Chapter 6, Problem 72GP

To determine

The spring constant of the spider webs.

Answer to Problem 72GP

Solution: $11.0N/m$

Explanation of Solution

Given

The initial speed of the train is:

  $\begin{array}{l}u=60\frac{\text{km}}{\text{h}}\\ u=60\times \frac{1000}{3600}\frac{\text{m}}{\text{s}}\\ u=16.6\frac{\text{m}}{\text{s}}\end{array}$

The mass of the train is $m={10}^4\text{kg}$

The final speed of the train is zero.

His web stretches by 500 m, so $x=500\text{m}$

Formula used

The kinetic energy of the train is $\frac{1}{2}m{v}^2$ .

Where, m is the mass and vis the speed.

The spring potential energy of the web is $\frac{1}{2}k{x}^2$ .

Here, k is the spring constant and x is the extension.

Calculations

Since the kinetic energy is equal to the spring potential energy,

  $\begin{array}{l}\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2\\ $ {10}^4\times {(16.6)}^2=k\times {(500)}^2$k=\frac{{10}^4\times {(16.6)}^2}{{(500)}^2}\frac{\text{N}}{\text{m}}\\ k=11.0\frac{\text{N}}{\text{m}}\end{array}$

Conclusion:

The spring constant of the spider web can be calculated by equating the spring potential energy with the kinetic energy of the train.