Concept:The elastic potential energy stored in a stretched spring follows Hooke's law, meaning it scales with the square of the displacement.
Formula:$$ PE = \frac{1}{2} k x^2 $$
Solution:- The initial potential energy is mathematically defined as \( V = \frac{1}{2} k x^2 \).
- If the new stretch distance becomes \( x' = nx \), the new potential energy is \( PE_{new} = \frac{1}{2} k (nx)^2 \).
- Distribute the square: \( PE_{new} = \frac{1}{2} k (n^2 x^2) = n^2 \left( \frac{1}{2} k x^2 \right) \).
- Substitute the initial energy \( V \) back into the equation: \( PE_{new} = n^2 V \).
Why other options are incorrect:Option A incorrectly assumes energy scales linearly with distance. Options C and D wrongly place the scaling factor in the denominator.
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.