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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

The potential energy of a long spring when stretched by 2 cm2\text{ cm} is UU. If the spring is stretched by 8 cm8\text{ cm}, the potential energy stored in it will be:

A

16U

B

2U

C

4U

D

8U

Step-by-Step Solution

  1. Identify the Formula: The potential energy (UU) stored in a spring of spring constant kk stretched by a distance xx is given by U=12kx2U = \frac{1}{2}kx^2 [Class 11 Physics, Ch 5, Sec 5.9, Eq 5.15].
  2. Establish Relationship: Since kk is constant for a given spring, the potential energy is directly proportional to the square of the extension (Ux2U \propto x^2).
  3. Calculate Ratio: Given U1=UU_1 = U for x1=2 cmx_1 = 2\text{ cm}. We need to find U2U_2 for x2=8 cmx_2 = 8\text{ cm}. U2U1=12kx2212kx12=(x2x1)2\frac{U_2}{U_1} = \frac{\frac{1}{2}kx_2^2}{\frac{1}{2}kx_1^2} = \left(\frac{x_2}{x_1}\right)^2 U2U=(82)2=(4)2=16\frac{U_2}{U} = \left(\frac{8}{2}\right)^2 = (4)^2 = 16
  4. Final Result: U2=16UU_2 = 16U

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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