Q69P - University Physics with Modern Physics

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 t does the box return to its initial position?

Step-by-Step Solution

Answer

(a) The velocity at t = 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 1: 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 $T\left(t\right)=(36.0\hspace{0.33em}\mathrm{N}/\mathrm{s} )t$.

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 = 1.00 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.

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

                                                                                                           

 

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-.

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

 

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.

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

The equation (i) has been recalled below:

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

  

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:

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

 

 

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

Step 5: (b) Determination of the maximum distance

The equation (i) has been recalled below:

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

 

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

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

 

 

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-.

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

 

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

Step 6: (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:

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

 

 

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