# When a 2.80-kg object is hung vertically on a certain light spring described by Hooke's law, the...

## Question:

When a 2.80-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.06 cm.

a) What is the force constant of the spring?

(Answer in N/m)

b) If the 2.80-kg object is removed, how far will the spring stretch if a 1.40-kg block is hung on it?

(Answer in cm)

c) How much work must an external agent do to stretch the same spring 6.60 cm from its unstretched position?

(Answer in J)

## Spring:

Spring is a mechanical device that stores energy during compression and releases energy during expansion. Different type of springs are:

1) Helical spring

2) Leaf spring

3) Torsion spring

4) Conical spring

Strain energy stored in a helical spring is the product of the average torque and the average displacement made by the spring.

## Answer and Explanation: 1

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

- The mass of the object {eq}m=2.80\ \text{kg} {/eq}

- Stretch of the spring {eq}x=3.06\ \text{cm} {/eq}

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

$$\begin{align} F&=kx&\left [\text{k is the spring constant} \right ]\\[0.3 cm] mg&=kx&\left [ \text{F=mg, m is the mass, g is the gravitational constant} \right ]\\[0.3 cm] 2.80\ \text{(kg})\times9.80\ ( \text{m/s}^{2})&=k\times3.06\ (\text{cm})\\[0.3 cm] k&=\dfrac{2.8\times9.80\ \text{(N)}}{3.06\ (\text{cm})}\\[0.3 cm] &=8.96\ \text{N/cm}\\[0.3 cm] &=8.96\times100\ \text{N/m}&\left [ \text{Converting N/cm into N/m} \right ]\\[0.3 cm] &=\boxed{\color{blue}{896\ \text{N/m}}} \end{align} $$

**b) When 2.80-kg object is removed and a 1.40-kg block is hung on it, the spring is stretched by:**

$$\begin{align} F&=kx\\[0.3 cm] mg&=kx&\left [ \text{F=mg, m is the mass, g is the gravitational constant} \right ]\\[0.3 cm] 1.40\ (\text{kg})\times9.80\ ( \text{m/s}^{2})&=896\ \text{(N/m}) \times x\\[0.3 cm] x&=\dfrac{1.40\times9.80\ (\text{N})}{896\ (\text{N/m})}\\[0.3 cm] &=0.015\ \text{m}\\[0.3 cm] &=0.015\times100\ \text{cm}&\left [ \text{Converting m into cm} \right ]\\[0.3 cm] &=\boxed{\color{blue}{1.50\ \text{cm}}} \end{align} $$

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

$$\begin{align} W&=\dfrac{1}{2}kx^{2}\\[0.3 cm] &=\dfrac{1}{2}\times896\ (\text{N/m})\times(0.066)^{2}\ (\text{m}^{2})&\left [ x=\dfrac{6.60}{100}=0.066\ \text{m} \right ]\\[0.3 cm] &=1.95\ \text{N.m}&\left [ \text{1 N.m=1 J} \right ]\\[0.3 cm] &=\boxed{\color{blue}{1.95\ \text{J}}} \end{align} $$

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.