91Ó°ÊÓ

The given data can be listed below as:

• The mass of the box is m = 2.00 kg .
• The box’s velocity is v = 9.00 m/s .

The tension is referred to as the force which gets transmitted in the axial direction due to the involvement of the rope or the string. The tension is also a pair of the reaction and the action forces.

The free body diagram of the box has been drawn below:

In the above diagram, the weight w of the box is acting in the downwards direction and the tension T is acting in the upwards direction. The velocity with respect to the time is also directed upwards.

The equation of the velocity of the box is expressed as:

$\mathrm{v}\left(\mathrm{t}\right)=\left(2.00\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}+\left(0.600\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2$ …(¾±)

Here, v (t) is the velocity with respect to the time t.

Differentiating the above equation with respect to the time, the acceleration of the box can be obtained.

$$\begin{gathered} \mathrm{a}\left(\mathrm{t}\right)=\frac{\mathrm{d}\left(\mathrm{v}\left(\mathrm{t}\right)\right)}{\mathrm{dt}} \\ =\frac{\mathrm{d}\left(\left(2.00\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}+\left(0.600\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2\right)}{\mathrm{dt}} \\ =2.0\mathrm{m}/{\mathrm{s}}^2+\left(1.20\mathrm{m}/{\mathrm{s}}^3\right)\mathrm{t} \end{gathered}$$

Here, a(t) is the acceleration with respect to the time.

The equation of the tension of the box is expressed as:

localid="1667643809029" $$\begin{gathered} T-mg=m\left(a\left(t\right)\right) \\ T=m\left[g+a\left(t\right)\right] \end{gathered}$$

Here, T is the tension and is the acceleration due to gravity.m is the mass of the box.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{T}=\left(2.00\mathrm{kg}\right)\left[9.8\mathrm{m}/{\mathrm{s}}^2+2.00\mathrm{m}/{\mathrm{s}}^2+\left(1.20\mathrm{m}/{\mathrm{s}}^3\right)\mathrm{t}\right] \\ =\left(2.00\mathrm{kg}\right)\left[11.8\mathrm{m}/{\mathrm{s}}^2+\left(1.20\mathrm{m}/{\mathrm{s}}^3\right)\mathrm{t}\right] \end{gathered}$$ …(¾±¾±)

Substitute 9.00 m/s for v in the equation (i).

$$\begin{gathered} 9.00\mathrm{m}/\mathrm{s}=\left(2.00\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}+\left(0.600\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2 \\ \left(0.600\mathrm{m}/{\mathrm{s}}^3\right){\mathrm{t}}^2+\left(2.00\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{t}-9.00\mathrm{m}/\mathrm{s}=0 \end{gathered}$$

The above equation can be solved as a quadratic equation.

$$\begin{gathered} \mathrm{t}=\frac{-2.00\mathrm{m}/{\mathrm{s}}^2Â\pm \sqrt{{\left(2.00\mathrm{m}/{\mathrm{s}}^2\right)}^2+4\left(0.600\mathrm{m}/{\mathrm{s}}^3\right)9.00\mathrm{m}/\mathrm{s}}}{2\left(0.600\mathrm{m}/{\mathrm{s}}^3\right)} \\ =\frac{-2.00\mathrm{m}/{\mathrm{s}}^2Â\pm \sqrt{\left(25.6{\mathrm{m}}^2/{\mathrm{s}}^4\right)}}{\left(1.2\mathrm{m}/{\mathrm{s}}^3\right)} \\ =\frac{-2.00\mathrm{m}/{\mathrm{s}}^2Â\pm \left(5.05\mathrm{m}/{\mathrm{s}}^2\right)}{\left(1.2\mathrm{m}/{\mathrm{s}}^3\right)} \\ =2.54\mathrm{s} \end{gathered}$$

The negative value of the time is neglected as the time cannot be negative.

Substitute for in the equation (ii).

$$\begin{gathered} \mathrm{T}=\left(2.00\mathrm{kg}\right)\left[11.8\mathrm{m}/{\mathrm{s}}^2+\left(1.20\mathrm{m}/{\mathrm{s}}^3\right)\left(2.54\mathrm{s}\right)\right] \\ =\left(2.00\mathrm{kg}\right)\left[11.8\mathrm{m}/{\mathrm{s}}^2+\left(3.05\mathrm{m}/{\mathrm{s}}^2\right)\right] \\ =\left(2.00\mathrm{kg}\right)\left[14.85\mathrm{m}/{\mathrm{s}}^2\right] \\ =29.7\mathrm{N} \end{gathered}$$

Thus, the tension in the rope is 29.7 N .