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For an ideal spring obeying Hooke's law, with force constant k, the force needed to stretch it a distance x grows linearly from zero to kx. Using the fact that the work done equals the area under the force-versus-distance graph, what is the potential energy stored in the spring?

APEs equals k times x, no more
BPEs equals k times x squared exactly
CPEs equals one half times k times x squared
DPEs equals one half times k times x
Answer & Solution
Correct answer: C. PEs equals one half times k times x squared
1. Since the force grows linearly from 0 to kx, the average force during the stretch is kx divided by 2. 2. The work done equals this average force multiplied by the total distance x, giving W = (kx/2)(x) = one half k x squared. 3. This work done in stretching the spring is stored as its potential energy, so PEs = one half k x squared. 4. Leaving out the factor of one half, or using only the first power of x, does not match the triangular area under the linearly rising force-versus-distance graph. _Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.4 Conservative Forces and Potential Energy_
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