Chapter 11, Problem 69GP

a

To determine

To Find: The stretch in fire net (when the same person lying in it).

Answer to Problem 69GP

Stretch in fire net when the same person lying in it is $3.16\text{cm}$ .

Explanation of Solution

Given Information:

Mass of person, $\mathrm{m}=65\text{kg}$

Acceleration due to gravity, $\text{g=9}\text{.8}{\text{m/s}}^2$

Distance between window and fire net is $18\text{m}$ .

Net stretched by $\text{x=1}\text{.1}\text{m}$

Concept used:

At the bottom relative to the point of stretch, the person has potential energy. $\mathrm{P}\mathrm{E}\text{ = mgh}\mathrm{...}\left(\text{1}\right)$

Here,

  $\mathrm{m}$ is mass of person,

  $\mathrm{h}$ is height from reference.

The energy stored as spring energy is

  $\mathrm{U}=\frac{1}{2}{\text{kx}}^{\text{2}}\mathrm{...}\left(2\right)$

Here,

  $\mathrm{k}$ is spring constant,

  $\mathrm{x}$ is compression in net.

Calculation:

From equation 1

  $\begin{array}{l}\mathrm{P}\mathrm{E}\text{ = mgh}\\ \text{=}\text{(65}\text{kg)(9}\text{.8}{\text{m/s}}^2)(18+1.1)\text{m}\\ \text{=}\text{(65}\text{kg)(9}\text{.8}{\text{m/s}}^2)(19.1\text{m)}\\ \text{=}\text{12166}\text{.7}\text{J}\end{array}$

From equation 2

  $\mathrm{U}=\frac{1}{2}{\text{kx}}^{\text{2}}$

As eventually this energy is stored as spring energy so,

  $\begin{array}{l}$ \text{12166}\text{.7}\text{J}=\frac{1}{2}\text{k}{(1.1\text{m)}}^2$\text{k}\text{=}\text{20110}\text{.2}\text{N/m}\end{array}$

Then, just lying in the fire net cause stretch

So,

  $\begin{array}{l}\text{x = }\frac{\text{mg}}{\text{k}}\\ \text{x}=\frac{(65\text{kg)}\times \text{(9}\text{.8}{\text{m/s}}^2)}{20110.2\text{N/m}}\\ \text{x}=0.0316\text{m}\\ \text{x = 3}\text{.16}\text{cm}\end{array}$

Conclusion:

Stretch in fire net is $\text{3}\text{.16}\text{cm}$ .

b

To determine

To Find: The stretch in fire net when a person jumps from $35\text{m}$ .

Answer to Problem 69GP

The stretch in fire net is $1.521\text{m}$ .

Explanation of Solution

Given Information:

Mass of person, $\mathrm{m}=65\text{kg}$ ,

Acceleration due to gravity, $\text{g=9}\text{.8}{\text{m/s}}^2$

Spring constant, $\mathrm{k}=20110.2\text{N/m}$

Distance between window and fire net is $35\text{m}$

Concept used:

Conservation of energy is used

  $\text{mgh = }\frac{1}{2}{\text{kx}}^{\text{2}}$

Here,

  $\mathrm{m}$ is mass of person,

  $\mathrm{h}$ is height from reference.

  $\mathrm{k}$ is spring constant,

  $\mathrm{x}$ is compression in net.

Calculation:

From equation 1

  $\begin{array}{l}\text{mgh = }\frac{1}{2}{\text{kx}}^{\text{2}}\\ (65\text{)(9}\text{.8})(35\text{+x)}\text{=}\frac{1}{2}(20110.2{\text{)x}}^2\\ (65\text{)(9}\text{.8})(35\text{+x)}\text{=}(10055.1{\text{)x}}^2\end{array}$

  $\mathrm{x}=1.521\text{m}$

Conclusion:

Stretch in fire net when a person jumps from $35\text{m}$ is $1.521\text{m}$ .