Question

A 0.340 Kg particle moves in an xyplane according to$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{15}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{t}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{00}{\mathit{t}}^{\mathbf{3}}$and$\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{00}\mathit{t}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{00}{\mathit{t}}^{\mathbf{2}}$ , with xand yin meters and tin seconds. Att = 0.700 s, (a) what is the magnitude and (b) What is the angle (relative to the positive direction of the xaxis) of the net force on the particle, and (c) what is the angle of the particle’s direction of travel?

Short answer

a) The magnitude of the net force on the particle at t = 0.700 s is 8.37 N .

b) The angle of the net force relative to the positive direction of x axis on the particle is $-{133}^{°}$.

c) The angle of the particle’s direction of travel is $-{125}^{°}$.

Step-by-step solution

The given data

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

Understanding the concept of Newton’s laws of motion

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

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

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,$\overrightarrow{a_x}=\left(\frac{d^{2}x}{d{t}^2}\right)\&\overrightarrow{a_y}=\left(\frac{d^{2}y}{d{t}^2}\right)$ (i)

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

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

Calculations for the magnitude of the net force on the particle at t = 0.700 s

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

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

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

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

Hence, the acceleration in vector notation is given as:

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

Again, the magnitude of acceleration is given as:

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

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

b) Calculations for angle of the net force relative to the positive direction of x axis on the particle

We have, the angle of force given as:

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

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

c) Calculations for angle of the particle’s direction of travel

We have velocity, using the position expression as:

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

Angle of particle’s direction is given as:

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

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