91Ó°ÊÓ

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 $T\left(t\right)=\left(36.0\text{N/s}\right)t$.

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.

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.

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

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-.

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

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 $3.8\text{m/s}$.

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-.

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

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 $-24.6\text{m/s}$.

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-.

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

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.

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

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

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

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

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-

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

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.