When a spring is stretched by 2 cm, it stores elastic potential energy based on Hooke's law, which states that the force exerted by the spring is proportional to its displacement from equilibrium. This energy is calculated using the formula U=12kx2U=\frac{1}{2}kx^2U=21​kx2, where kkk is the spring constant and xxx is the stretch distance (here, 0.02 m).

Physics Behind Spring Stretch

Springs follow Hooke's law (F=−kxF=-kxF=−kx) up to their elastic limit, converting mechanical work into stored potential energy. Stretching by 2 cm (a small deformation) typically remains within this linear range for most springs, avoiding permanent damage. This concept appears frequently in physics problems, like those calculating total energy after further extension.

Common Problem Example

A classic question asks: If a spring stretched by 2 cm stores 100 J, what happens if stretched further by another 2 cm (total 4 cm)?

  • Initial energy: U1=12k(0.02)2=100U_1=\frac{1}{2}k(0.02)^2=100U1​=21​k(0.02)2=100 J, so k=5×105k=5\times 10^5k=5×105 N/m.
  • Final energy: U2=12k(0.04)2=400U_2=\frac{1}{2}k(0.04)^2=400U2​=21​k(0.04)2=400 J.
  • Increase: ΔU=400−100=300\Delta U=400-100=300ΔU=400−100=300 J, since energy scales with x2x^2x2.

This quadratic relationship means doubling the stretch quadruples the energy.

Real-World Applications

  • Shock Absorbers : Car suspensions use springs stretched/compressed by centimeters to store and release energy smoothly.
  • Engineering Tests : Forums discuss how repeated stretching affects spring constants, noting coil diameter changes under high loads.
  • Trending Discussions : Recent Reddit threads (as of late 2025) explore spring behavior in RC cars and physics homework, confirming energy calculations.

Variations in Problems

Different scenarios alter outcomes:

Stretch Scenario| Initial Energy| New Stretch| New Total Energy| Increase
---|---|---|---|---
2 cm to 4 cm| 100 J 1| +2 cm| 400 J 1| 300 J
2 cm to 10 cm| U 10| +8 cm| 25U 10| 24U
Long spring (2 cm)| U 6| 10 cm| 25U 6| Varies

TL;DR : Stretching a spring by 2 cm stores energy per 12kx2\frac{1}{2}kx^221​kx2; further stretching increases it quadratically, often by 300 J in standard problems from 100 J baseline.

Information gathered from public forums or data available on the internet and portrayed here.