91Ó°ÊÓ

Position of particle of mass 0.340 kg in x-y plane:

$$\begin{gathered} x\left(t\right)=-15.00+2.00t-4.00t^3 \\ y\left(t\right)=25.00+7.00t-9.00t^2 \end{gathered}$$

Where, x and y in m and t in s

The acceleration is rate of change of velocity with respect to time. The force is equal to mass times the acceleration. The force ${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{Net}}$ can be written as the product of mass m and acceleration $\stackrel{\mathbf{→}}{\mathbf{a}}$.

${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{Net}}\mathbf{=}\mathbf{m}\stackrel{\mathbf{→}}{\mathbf{a}}$

Using the formula for Newton’s second law, we can find the magnitude and angle of the net force relative to the positive direction of the x-axis on the particle at t = 0.700 s

Formulae:

Acceleration of a body along with x and y directions,$\stackrel{→}{a_x}=\left(\frac{d^{2}x}{d{t}^2}\right)\&\stackrel{→}{a_y}=\left(\frac{d^{2}y}{d{t}^2}\right)$ (i)

The net force on a body according to Newton’s second law,${\stackrel{→}{F}}_{Net}=M\stackrel{→}{a}$ (ii)

Where, ${\stackrel{→}{F}}_{Net}$is net force, Mis mass of the object, and $\stackrel{→}{a}$is the acceleration.

Using equation (i) and the position expression, we get the x-component of acceleration as:

$$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=-15.00+2.00\mathrm{t}-4.00{\mathrm{t}}^3 \\ \stackrel{→}{{\mathrm{a}}_{\mathrm{x}}}=\left(\frac{{\mathrm{d}}^2\left[-15.00+2.00\mathrm{t}-4.00{\mathrm{t}}^2\right]}{{\mathrm{dt}}^2}\right) \\ =(-24.0\mathrm{t})\hat{\mathrm{i}} \end{gathered}$$

Now, the y-component of acceleration using equation (i) and the position expression is given as:

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)=25.00+7.00\mathrm{t}-9.00{\mathrm{t}}^2 \\ \stackrel{→}{{\mathrm{a}}_{\mathrm{x}}}=\left(\frac{{\mathrm{d}}^2\left[25.00+7.00\mathrm{t}-9.00{\mathrm{t}}^2\right]}{{\mathrm{dt}}^2}\right) \\ =(-18.\mathrm{m}/{\mathrm{s}}^2)\hat{\mathrm{j}} \end{gathered}$$

Hence, the acceleration in vector notation is given as:

$$\begin{gathered} \stackrel{→}{a}=\stackrel{→}{a_x}+\stackrel{→}{a_y} \\ =(-24.0t)\hat{i}+(18.0)\hat{j} \\ =(-16.8m/s^2)\hat{i}+(-18.0m/s^2)\hat{j}(fort=0.700s) \end{gathered}$$

Again, the magnitude of acceleration is given as:

$$\begin{gathered} \mathrm{a}=\left|\stackrel{→}{\mathrm{a}}\right| \\ =\sqrt{{(-16.8)}^2+{(-18.0)}^2} \\ =24.62\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the magnitude of force is:

$$\begin{gathered} \mathrm{F}=\mathrm{Ma} \\ =(0.34\mathrm{kg})(24.62\mathrm{m}/{\mathrm{s}}^2) \\ =8.37\mathrm{N} \end{gathered}$$

Hence, the value of force is 8.37 N

We have, the angle of force given as:

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{{\mathrm{a}}_{\mathrm{y}}}{{\mathrm{a}}_{\mathrm{x}}}\right) \\ ={\tan }^{-1}\left(\frac{-18.0\mathrm{m}/{\mathrm{s}}^2}{-16.8\mathrm{m}/{\mathrm{s}}^2}\right) \\ ={47}^{°}\mathrm{or}-{133}^{°} \\ \mathrm{θ}=-{133}^{°}\mathrm{Because},\stackrel{→}{\mathrm{F}}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant} \end{gathered}$$

Hence, the value of angle is $-{133}^{°}$

We have velocity, using the position expression as:

$$\begin{gathered} \stackrel{→}{\mathrm{v}}=\left({\mathrm{v}}_{\mathrm{x}}\right)\hat{\mathrm{i}}+\left({\mathrm{v}}_{\mathrm{x}}\right)\hat{\mathrm{j}} \\ =\frac{\mathrm{dx}}{\mathrm{dt}}\hat{\mathrm{i}}+\frac{\mathrm{dy}}{\mathrm{dt}}\hat{\mathrm{j}} \\ =(2.00-12.0{\mathrm{t}}^2)\hat{\mathrm{i}}+(7.00-18\mathrm{t})\hat{\mathrm{j}} \\ =(-3.88\mathrm{m}/\mathrm{s})\hat{\mathrm{i}}+(-5.60\mathrm{m}/\mathrm{s})\hat{\mathrm{j}}(\mathrm{for}\mathrm{t}=0.700\mathrm{s}) \end{gathered}$$

Angle of particle’s direction is given as:

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{-5.60\mathrm{m}/\mathrm{s}}{-3.80\mathrm{m}/\mathrm{s}}\right) \\ =55.3^{°}\mathrm{or}-{125}^{°} \\ =-{125}^{°}\mathrm{because},\stackrel{→}{\mathrm{v}}\mathrm{is}\mathrm{in}\mathrm{third}\mathrm{quadrant} \end{gathered}$$

Hence, the value of the angle is $-{125}^{°}$.