Potential Energy Physics Example 2

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Example 2

medium
A spring (k=500 N/mk = 500 \text{ N/m}) is stretched 0.2 m0.2 \text{ m}. How much work was done to stretch it?

Solution

  1. 1
    The work done equals the elastic PE stored.
  2. 2
    W=12kx2=12(500)(0.04)=10 JW = \frac{1}{2}kx^2 = \frac{1}{2}(500)(0.04) = 10 \text{ J}

Answer

W=10 JW = 10 \text{ J}
The work done on a spring is stored as elastic potential energy. This energy can be released when the spring returns to its natural length.

About Potential Energy

Energy stored in a system due to the position or configuration of its parts, ready to be converted into kinetic or other forms of energy.

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