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

★★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:-

$\sf\:V=25u$✔️✔️

★★Concept:-

• Potential Energy is equal to the work done or needed to reach a particular state.

• Work done against a force is the force multiplied by the distance moved against it.

• As the force (by Hooke’s Law) is itself proportional to the extension, we can see that the potential energy of an extended chain is going to be proportional to the square of the extension.

★★Explanation:-

potential energy in a string(spring) can be given by,

$\sf\:v = \frac{1}{2} k \: {x}^{2} \\\\ \sf\:v =\frac{1}{2} k {(2)}^{2} \\\\ \sf\:v = 2k$

Let potential energy for 10cm stretch is v'

$\sf\:v' =\frac{1}{2} k \: {y}^{2} \\\\ \sf\:v' = \frac{1}{2} k {(10)}^{2} \\\\ \sf\:v' =50k \\\\ \sf\:v' =25(2k) \\\\ \sf\:v' = 25u$

Hence,the potential energy will be 25 times of initial stretch.

Answer 2

Aɴꜱᴡᴇʀ

$\bf{\green{U'=25U}}$

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Gɪᴠᴇɴ

• Spring stores U potential energy when stretched by 2cm

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ᴛᴏ ꜰɪɴᴅ

• Potential energy stored in spring when stretched by 10cm

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Sᴛᴇᴘꜱ

Formula of stored potential energy in spring is

$\dashrightarrow{ \pink{\tt\:U=\dfrac{1}{2}kx^2}}$

• U denotes potential energy

• k denotes spring constant

• x denotes elongation

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

$\begin{lgathered}\sf\bigstar\red\:U\propto x^2\\ \\ \leadsto\sf\:\dfrac{U}{U'}=\dfrac{{x_1}^2}{{x_2}^2}\\ \\ \leadsto\sf\:\dfrac{U}{U'}=\dfrac{4}{100}\\ \\ \leadsto\sf\:\dfrac{U}{U'}=\dfrac{1}{25}\\ \\ \dashrightarrow\bf{\red{U'=25U}}\end{lgathered}$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$