Question

A wire of natural length l young's modulus y and area of cross section a is extended by x then what is the energy stored in the wire

Answer 1

Answer:

Explanation:

Energy stored 1/2 F Delta l

So,as given in the question in terms of y

Answer 2

Answer:

$U=\dfrac{1}{2}\times \dfrac{\gamma A x^2}{l}$

Explanation:

Given that

Length = l

Young's modulus = y

Area of cross section = A

Extended length (Δl)= x

Force = F

From Hooks law ,change in length given as

$\Delta l=\dfrac{Fl}{A\gamma }$

The energy stored in the wire U

$U=\dfrac{1}{2}F.\Delta$

Now by putting the values

F= σ A = ε γ .A

ε =x/l

F=  γ .A. x/l

$U=\dfrac{1}{2}\times \dfrac{\gamma A x}{l}\times x$

$U=\dfrac{1}{2}\times \dfrac{\gamma A x^2}{l}$