}\left(118°\right) \\ =-3.75\mathrm{N} \end{gathered}$$

For y-direction,

$$\begin{gathered} {\overrightarrow{F}}_{2y}=8N\sin 118 \\ =7.06\mathrm{N} \end{gathered}$$

The total force in x-direction is given as:

$$\begin{gathered} {\overrightarrow{F}}_x=\overrightarrow{F_{1x}}+{\overrightarrow{F}}_{2x} \\ =9\mathrm{N}-3.75\mathrm{N} \\ =5.25\mathrm{N} \end{gathered}$$

The total force in y-direction is given as:

$$\begin{gathered} {\overrightarrow{F}}_x=\overrightarrow{F_{1y}}+{\overrightarrow{F}}_{2y} \\ =0+7.06\mathrm{N} \\ =7.06\mathrm{N} \end{gathered}$$

The magnitude of the net force is given as:

$$\begin{gathered} F=\sqrt{{\left(5.25\mathrm{N}\right)}^2+{\left(7.06\mathrm{N}\right)}^2} \\ =8.79\mathrm{N} \end{gathered}$$

Therefore, the magnitude value of acceleration is given as:

$a=\frac{F}{m}$

Substitute the 8.79 N for F, and 3 kg for m in the above equation.

role="math" localid="1657015232527" $$\begin{gathered} =\frac{8.79}{3\mathrm{kg}} \\ =2.9\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of body’s acceleration is $2.9\mathrm{m}/{\mathrm{s}}^2$.