A 2.00-kg box is moving to the right with speed on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude $\mathit{F}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\left(6.00N/s^2\right){\mathit{t}}^{\mathbf{2}}$. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?

The given data can listed below as,

• Mass of the box is, 2.00 kg.
• Velocity of the box is, 9.00 m/s.
• Magnitude of the force is, $F\left(t\right)=\left(6.00\mathrm{N}/{\mathrm{s}}^2\right)t^2$.

Newton’s second law of motion describe the relationship between the motion and the force, with the help of this we can also find out the acceleration, velocity and position of an object.

The expression of force can be expressed as,

$\begin{array}{l}F=ma\\ a=\frac{F}{m}\end{array}$

Here, m is the mass of the box, a is the acceleration, and F is the horizontal force applied to the box.

Substitute $(600N/s^2)t^2$ for F, and 2.00 kg for m in the above equation.

$$\begin{gathered} \mathrm{a}=\frac{-(6.00\mathrm{N}/{\mathrm{s}}^2){\mathrm{t}}^2}{2.00\hspace{0.33em}\mathrm{kg}} \\ =-3\hspace{0.33em}{\mathrm{t}}^2\frac{\mathrm{N}/{\mathrm{s}}^2}{\mathrm{kg}} \end{gathered}$$

The expression for the acceleration is expressed as,

$\begin{array}{l}a=\frac{dv}{dt}\\ dv=adt\end{array}$

Substitute$-3t^2\frac{N/s^2}{\mathrm{kg}}$ for and integrate the above equation.

$$\begin{gathered} v=\int (-3\hspace{0.33em}{t}^2\frac{N/s^2}{kg})\hspace{0.33em}dt \\ v=-t^3\frac{N/s^2}{kg}+c \end{gathered}$$

Here, v is the velocity, t is the time and c is the constant.

Substitute 9.00 m/s for v and 0s for t in the above equation.

$$\begin{gathered} (9.00\hspace{0.33em}\mathrm{m}/\mathrm{s})=-(0\hspace{0.33em}\mathrm{s}{)}^3+\mathrm{c} \\ \mathrm{c}=9.00\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

The expression for the speed of the box is expressed as,

$\mathrm{v}=-{\mathrm{t}}^3\frac{\mathrm{N}/{\mathrm{s}}^2}{\mathrm{kg}}+9.00\mathrm{m}/\mathrm{s}$

When speed is zero

Substitute 0 m/s for v in the above equation.

$$\begin{gathered} 0=-{\mathrm{t}}^3\frac{\mathrm{N}/{\mathrm{s}}^2}{\mathrm{kg}}+9.00\hspace{0.33em}\mathrm{m}/\mathrm{s} \\ \mathrm{t}=2.08\hspace{0.33em}\mathrm{s} \end{gathered}$$

The expression of the velocity can be expressed as,

$\begin{array}{ll}v=\frac{dx}{dt}& \\ dx=vdt& \end{array}$

Integrate the above equation both side and can be expressed as,

$$\begin{gathered} x={\int }_0^{2.08} -t^3\frac{\mathrm{N}/{\mathrm{s}}^2}{\mathrm{kg}}+9.00\mathrm{m}/\mathrm{s} \\ =\frac{-{\mathrm{t}}^4}{4}\frac{\mathrm{N}/{\mathrm{s}}^2}{\mathrm{kg}}+{\overline{)(9.00\mathrm{m}/\mathrm{s})\mathrm{t}}}_0^{2.08} \\ =\frac{-(0.28\mathrm{s}{)}^4}{4}\frac{\mathrm{N}/{\mathrm{s}}^2}{\mathrm{kg}}+9.00\left(2.08\mathrm{s}\right) \\ =14.04\mathrm{m} \end{gathered}$$

Hence, the distance moved by object is, 14.04 m.

The expression for the speed is expressed as,

$v=-t^3+9.00\hspace{0.33em}m/s$

Substitute 3.00 s for in the above equation.

$$\begin{gathered} v=-(3.00\hspace{0.33em}s{)}^3+9.00\hspace{0.33em}m/s \\ =-18\hspace{0.33em}m/s \end{gathered}$$

Hence, the required speed is -18 m/s.