Chapter 11, Problem 71GP

a

To determine

To Find: The spring stiffness constant of the spring.

Answer to Problem 71GP

The spring stiffness constant of the spring is $1.839\times {10}^4\text{N/m}$ .

Explanation of Solution

Given Information:

Mass of the car, $\mathrm{m}=950\text{kg}$

Velocity, ${\mathrm{v}}_1=22\text{m/s}$

Compression of the spring, ${\mathrm{x}}_2=5.0\text{m}$

Concept used:

According to the law of conservation of energy

  $\begin{array}{l}{\mathrm{E}}_1={\mathrm{E}}_2\\ \frac{1}{2}\mathrm{m}{\mathrm{v}}_1{}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}_1{}^2=\frac{1}{2}\mathrm{m}{\mathrm{v}}_2{}^2+\frac{1}{2}\mathrm{k}{\mathrm{x}}_2{}^2\left[\begin{array}{l}{\mathrm{x}}_1=0\\ {\mathrm{v}}_2=0\end{array}\right]\\ \frac{1}{2}\mathrm{m}{\mathrm{v}}_1{}^2=\frac{1}{2}\mathrm{k}{\mathrm{x}}_2{}^2\\ \mathrm{m}{\mathrm{v}}_1{}^2=\mathrm{k}{\mathrm{x}}_2{}^2\\ \mathrm{k}=\frac{\mathrm{m}{\mathrm{v}}_1{}^2}{{\mathrm{x}}_2{}^2}\end{array}$

Here,

  $\mathrm{k}$ is spring stiffness constant,

  ${\mathrm{x}}_1,{\mathrm{x}}_2$ is compression of spring,

  $\mathrm{m}$ is the mass of car,

  ${\mathrm{v}}_1,{\mathrm{v}}_2$ is the speed of car.

Calculation:

Substitute the values and solve:

  $\begin{array}{l}\mathrm{k}=\frac{(950\text{kg}){(22\text{m/s})}^2}{{(5\text{m})}^2}\\ \mathrm{k}=1.839\times {10}^4\text{N/m}\end{array}$

Conclusion:

The spring stiffness constant of the spring is $1.839\times {10}^4\text{N/m}$ .

b

To determine

To Find: The contact time of car with the spring.

Answer to Problem 71GP

The contact time of car with spring is $0.7140\text{s}$.

Explanation of Solution

Given Information:

Mass of the car =

  $950\text{kg}$

Spring stiffness constant, $\mathrm{k}=1.839\times {10}^4\text{n/m}$

Formula used:

Here, time period of oscillation is given by:

  $\mathrm{T}=(2\mathrm{\pi })\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

The car will be in contact with the string for half a cycle. Therefore, time of contact is half of time of oscillation.

Calculation:

Time of contact is:

  $\frac{\mathrm{T}}{2}=\frac{1}{2}(2\mathrm{\pi })\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

  $=\sqrt{\frac{950}{1.839\times {10}^4\text{N/m}}}\times \mathrm{\pi }$

  $=0.7140\text{s}$

Conclusion:

The contact time of car with spring is $0.7140\text{s}$.