Question

A uniform chain is stretched by 2 cm , its potential energy is U . if the spring is stretched by 10 cm ,the potential energy stored in it will be ?

Answer 1

Answer:

Given:

Uniform chain stores U amount of Potential Energy when stretched by 2 cm

To find:

Potential Energy stored when chain is stretched by 10 cm

Concept:

Whenever a person stretches a spring or chain , a certain amount of work is done to stretch the molecules against the inter-molecular force of attraction. That work actually gets stored in the chain as the Potential Energy.

Spring Energy (present in Chains as potential energy) is a Conservative energy and doesn't dissipate.

Calculation:

Potential energy be denoted as U , spring constant is "k" and elongation is "x"

$\bigstar \: \: U_{1} = \frac{1}{2} k {x}^{2}$

$= > \: \: U_{1} = \frac{1}{2} \times k \times {(2)}^{2}$

$= > \: \: U_{1} = 2 {k}^{2} = U$

In the 2nd case :

$\bigstar \: \: U_{2} = \frac{1}{2} k {(x2)}^{2}$

$= > \: \: U_{2} = \frac{1}{2} \times k \times {(10)}^{2}$

$= > \: \: U_{2} = 50 {k}^{2}$

Dividing the equations :

$\bigstar \: \: \dfrac{U_{2}}{U} = \dfrac{50 {k}^{2} }{2 {k}^{2} }$

$= > U_{2} = 25 \: U$

So final answer :

$\boxed{ \red{ \huge{ \bold{U_{2} = 25 \: U}}}}$

Answer 2

Solution :

✴ Given :-

▪ Spring stores U potential energy when stretched by 2cm

✴ To Find :-

▪ Potential energy stored in spring when stretched by 10cm

✴ Formula :-

▪ Formula of stored potential energy in spring is given by

$\bf\:U=\dfrac{1}{2}kx^2$

• U denotes potential energy
• k denotes spring constant
• x denotes elongation

✴ Calculation :-

✏ As per formula of energy stored in spring, It is clear that...

$\sf\circ\red\:U\propto x^2\\ \\ \implies\sf\:\dfrac{U}{U'}=\dfrac{{x_1}^2}{{x_2}^2}\\ \\ \implies\sf\:\dfrac{U}{U'}=\dfrac{4}{100}\\ \\ \implies\sf\:\dfrac{U}{U'}=\dfrac{1}{25}\\ \\ \implies\bf{\orange{U'=25U}}$