Question

A wire is suspended by one end. At the other end a weight equivalent to 20 N force is applied. If the increase in length is 1.0 mm, the increase in energy of the wire will be

Answer 1

Solution:

Energy increased in wire per unit volume is given by,

$U = \frac{1}{2} * Stress * Strain$

Energy increased in wire is given by,

$E = \frac{1}{2} * Stress * Strain * Volume$

$Stress = \frac{Force(F)}{Area(A)}$

$Volume = Area(A) * Length(L)$

$Strain = \symup\Delta L/L$

Energy increased in wire ,

$E = \frac{1}{2} * \frac{F}{A} * \frac{\symup\Delta L}{L} * (A * L) = \frac{1}{2} * F * \symup\Delta L$

Given => F = 20 N and ∆L = 1.0 mm = 10^-3 m

Energy increased in wire ,

$E = \frac{1}{2} * F * \symup\Delta L = \frac{1}{2} * 20 * 10^{-3} = 0.01 J$

Thus, the energy increased in wire will be 0.01 J

Answer 2

The increase in energy of the wire is 0.001 Joule.

Explanation:

The energy increased by the wire is given by the formula:

E = 1/2 × stress × strain × volume

Where,

Stress = Force/Area = 20/Area

Strain = Δl/l = (1.0 × 10⁻³)/l

Volume = Area × Length = Area × l

On substituting the values, we get,

E = 1/2 × 20/Area × (1.0 × 10⁻³)/l × Area × l

E = 10 × (1.0 × 10⁻³)

∴ E = 0.001 J