Question

Find the elastic potential energy in a system shownbelow if the material of wires is same (Y = Young'smodulus)​

Answer 1

Answer:

17W^2L÷4πR^2Y

Explanation:

Answer 2

$= \bold{\frac{17\times W^2 \times L}{4\pi^2\times R^2 \times Y}}$

Explanation:

$Y = \frac{F}{A}\times \frac{L}l}$

l = elongation

$l = \frac{F}{A}\times \frac{l}{Y}$

$U = \frac{1}{2}\times F\times L$

$U = \frac{1}{2}\times F\times \frac{F}{A}\times \frac{L}{Y}$

$U = \frac{ F^2\times L}{2\times A\times Y}$

∵ F = W

Then,

$U = \frac{W^2 \times \frac{L}{2}}{2\times\pi\times R^2 \times y} + \frac{W^2\times 2\times L}{2\times \frac{\pi\times R^2}{4}\times Y}$

$= \frac{W^2\times L}{4\times \pi\times R^2\times Y} + \frac{W^2\times 4\times L}{\pi\times R^2\times Y}$

$= \frac{W^2\times L + 16\times W^2\times L}{4\times \pi\times R^2\times Y}$

$= \frac{17\times W^2 \times L}{4\pi\times R^2 \times Y}$