Question

A0.150 Kgparticle moves along an xaxis according to $\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{00}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{t}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{00}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}{\mathbf{t}}^{\mathbf{3}}$, with xin meters and tin seconds. In unit-vector notation, what is the net force acting on the particle at t= 3.40 s?

Short answer

The net force acting on the particle in the unit-vector notation at t = 3.40s is $\left(-7.98\mathrm{N}\right)\hat{\mathrm{i}}$.

Step-by-step solution

The given data

• Mass of the particle, m = 0.150 kg.
• Time at which we need to calculate the force, t = 3.40s .
• Position of the particle of mass 0.150 kg along x axis is$\mathrm{x}\left(\mathrm{t}\right)=-13.00+2.00\mathrm{t}+4.00{\mathrm{t}}^2-3.00{\mathrm{t}}^3$

Understanding the concept of Newton’s law of motion

The net force ${\overrightarrow{\mathbf{F}}}_{\mathbf{Net}}$on a body is a product of mass m and acceleration $\overrightarrow{\mathbf{a}}$of the body and is given by,

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

The acceleration can be differentiating the displacement vector twice with respect to time.

Using the formula for Newton’s second law, we can find the net force acting on the particle at t = 3.40 s

Formulae:

The net force acting on the body,

${\overrightarrow{F}}_{Net}=M\overrightarrow{a}$ (i)

Here, ${\overrightarrow{F}}_{net}$is the net force, M is mass of the object, and $\overrightarrow{a}$is the acceleration.

The acceleration of a body in motion, $\overrightarrow{a}=\left(\frac{d^{2}x}{d{t}^2}\right)$ (ii)

Calculation of net force

From equation (ii) and the given position equation,

$x\left(t\right)=-13.00+2.00t+4.00t^2-3.00t^3$

we get the acceleration of the particle from equation (ii) by substituting the displacement in the equation.

role="math" localid="1657015679628" $$\begin{gathered} \overrightarrow{\mathrm{a}}=\left(\frac{{\mathrm{d}}^2\left[-13.00+2.00\mathrm{t}+4.00{\mathrm{t}}^2-3.00{\mathrm{t}}^3\right]}{{\mathrm{dt}}^2}\right) \\ =8.00-18.0\mathrm{t} \\ =53.2\mathrm{m}/{\mathrm{s}}^2(\because \mathrm{t}=3.40\mathrm{s}) \end{gathered}$$

Now using equation (i) and the derived and given values, we get the net force as:

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{Net}}=(0.150\mathrm{kg})(8.00-18.0\mathrm{t}) \\ =(-7.98\mathrm{N})\hat{\mathrm{i}}(\because \mathrm{t}=3.40\mathrm{s}) \end{gathered}$$

Hence, the net force in unit-vector notation is $(-7.98\mathrm{N})\hat{\mathrm{i}}$.