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A 20 m long steel wire (cross-section 1 cm², Young's modulus 2x10¹¹ N/m²), is subjected to a load of 25,000 N. How much will the wire stretch under this load?

a. .25 cm
b. 2.5 cm
c. 12.5 cm
d. 25 cm
e. 1.25 cm

1 Answer

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Final answer:

The steel wire subjected to a 25,000 N load will stretch by 2.5 cm under the given conditions, which corresponds to option (b).

Step-by-step explanation:

The student is asking about the stretching of a steel wire under a given load, a problem which can be solved using Hooke's Law and the concept of Young's modulus.

To determine the amount by which the wire stretches, we use the formula ΔL = (F × L) / (A × E), where ΔL is the change in length, F is the force applied, L is the original length, A is the cross-sectional area, and E is Young's modulus of the material. Substituting the given values: ΔL = (25,000 N × 20 m) / (1 cm² × 2×10¹¹ N/m²).

Converting the cross-sectional area to meters squared (1 cm² = 1×10¹´ m²) gives us: ΔL = (25,000 N × 20 m) / (1×10¹´ m² × 2×10¹¹ N/m²) = 2.5 cm, which is option (b).

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