Chapter 14, Problem 95A

(a)

To determine

Spring constant for each spring.

Answer to Problem 95A

  $\text{22050}\text{N/m}$

Explanation of Solution

Given:

Load, $m=45\mathrm{kg}$

Compression in spring, $x=1.0\text{cm}$

Formula used:

  $F=mg$

  $F=kx$

Where $F$ is force, $m$ is mass, k is the spring constant, x is the length and $g$ denotes acceleration due to gravity.

Calculation:

  $\begin{array}{l}F=mg\\ =\text{(45kg)(9}\text{.8N/kg)}\\ =441\text{N}\end{array}$

Therefore, force per string is $\text{220}\text{.5}\text{N}$ .

  $\text{1}\text{.0}\text{cm = 0}\text{.010 m}$

For spring constant,

  $F=kx$

Where $F$ stands for force, $k$ for spring constant and $x$ for displacement.

So,

  $\begin{array}{l}k=\frac{F}{x}\\ k=\frac{\text{220}\text{.5}\text{N}}{\text{0}\text{.010}\text{m}}\\ k\text{=22050}\text{N/m}\end{array}$

Conclusion:

Hence, the spring constant for each of the springs is $\text{22050}\text{N/m}$ .

(b)

To determine

Increase in potential energy.

Answer to Problem 95A

Increase in potential energy is $\text{1}\text{.1J}$ .

Explanation of Solution

Given:

Compression in spring, $x=1.0\text{cm}$

Spring constant, $k=22050\text{N/m}$

Formula used:

Potential energy stored in spring:

  $PE=\frac{1}{2}k{x}^2$

Where, $k$ is spring constant and $x$ islength of spring.

Calculation:

  $\begin{array}{l}PE=\frac{1}{2}k{x}^2\\ =\frac{1}{2}\text{(22050N/m)(0}\text{.010m)}\\ =1.1\text{J}\end{array}$

Conclusion:

Hence, increase in each of the spring’s potential energy after loading the trunk is $\text{1}\text{.1J}$ .