Question

A 2.00 - kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to$\mathbf{v}\left(t\right)\mathbf{=}\left(2.00m/s^2\right)\mathbf{t}\mathbf{+}\left(0.600m/s^3\right){\mathbf{t}}^{\mathbf{2}}$. What is the tension in the rope when the velocity of the box is ?

Short answer

The tension in the rope is 29.7 N .

Step-by-step solution

Identification of the given data

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 .

Significance of the tension

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.

Determination of the tension in the rope

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:

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

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.

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

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.

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

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 .