If a long spring is stretched by x cm, its P.E is U. If the spring is stretched by Nx cm, the P.E stored in it will be:
Correct answer: C. N2U
- A. U/N
- B. NU
- C. N2U
- D. U /N3
Explanation
Understanding the ConceptsPotential Energy of a Spring: The potential energy (PE) stored in a spring is given by:PE=1/2kx²Where:PE is the potential energyk is the spring constantx is the displacement (stretch or compression) from the equilibrium position.Applying the FormulaInitial State:Stretch: x cmPotential Energy: UTherefore: U=1/2kx²New State:Stretch: Nx cmPotential Energy: PE′ (what we want to find)Therefore: PE′=1/2 k(Nx)²Relating the Two States:PE′=1/2kN²x²We know that U=1/2kx², so we can substitute:PE′=N²(1/2kx²)PE′=N2U
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