Question

A 3.00-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 $\mathbf{T}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{36}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{t}$ . 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) t = 1.00 s and (ii) t = 3.00 s ? (b) What is the maximum distance that the box descends below its initial position? (c) At what value of tdoes the box return to its initial position?

Short answer

(a) The velocity att = 1.00 s is 3.8 m/s and the velocity at t = 3.00 s is -24.6 m/s .

(b) The maximum distance that the box descends below is 0.3 m.

(c) The box returns to its initial position at 1.63 s.

Step-by-step solution

Identification of the given data

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 $T\left(t\right)=(36.0\hspace{0.33em}\mathrm{N}/\mathrm{s})t$.

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.

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

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.

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

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

for $g=9.8\hspace{0.33em}m/s^2,t=1.00\hspace{0.33em}s,T=(36.0\hspace{0.33em}N/s)(1.00\hspace{0.33em}s)$and m=3.00 kg ; equation becomes-.

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

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.

(a) Determination of the velocity at (ii) t =3.00 s  

The equation (i) has been recalled below:

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

for $g=9.8m/s^2,t=3.00\hspace{0.33em}s,T=(36.0\hspace{0.33em}\mathrm{N}/\mathrm{s})(3.00\hspace{0.33em}\mathrm{s})$ and m =3.00 kg ; equation (1) becomes-.

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

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 .

(b) Determination of the maximum distance

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 $g=9.8\hspace{0.33em}m/s^2,V\left(t\right)=0,T=(36.0\hspace{0.33em}\mathrm{N}/\mathrm{s})(3.00\hspace{0.33em}\mathrm{s})$and m =3.00 kg ; equation (1) becomes-.

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

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.

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

for $g=9.8\hspace{0.33em}m/s^2,t=1.63\hspace{0.33em}s$ and m =3.00 kg ; equation (1) becomes-.

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

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

(c) Determination of the time  

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

for $g=9.8\hspace{0.33em}m/s^2,d\left(t\right)=0,T=(36.0\hspace{0.33em}\mathrm{N}/\mathrm{s})\left(t\right)$and m=3.00 kg; equation (1) becomes-.

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

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.