A steel wire of length 2 m and Young's modulus 2.0 × 10¹¹ N m⁻² is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10⁻³ respectively, then the elastic potential energy density of the wire is __________ × 10⁵ (in SI units).

Solution

Physics sir · 4 steps

Turn the sideways strain into the stretching strain first, then use half stress times strain.

1. Longitudinal strain. Poisson ratio is transverse strain divided by longitudinal strain. So longitudinal strain = 10⁻³/0.2 = 5 × 10⁻³.
2. Energy density formula. Energy stored per unit volume is 1/2 × stress × strain. Since stress = Y × strain, this becomes 1/2 Y (strain)². The length of 2 m is not needed.

   u = 1/2 Y ε² = 1/2 × 2.0 × 10¹¹ × (5 × 10⁻³)²

3. Put in numbers. (5 × 10⁻³)² = 25 × 10⁻⁶. Then u = 10¹¹ × 25 × 10⁻⁶ = 25 × 10⁵ J m⁻³.
4. So. The elastic potential energy density is 25 × 10⁵ J m⁻³, so the answer is 25.

The common slip: 1. Putting the transverse strain 10⁻³ straight into 1/2 Y ε² gives 1 × 10⁵, so 1 instead of 25. Convert to longitudinal strain first.

The idea to keep. Young's modulus works only with longitudinal strain. If you are given transverse strain, divide by the Poisson ratio first, then use u = 1/2 Y ε².

Answer from the NTA final answer key. Solution by PrepWiser.