# 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?

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?

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

## 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.

(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}

Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
202K

After 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.