Question

A steel wire (y = 200 gpa) of cross-sectional area = 1 mm2 is stretched by a force = 1000 n, if the length of wire is 2 m, the elastic potential energy stored in wire is :

Answer 1

Answer:

Elastic potential energy stored in the wire is 5 J

Explanation:

As we know by law of elasticity that stress and strain is proportional to each other and the proportionality constant is known as Young's Modulus

So we will have

$\frac{T}{A} = Y\times (\frac{\Delta L}{L})$

here we know that

T = 1000 N

area of cross-section of wire is

$A = 1 mm^2$

also we know that Young's Modulus of the wire is

$Y = 200 \times 10^9 Pa$

Length of the wire is

L = 2 m

Now we have

$\frac{1000}{1\times 10^{-6}} = (200 \times 10^9)(\frac{\Delta L}{2})$

$10^9 = (10^{11})\Delta L$

$\Delta L = 0.01 m$

now the elastic potential energy stored in the wire is given as

$U = \frac{1}{2}T \Delta L$

$U = \frac{1}{2}(1000)(0.01)$

$U = 5 J$

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Topic : Elastic potential energy

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