Question

13. Find the elastic potential energy in a system shown
below if the material of wires is same (Y = Young's
modulus)
w​

Answer 1

Answer:

$\dfrac{17W^2L}{4\pi R^2Y}$

Explanation:

given,

wire have same young's modulus = Y

elongation

$d_1 = \dfrac{WL_1}{A_1Y}$

$d_1 = \dfrac{WL}{2\pi R^2 Y}$

$d_2 = \dfrac{WL_2}{A_2Y}$

$d_2 = \dfrac{2WL}{\pi (R/2)^2 Y}$

$d_2 = \dfrac{8WL}{\pi (R)^2 Y}$

total elastic potential energy

= $\dfrac{1}{2}W d_1 + \dfrac{1}{2}W d_2$

= $\dfrac{1}{2}W\times (\dfrac{WL}{2\pi R^2 Y}) + \dfrac{1}{2}W \times (\dfrac{8WL}{\pi (R)^2 Y})$

=$\dfrac{W^2L}{4\pi R^2Y} +\dfrac{8W^2L}{2\pi R^2Y}$

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

Answer 2

Answer:

Explanation:

DeltaL1=fl/ya

DeltaL2=fl/ya

Potatial energy =1/2*k*deltaL1+1/2*k*deltaL2

Where K=f/x

F=mg=w