Question

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200n to the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is

Answer 1

Energy = 1/2 × Force × Extension

E = 1/2 × 200N × 1 mm

E = 100N × 0.001m

Therefore, E = 0.1J

Answer 2

answer : 0.1 J

explanation : we know, energy stored in wire per unit volume is given by, U = 1/2 × stress × strain

so, energy stored in wire is given by, E = 1/2 × stress × strain × volume

we also know, stress = F/A

and volume = A × L

and strain = ∆L/L,

then, energy stored in wire , E = 1/2 × (F/A) × (∆L/L) × (A × L)

= 1/2 × F × ∆L

given, F = 200N and ∆L = 1mm = 10^-3 m

so, energy stored in wire , E = 1/2 × 200 × 10^-3 = 0.1J

hence, answer should be 0.1J