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Spring Potential Energy

Easyphysics

A spring with force constant 400 N/m is compressed by 0.1 m and used to launch a small block on a frictionless table. How much elastic potential energy is stored in the compressed spring before release?

Select the correct option:

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About This Question

Subject
physics
Chapter
work, energy and power
Topic
spring potential energy
Difficulty
Easy
Year
2025
Tags
spring potential energyHooke's lawhalf k x squaredelastic energyspring constant

Solution

Correct Answer:

2 J

As given in NCERT Class 11, Chapter 6 (Work, Energy and Power), the elastic potential energy stored in a spring obeying Hooke's law is \cup = (1/2)kx², where k is the spring constant and x is the deformation from its natural length. Substituting, \cup = (1/2)(400)(0.1²) = (1/2)(400)(0.01) = 2 J. This stored energy will be converted into kinetic energy of the block once the spring is released. The option 4 J omits the factor of one-half. The option 40 J mistakenly uses x rather than x squared. The option 0.4 J misplaces the decimal in squaring 0.1. A plausibility check: squaring the small compression of 0.1 m gives 0.01, which keeps the stored energy modest at 2 J, consistent with a gently compressed spring on a tabletop.

This easy difficulty physics question is from the chapter work, energy and power, covering the topic of spring potential energy. It appeared in the 2025 exam.

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