Question

A uniform steel wire of length 3 m and area of cross section 2 mm² is extended through 3 mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. ( $Y_{steel}$ = 20 x 10¹⁰ N / m²) (Ans 0.6 J)

Answer 1

Given, area , A = 2mm² = 2 × 10^-6 m²

length of uniform steel wire, l = 3m

extension of steel wire , ∆l = 3mm = 3 × 10^-3m

Young's modulus of steel , Y = 20 × 10^10 N/m²

use formula , F = YA∆l/l

= 20 × 10^10 × 2 × 10^-6 × 3 × 10^-3/3

= 400 N

now , energy stored in the wire, dU = 1/2 × F × ∆l

= 1/2 × 400 × 3 × 10^-3

= 600 × 10^-3 J = 0.6 J

hence, energy stored in wire = 0.6J

Answer 2

Answer:

Given, area , A = 2mm² = 2 × 10^-6 m²

length of uniform steel wire, l = 3m

extension of steel wire , ∆l = 3mm = 3 × 10^-3m

Young's modulus of steel , Y = 20 × 10^10 N/m²

use formula , F = YA∆l/l

= 20 × 10^10 × 2 × 10^-6 × 3 × 10^-3/3

= 400 N

now , energy stored in the wire, dU = 1/2 × F × ∆l

= 1/2 × 400 × 3 × 10^-3

= 600 × 10^-3 J = 0.6 J

hence, energy stored in wire = 0.6J