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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

The potential energy of a long spring when stretched by 2 cm is UU. If the spring is stretched by 8 cm, the potential energy stored in it is:

A

4U

B

8U

C

16U

D

U/4

Step-by-Step Solution

The elastic potential energy (VV) stored in a spring is proportional to the square of its displacement (xx) from its natural length, as defined by the formula V(x)=12kx2V(x) = \frac{1}{2}kx^2, where kk is the spring constant .

  1. Initial State: For a stretch of 2 cm, the energy is U=12k(2)2U = \frac{1}{2}k(2)^2.
  2. Final State: For a stretch of 8 cm, the energy is U=12k(8)2U' = \frac{1}{2}k(8)^2.
  3. Comparison: Dividing the two equations to find the relationship: UU=12k(8)212k(2)2=(82)2=42=16\frac{U'}{U} = \frac{\frac{1}{2}k(8)^2}{\frac{1}{2}k(2)^2} = \left(\frac{8}{2}\right)^2 = 4^2 = 16

Thus, U=16UU' = 16U. Doubling the stretch quadruples the energy, and in this case, quadrupling the stretch (from 2 cm to 8 cm) increases the energy by a factor of 42=164^2 = 16.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERpotentialenergyspringstretchedspring

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