# A biologist hangs a sample of mass 0.515 kg on a pair of identical, vertical springs in parallel...

## Question:

A biologist hangs a sample of mass 0.515 kg on a pair of identical, vertical springs in parallel and slowly lowers the sample to equilibrium, stretching the springs by 0.250 m.

Calculate the value of the spring constant (in N/m) of one of the springs. (The acceleration of gravity is g = 9.80 m/s2.)

## Spring Constant:

For spring, the spring constant is the total force required to pull the spring by the unit length. It also defines the stiffness of a spring. It can be determined by taking the ratio of the total force applied to a spring, and the displaced length of the spring. The spring constant is shown by the unit Newton per meter.

## Answer and Explanation: 1

Given data

• The mass of the sample is {eq}m=0.515\ \text{kg} {/eq}
• The number of samples is {eq}n = 2 {/eq}
• The stretched displacement is {eq}x=0.25\ \text{m} {/eq}
• The acceleration due to gravity is {eq}g=9.8\ \text{m/}{{\text{s}}^{2}} {/eq}

The weight of the sample is calculated as,

{eq}\begin{align} W &=mg \\ &=\left( 0.515\ \text{kg} \right)\left( 9.8\ \text{m/}{{\text{s}}^{2}} \right) \\ &=5.047\ \text{N} \end{align} {/eq}

The spring force of one spring is calculated as,

{eq}\begin{align} {{F}_{1}} &=kx\ \ \ \ \ \ \ \ \ \ \left[ \text{Here,}\ k\ \text{is}\ \text{the}\ \text{spring}\ \text{constant}\ \text{of}\ \text{the}\ \text{spring} \right] \\ &=k\left( 0.25\ \text{m} \right) \end{align} {/eq}

The total spring force of spring is calculated as,

{eq}\begin{align} F &=2{{F}_{1}} \\ &=2\left[ k\left( 0.25\ \text{m} \right) \right] \\ &=k\left( 0.5\ \text{m} \right) \end{align} {/eq}

In the given question we can see that two similar spring forces area acting vertically in the upward direction on the given sample of mass, and the weight of the sample of mass acting vertically but in the downward direction. Now, use the equilibrium and calculate the spring constant as shown below,

{eq}\begin{align} F-W &=0 \\ k\left( 0.5\ \text{m} \right)-\left( 5.047\ \text{N} \right) &=0 \\ k\left( 0.5\ \text{m} \right) &=\left( 5.047\ \text{N} \right) \\ k &=\dfrac{\left( 5.047\ \text{N} \right)}{\left( 0.5\ \text{m} \right)} \\ k &=10.094\ \text{N/m} \end{align} {/eq}

Thus, the spring constant of each spring is 10.094 Newton per meter.

Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.4K

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.