# When a 2.50-kg object is hung vertically on a certain light spring described by Hooke's law, the...

## Question:

When a 2.50-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.69 cm.

(a) What is the force constant of the spring?

(b) If the 2.50-kg object is removed, how far will the spring stretch if a 1.25-kg block is hung on it?

(c) How much work must an external agent do to stretch the same spring 8.40 cm from its unstretched position?

## Spring:

A spring is a mechanical device used to store energy on compression. The energy stored in the spring is called the strain energy. It is proportional to the stiffness of the spring. The strain energy stored in the spring is mathematically represented by the equation:

{eq}W=\dfrac{1}{2}k(x)^{2} {/eq}

{eq}k{/eq} is the stiffness of the spring.

{eq}x{/eq} is the compression of the spring.

## Answer and Explanation: 1

**(**a**)The force constant of the spring:**

- The mass of the object {eq}m=2.50\ \text{kg} {/eq}

- The stretch of the spring {eq}x=2.69\ \text{cm} {/eq}

By using Hooke's law, the force constant of the spring is:

$$\begin{align} F&=kx&\left [\text{k is the spring constant} \right ]\\[0.3 cm] mg&=kx&\left [ \text{F=mg, m is the mass, g is the gravitational constant} \right ]\\[0.3 cm] 2.50\ \text{(kg})\times9.81\ ( \text{m/s}^{2})&=k\times2.69\ (\text{cm})\\[0.3 cm] k&=\dfrac{2.50\times9.81\ \text{(N)}}{2.69\ (\text{cm})}\\[0.3 cm] &=9.12\ \text{N/cm}\\[0.3 cm] &=9.12\times100\ \text{N/m}&\left [ \text{Converting N/cm into N/m} \right ]\\[0.3 cm] &=\boxed{\color{blue}{912\ \text{N/m}}} \end{align} $$

**b) When the 2.50-kg object is removed and a 1.25-kg block is hung on it, the spring is stretched by:**

{eq}x' {/eq} is the stretch of the spring due to the 1.25-kg mass.

$$\begin{align} F'&=k x' \\[0.3 cm] m' g&=kx' &\left [ \text{F'=m' g, m' is the mass, g is the gravitational constant} \right ]\\[0.3 cm] 1.25\ (\text{kg})\times9.81\ ( \text{m/s}^{2})&=912 \text{(N/m}) \times x' \\[0.3 cm] x' &=\dfrac{1.25\times9.81\ (\text{N})}{912\ (\text{N/m})}\\[0.3 cm] &=0.0134\ \text{m}\\[0.3 cm] &=0.0134\times100\ \text{cm}&\left [ \text{Converting m into cm} \right ]\\[0.3 cm] &=\boxed{\color{blue}{1.34\ \text{cm}}} \end{align} $$

**c) When an external agent stretches the spring, the work done by the agent is:**

{eq}x''=8.40 \ {\rm cm} {/eq}, is the stretch of the spring.

$$\begin{align} W&=\dfrac{1}{2}k(x'')^{2}\\[0.3 cm] &=\dfrac{1}{2}\times912\ (\text{N/m})\times(0.084)^{2}\ (\text{m}^{2})&\left [ x'' =\dfrac{8.40}{100}=0.084\ \text{m} \right ]\\[0.3 cm] &=3.22\ \text{N.m}&\left [ \text{1 N.m=1 J} \right ]\\[0.3 cm] &=\boxed{\color{blue}{3.22\ \text{J}}} \end{align} $$

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Chapter 17 / Lesson 11In 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.