Question

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" $\stackrel{\mathbf{\rightharpoonup }}{{\mathit{F}}_{\mathit{a}}}$?

Short answer

1. Magnitude of force$11.66\mathrm{N}$.
2. Direction of force$-59.0^{°}$. The negative angle implies that the force is applied below the horizontal axis.

Step-by-step solution

Given data

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 ${\mathrm{a}}_{\mathrm{y}}=-5.0\mathrm{m}/{\mathrm{s}}^2$

Understanding the concept of the net force

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:

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

(a) Calculate the magnitude of the force

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

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

$\mathrm{Now},$

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

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

(b) Calculate the direction of force

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