Question

the extension developed on a wire by 300N force is 2mm. the elastic potiential energy is​

Answer 1

swe your answer in the attachment above

Answer 2

Given:

F = 300 N

l = 2 mm

To find:

elastic potential energy E =?

Explanation:

In stretching a wire work is done against internal restoring forces.

This work is kept as elastic mechanical energy or strain energy.

This work done is given by

$\displaystyle W = \frac{1}{2}\frac{YAl^2}{L}$

where

W = work done as a potential energy

Y = Young's modulus

A = area

l = increase in length

L = Length of wire

But the force is required to increase the length of the wire.

This force is given by

$\displaystyle F = \frac{ YAl}{L}$

By putting this value in eq (1) we get,

$\displaystyle W= \frac{1}{2} Fl$

So put given values in the above equation

$\displaystyle W = 0.5 \times 300 \times 2 \times10^{-3$

    = 300 $\times 10^{-3$

W = 0.3 J

Hence required elastic potential energy is​ 0.3 J.