Question

derive an expression for potential energy of an elastic stretched spring.if the length of a steel wire increases by 0.5cm when it is loaded with a weight of 5kg . calculate the (i) force constant of wire (ii) work done in stretching the wire. *

Answer 1

Let a spring of spring constant $\sf{k}$ get stretched from normal state by an extension $\sf{x.}$

Due to this extension, a restoring force is developed in the spring which is given by,

$\sf{\longrightarrow F=-kx\quad\quad\dots(1)}$

Negative sign shows that the restoring force is opposite to the extension.

We can see this force vary according to the extension $\sf{x.}$

The work done in stretching a spring by a small distance $\sf{dx}$ is,

$\sf{\longrightarrow dW}=\bf{F}\cdot\bf{dx}$

Since force and extension are in opposite direction,

$\sf{\longrightarrow dW=F\,dx\,\cos180^o}$

$\sf{\longrightarrow dW=-F\,dx}$

Hence the total work done will be,

$\displaystyle\sf{\longrightarrow \int dW=\int-F\,dx}$

From (1),

$\displaystyle\sf{\longrightarrow \int dW=\int-(-kx)\,dx}$

$\displaystyle\sf{\longrightarrow W=\int kx\,dx}$

$\displaystyle\sf{\longrightarrow W=k\int x\,dx}$

$\displaystyle\sf{\longrightarrow W=k\cdot\dfrac{x^2}{2}}$

$\displaystyle\sf{\longrightarrow W=\dfrac{1}{2}\,kx^2}$

This work is stored in the spring as its potential energy due to elasticity.

$\displaystyle\sf{\longrightarrow\underline{\underline{U=\dfrac{1}{2}\,kx^2}}}$

Here a steel wire gets extended due to a load of mass,

• $\sf{m=5\ kg.}$

Extension produced is,

• $\sf{x=0.5\ cm=0.5\times10^{-2}\ m}$

The restoring force developed in the wire is equal to weight of the load.

$\sf{\longrightarrow kx=mg}$

Hence force constant of wire is,

$\sf{\longrightarrow k=\dfrac{mg}{x}}$

$\sf{\longrightarrow k=\dfrac{5\times10}{0.5\times10^{-2}}\ N\,m^{-1}}$

$\sf{\longrightarrow\underline{\underline{k=10^4\ N\,m^{-1}}}}$

And work done in stretching the wire is,

$\sf{\longrightarrow W=\dfrac{1}{2}\,kx^2}$

$\sf{\longrightarrow W=\dfrac{1}{2}\times10^4\times\left(0.5\times10^{-2}\right)^2}$

$\sf{\longrightarrow\underline{\underline{W=0.125\ J}}}$

Answer 2

Answer

Δx = stretch in the length of wire = 0.5 cm = 0.005 m

m = mass attached to the wire = 5 kg

g = acceleration due to gravity = 9.8 m/s²

W = weight attached to the wire

weight of mass hanged is given as

W = mg

W = 5 x 9.8

W = 49 N

k = force constant of the wire

the spring force by the wire balances the weight here

hence  , Spring force = weight

k Δx = mg

k (0.005) = 49

k = 9800 N/m