The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is
Potential energy is the stored energy in an object due to its position or configuration. It is primarily associated with gravitational force, elastic materials, and electric charges. The higher an object is positioned or the more it is stretched, the greater its potential energy, allowing it to do work when released.
Class 11 Physics covers Chapter 5 on Work, Energy and Power, focusing on the definitions and principles of work, energy transformations and the concept of power. Students learn about kinetic and potential energy, conservation laws and real-world applications, preparing them for examinations and enhancing their understanding of fundamental physics concepts.
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To determine the potential energy in a stretched spring, we have the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension (or compression) of the spring from its equilibrium position
Given:
– When the spring is stretched by x₁ = 2 cm:
U = (1/2) k (2 cm)² = (1/2) k (0.02 m)²
– When the spring is stretched by x₂ = 8 cm:
U’ = (1/2) k (8 cm)² = (1/2) k (0.08 m)²
Step 1: Calculate the ratio of potential energies
We are required to find the ratio of the potential energies when stretched by 8 cm compared to when stretched by 2 cm:
(U’) / (U) = [(1/2) k (0.08 m)²] / [(1/2) k (0.02 m)²] = [(0.08)²] / [(0.02)²]
Step 2: Simplify the ratio
Compute the squares:
(U’) / (U) = [(0.08)²] / [(0.02)²] = [0.0064] / [0.0004] = 16
Conclusion
The potential energy in the spring stretched by 8 cm is given by:
U’ = 16 U
For more solutions:
https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/