91Ó°ÊÓ

The given data can be listed below as:

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

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(t)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.

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

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

for $g=9.8 m/s^2,t=1.00 s,T=(36.0 N/s)(1.00 s)$and m=3.00 kg ; equation becomes-.

role="math" localid="1667645538886" $$\begin{gathered} V\left(t\right)=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.00\mathrm{s}\right)−\frac{(36.0\mathrm{N}/\mathrm{s})(1.00\mathrm{s}{)}^2}{2×3.00\mathrm{kg}} \\ =(9.8\mathrm{m}/\mathrm{s})−\frac{\left(36.0\mathrm{N}/\mathrm{s}×\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)\left(1.00{\mathrm{s}}^2\right)}{6.00\mathrm{kg}} \\ =(9.8\mathrm{m}/\mathrm{s})−\frac{\left(36.0\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^3\right)\left(1.00{\mathrm{s}}^2\right)}{6.00\mathrm{kg}} \\ =(9.8\mathrm{m}/\mathrm{s})−\frac{(36.0\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s})}{6.00\mathrm{kg}} \end{gathered}$$

On further solving:

$$\begin{gathered} V\left(t\right)=\left(9.8\mathrm{m}/\mathrm{s}\right)-\left(6.0\mathrm{m}/\mathrm{s}\right) \\ =3.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity at t = 1.00 s is 3.8 m/s.

The equation (i) has been recalled below:

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

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

$$\begin{gathered} V\left(t\right)=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(3.00\mathrm{s})−\frac{(36.0\mathrm{N}/\mathrm{s})(3.00\mathrm{s}{)}^2}{2×3.00\mathrm{kg}} \\ =(29.4\mathrm{m}/\mathrm{s})−\frac{\left(36.0\mathrm{N}/\mathrm{s}×\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)\left(9.00{\mathrm{s}}^2\right)}{6.00\mathrm{kg}} \\ =(29.4\mathrm{m}/\mathrm{s})−\frac{\left(36.0\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^3\right)\left(9.00{\mathrm{s}}^2\right)}{6.00\mathrm{kg}} \\ =(29.4\mathrm{m}/\mathrm{s})−\frac{(324\mathrm{kg}â‹\dots \mathrm{m}/\mathrm{s})}{6.00\mathrm{kg}} \end{gathered}$$

On further solving:

role="math" localid="1667645685950" $$\begin{gathered} V\left(t\right)=\left(29.4\mathrm{m}/\mathrm{s}\right)-\left(54\mathrm{m}/\mathrm{s}\right) \\ =-24.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity at (i) t =3.00 s is -24.6 m/s .

The equation (i) has been recalled below:

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

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

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

$$\begin{gathered} 0=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t−\frac{\left((36.0\mathrm{N}/\mathrm{s})(t{)}^2\right)}{2\left(3.00\mathrm{kg}\right)} \\ \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t−\frac{\left(\left(36.0\mathrm{N}/\mathrm{s}×\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)(t{)}^2\right)}{\left(6.00\mathrm{kg}\right)}=0 \\ \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)t−\left(6.0\mathrm{m}/{\mathrm{s}}^3\right)\left(t^2\right)=0 \end{gathered}$$

On further solving

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

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

$$\begin{gathered} d\left(t\right)=∫V\left(t\right)dt \\ =∫\left(gt−\frac{(36.0\mathrm{N}/\mathrm{s})t^2}{2m}\right)dt \\ =g\frac{t^2}{2}−\frac{(36.0\mathrm{N}/\mathrm{s})t^3}{3×2m} \\ =\frac{g{t}^2}{2}−\frac{(36.0\mathrm{N}/\mathrm{s})t^3}{6m}.....................................\left(2\right) \end{gathered}$$

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

localid="1667646508016" $$\begin{gathered} d\left(t\right)=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\frac{(1.63\mathrm{s}{)}^2}{2}−\frac{(36.0\mathrm{N}/\mathrm{s})(1.63\mathrm{s}{)}^3}{4\left(3.00\mathrm{kg}\right)} \\ =\left(13.02\mathrm{m}\right)−\frac{\left(36.0\mathrm{N}/\mathrm{s}×\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)(1.63\mathrm{s}{)}^3}{\left(12.00\mathrm{kg}\right)} \\ =\left(13.02\mathrm{m}\right)−\left(12.99\mathrm{m}\right) \\ =0.03\mathrm{m} \end{gathered}$$.

Thus, the maximum distance that the box descends below is 0.03 m.

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

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

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

Hence, further as:

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

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