A constant horizontal force ${\overrightarrow{\mathit{F}}}_{\mathit{a}}$pushes a $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{kg}$FedEx package across a frictionless floor on which an xycoordinate system has been drawn. The figure gives the package’s xand yvelocity components versus time t. (a) What is the magnitude and (b) What is the direction of localid="1657016170500" $\stackrel{\mathbf{\rightharpoonup }}{{\mathit{F}}_{\mathit{a}}}$?

Graphs of velocity components versus time, that interpret the acceleration along X and Y-axis. That graph has slop 3 and 5 respectively. Therefore, we can say${\mathrm{a}}_{\mathrm{x}}=3\mathrm{m}/{\mathrm{s}}^2$ and ${\mathrm{a}}_{\mathrm{y}}=-5.0\mathrm{m}/{\mathrm{s}}^2$

Newton’s second law states that the net force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$ on a body with mass mis related to the body’s acceleration $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$by,

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}\mathbf{}}\mathbf{=}\mathit{m}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$

By using the graphs, the components of acceleration can be found, and using which the magnitude of acceleration can be found. The angle of acceleration with the axis can be determined by the components of acceleration.

Formulae:

$$\begin{gathered} a=\sqrt{a_x^2+a_y^2} \\ \theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right) \end{gathered}$$

$$\begin{gathered} a=\sqrt{a_x^2}+a_y^2 \\ =\sqrt{{(3.0)}^2+{(-5.0)}^2} \end{gathered}$$

$=5.83\mathrm{m}/{\mathrm{s}}^2$

$$\begin{gathered} F=ma \\ =2.0\mathrm{kg}\times 5.83\mathrm{m}/{\mathrm{s}}^2 \\ =11.66\mathrm{N} \end{gathered}$$

$\mathrm{Hence},\mathrm{the}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{force}\mathrm{is}11.66\mathrm{N}.$

The direction of force is same as direction of acceleration.

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right) \\ ={\tan }^{-1}\left(-\frac{5.0}{3.0}\right) \\ =-59.0° \end{gathered}$$

Force makes an angle$-59.0°$ with the positive x axis