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 × 10¹⁰ N/m²)

Answer 1

Given: L = 3 m, A = 2 mm² = 2 × 10^-6 m²,

l = 3 mm = 3 × 10^-3 m,

Y_steel = 20 × 10¹⁰ N/m²

To find: Energy stored (U)

Formula: W = \(\frac{1}{2}\) × F × l

Calculation: Since, Y = \(\frac{FL}{Al}\)

∴ F = \(\frac{YAl}{L}\)

From formula,

The energy stored in the steel wire is 0.6 J.