Q 69P - University Physics with Modern Physics

Question

Question: A $\mathbf{3}.\mathbf{00}-\mathbf{kg}$ box that is several hundred meters above the earth’s surface is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope and results in a tension in the rope of $\mathbf{T}\left(\mathbf{t}\right)=\left(\mathbf{36}.\mathbf{0} \mathbf{N}/\mathbf{s}\right)\mathbf{t}$. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at (i) $\mathbf{t}=\mathbf{1}.\mathbf{00} \mathbf{s}$ and (ii) $\mathbf{t}=\mathbf{3}.\mathbf{00} \mathbf{s}$? (b) What is the maximum distance that the box descends below its initial position? (c) At what value of   does the box return to its initial position?

Step-by-Step Solution

Answer

(a) The velocity at $t=1.00 \text{s}$ is $3.8 \text{m/s}$ and the velocity at $t=3.00 \text{s}$ is $-24.6 \text{m/s}$.

(b) The maximum distance that the box descends below is $0.03 \text{m}$.

(c) The box returns to its initial position at $1.63 \text{s}$.

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of the box is $m=3.00 \text{kg}$.
• The equation of the tension in the rope is $T\left(t\right)=\left(36.0 \text{N/s}\right)t$.

Step 2: Significance of the tension

Tension is described as the force that acts on the body by virtue of its elastic nature. It always acts in an action-reaction pair.

Step 3: (a) Determination of the velocity at (i) t = 1.00s

The equation of the acceleration of the box can be obtained from the relation-

 $$\begin{gathered} ma\left(t\right)=mg-T\left(t\right) \\ a\left(t\right)=g-\frac{T\left(t\right)}{m} \end{gathered}$$

Here, $a\left(t\right)$ is the acceleration of the box at time t, g is the acceleration due to gravity, T(t) is the tension in the rope at time t, and m is the mass of the box.

 

Integrating the above equation with respect to the time t, the velocity of the box can be obtained.

$\begin{array}{rcl}V\left(t\right)& =& \int a\left(t\right)dt\\ & =& \int \left(g-\frac{T\left(t\right)}{m}\right)dt\\ & =& \int \left(g-\frac{\left(36.0 \text{N/s}\right)t}{m}\right)dt\\ & =& gt-\frac{\left(36.0 \text{N/s}\right)t^2}{2m} ························\left(1\right)\end{array}$

Here, V(t) is the velocity of the box with respect to the time  .

 

for  $g=9.8 {\text{m/s}}^{\text{2}}$,  $t=1.00 \text{s}$, $T=\left(36.0 \text{N/s}\right)\left(1.00 \text{s}\right)$ and $m=3.00 \text{kg}$; equation (1) becomes-.

 $\begin{array}{rcl}V\left(t\right)& =& \left(9.8 {\text{m/s}}^{\text{2}}\right)\left(1.00 \text{s}\right)-\frac{\left(36.0 \text{N/s}\right){\left(1.00 \text{s}\right)}^2}{2\times 3.00 \text{kg}}\\ & =& \left(9.8 \text{m/s}\right)-\frac{\left(36.0 \text{N/s}\times \frac{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}{\text{1} \text{N}}\right)\left(1.00 {\text{s}}^2\right)}{6.00 \text{kg}}\\ & =& \left(9.8 \text{m/s}\right)-\frac{\left(36.0 \text{kg}·{\text{m/s}}^3\right)\left(1.00 {\text{s}}^2\right)}{6.00 \text{kg}}\\ & =& \left(9.8 \text{m/s}\right)-\frac{\left(36.0 \text{kg}·\text{m/s}\right)}{6.00 \text{kg}}\end{array}$

On further solving:

 $\begin{array}{rcl}V\left(t\right)& =& \left(9.8 \text{m/s}\right)-\left(6.0 \text{m/s}\right)\\ & =& 3.8 \text{m/s}\end{array}$

Thus, the velocity at $t=1.00 \text{s}$ is $3.8 \text{m/s}$.

Step 4: (a) Determination of the velocity at (ii) t = 3.00s.

The equation (i) has been recalled below:

 $V\left(t\right)=gt-\frac{\left(36.0 \text{N/s}\right)t^2}{2m}$

for $g=9.8 {\text{m/s}}^{\text{2}}$, t = 3.00s, $T=\left(36.0 \text{N/s}\right)\left(3.00 \text{s}\right)$ and $m=3.00 \text{kg}$; equation (1) becomes-.

 $\begin{array}{rcl}V\left(t\right)& =& \left(9.8 {\text{m/s}}^{\text{2}}\right)\left(3.00 \text{s}\right)-\frac{\left(36.0 \text{N/s}\right){\left(3.00 \text{s}\right)}^2}{2\times 3.00 \text{kg}}\\ & =& \left(29.4 \text{m/s}\right)-\frac{\left(36.0 \text{N/s}\times \frac{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}{\text{1} \text{N}}\right)\left(9.00 {\text{s}}^2\right)}{6.00 \text{kg}}\\ & =& \left(29.4 \text{m/s}\right)-\frac{\left(36.0 \text{kg}·{\text{m/s}}^3\right)\left(9.00 {\text{s}}^2\right)}{6.00 \text{kg}}\\ & =& \left(29.4 \text{m/s}\right)-\frac{\left(324 \text{kg}·\text{m/s}\right)}{6.00 \text{kg}}\end{array}$

 

On further solving:

$\begin{array}{rcl}V\left(t\right)& =& \left(29.4 \text{m/s}\right)-\left(54 \text{m/s}\right)\\ & =& -24.6 \text{m/s}\end{array}$

Thus, the velocity at (i) $t=3.00 \text{s}$ is  $-24.6 \text{m/s}$.

Step 5: (b) Determination of the maximum distance

The equation (i) has been recalled below:

 $V\left(t\right)=gt-\frac{\left(36.0 \text{N/s}\right)t^2}{2m}$

At maximum distance, the velocity of the box should be zero.

 

for $g=9.8 {\text{m/s}}^{\text{2}}$,  V(t) = 0, $T=\left(36.0 \text{N/s}\right)\left(3.00 \text{s}\right)$ and $m=3.00 \text{kg}$; equation (1) becomes-.

 $$\begin{gathered} 0=\left(9.8 {\text{m/s}}^{\text{2}}\right)t-\frac{\left(\left(36.0 \text{N/s}\right){\left(t\right)}^2\right)}{2\left(3.00 \text{kg}\right)} \\ \left(9.8 {\text{m/s}}^{\text{2}}\right)t-\frac{\left(\left(36.0 \text{N/s}\times \frac{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}{\text{1} \text{N}}\right){\left(t\right)}^2\right)}{\left(6.00 \text{kg}\right)}=0 \\ \left(9.8 {\text{m/s}}^{\text{2}}\right)t-\left(6.0 {\text{m/s}}^3\right)\left(t^2\right)=0 \end{gathered}$$

On further solving

 $$\begin{gathered} \left(9.8 {\text{m/s}}^{\text{2}}\right)t=\left(6.0 {\text{m/s}}^3\right)\left(t^2\right) \\ t=\frac{\left(9.8 {\text{m/s}}^{\text{2}}\right)}{\left(6.0 {\text{m/s}}^3\right)} \\ t=1.63 \text{s} \end{gathered}$$

Integrating equation (i) with respect to the time t, the distance moved by the box can be obtained.

$\begin{array}{rcl}d\left(t\right)& =& \int V\left(t\right)dt\\ & =& \int \left(gt-\frac{\left(36.0 \text{N/s}\right)t^2}{2m}\right)dt\\ & =& g\frac{t^2}{2}-\frac{\left(36.0 \text{N/s}\right)t^3}{3\times 2m}\\ & =& \frac{g{t}^2}{2}-\frac{\left(36.0 \text{N/s}\right)t^3}{6m} ·····················\left(2\right)\end{array}$

for  $g=9.8 {\text{m/s}}^{\text{2}}$,  $t=1.63 \text{s}$, and $m=3.00 \text{kg}$; equation (1) becomes-

$\begin{array}{rcl}d\left(t\right)& =& \left(9.8 {\text{m/s}}^{\text{2}}\right)\frac{{\left(1.63 \text{s}\right)}^2}{2}-\frac{\left(36.0 \text{N/s}\right){\left(1.63 \text{s}\right)}^3}{4\left(3.00 \text{kg}\right)}\\ & =& \left(13.02 \text{m}\right)-\frac{\left(36.0 \text{N/s}\times \frac{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}{\text{1} \text{N}}\right){\left(1.63 \text{s}\right)}^3}{\left(12.00 \text{kg}\right)}\\ & =& \left(13.02 \text{m}\right)-\left(12.99 \text{m}\right)\\ & =& 0.03 \text{m}\end{array}$

Thus, the maximum distance that the box descends below is $0.03 \text{m}$.

Step 6: (c) Determination of the time t.

When the box returns to its initial position, then the distance covered by the box is zero.

for $g=9.8 {\text{m/s}}^{\text{2}}$, d(t) = 0, $T=\left(36.0 \text{N/s}\right)\left(t\right)$ and $m=3.00 \text{kg}$; equation (1) becomes-

$$\begin{gathered} 0=\left(9.8 {\text{m/s}}^{\text{2}}\right)\frac{{\left(t\right)}^2}{2}-\frac{\left(36.0 \text{N/s}\right){\left(t\right)}^3}{4\left(3.00 \text{kg}\right)} \\ \left(4.9 {\text{m/s}}^{\text{2}}\right)t^2-\frac{\left(36.0 \text{N}\times \frac{\text{1} \text{kg}·{\text{m/s}}^{\text{2}}}{\text{1} \text{N}}\right)}{\left(12.00 \text{kg}\right)}{t}^3=0 \\ \left(4.9 {\text{m/s}}^{\text{2}}\right)t^2-\frac{\left(36.0 \text{kg}·{\text{m/s}}^3\right)}{\left(12.00 \text{kg}\right)}{t}^3=0 \\ \left(4.9 {\text{m/s}}^{\text{2}}\right)t^2=\left(3 {\text{m/s}}^3\right)t^3 \\ t=1.63 \text{s} \end{gathered}$$

Hence, further as:

$$\begin{gathered} \left(4.9 {\text{m/s}}^{\text{2}}\right)t^2=\left(3 {\text{m/s}}^3\right)t^3 \\ t=1.63 \text{s} \end{gathered}$$

Thus, the box returns to its initial position at 1.63.