Question

A steel wire of length 2.0 m/s is stretched through 2.0 mm. The cross sectional area of the wire is 4.0 mm^2. Calculate the elastic potential energy stored in the wire in the stretched condition. Young modulus of steel =2.0x10^11Nm^-2

Answer 1

Answer:

Hey Buddy here's ur answer

Answer 2

Answer:

Elastic potential energy stored in the wire in stretched condition is 0.8 J.

Explanation:

Given,

A steel wire of length 2.0 m/s is stretched through 2.0 mm. The cross sectional area of the wire is 4.0 mm^2.

To find: Calculate the elastic potential energy stored in the wire in the stretched condition.

Solution:

Strain can be given as change in dimension / original dimension.

or, Strain = Δl / l

here, Δl = 2.0 mm = 2 * 10^-3 m , l = 2.0 m

or, Strain = 2 * 10^-3  / 2.0 = 10⁻³

Young's modulus can be given as:

Y = stress / strain

or, stress = Y * strain = 2.0 * 10^11 * 10^-3 = 2.0*10⁸ Nm⁻²

Volume of the wire can be calculated as:

V = area * length = 4.0*10⁻⁶ * 2.0 = 8 * 10⁻⁶ m³

Now, Elastic potential energy is given as:

Ue = (1 / 2) * stress* strain * volume'

or, Ue = (1/2) * 2.0*10⁸ * 10⁻³ * 8*10⁻⁶

or, Ue = 0.8 J

Therefore, Elastic potential energy stored in the wire in stretched condition is 0.8 J.