Work, Energy and Power · Lesson 7 of 16
Spring Force and Elastic Potential Energy
“Derive spring work and energy for both extension and compression.”
• Interpret the sign of Hooke’s law. • Integrate the spring force. • Calculate energy exchange and maximum compression.
The restoring force
For a light ideal spring within its elastic range, displacement from natural length produces a restoring force. Choose +x along extension. If the spring is extended, its pull on the attached body points toward −x; if compressed, its push points toward +x. The spring constant measures stiffness: a larger constant requires more force for the same deformation. Hooke’s law is a model valid over a limited deformation range, not a rule for every real spring at any extension.
Why final force times extension is wrong
Starting at natural length, the force magnitude rises from zero to kx. Its graph is a straight line, so the magnitude of spring work is the triangular area, half of the rectangle kx times x. A slow external agent applies a force opposite to the spring force, storing energy as it stretches or compresses the spring. Slow motion means negligible kinetic-energy change; without that condition external work need not equal stored energy alone.
Problem
A spring has k = 200 N/m. Find the energy at +0.10 m and −0.10 m, then the spring work from +0.10 m to −0.10 m.
- 1.At either deformation, U = ½×200×0.10² = 1 J.
- 2.The spring work between these endpoints is Ui − Uf = 0 J.
- 3.The spring first releases 1 J on reaching natural length and then takes back 1 J during compression. Zero total work does not mean the force or energy transfer was zero throughout.
Motion between turning points
At the endpoints x = ±A, all the energy is elastic and speed is zero. At x = 0, elastic energy is minimum and speed is maximum: vmax = A√(k/m). At intermediate positions the two forms share the energy. A turning point can have zero velocity but nonzero force and acceleration. The energy equation supplies speed as a function of position; finding the period or time dependence requires additional dynamics.
A car compressing a spring
Problem
A 1000 kg car moving at 18.0 km/h meets an ideal horizontal spring with k = 5.25×10³ N/m on a smooth surface. Find maximum compression.
- 1.Convert speed: 18.0 km/h = 5.00 m/s. Initial kinetic energy is ½×1000×25 = 12500 J.
- 2.At maximum compression the car is momentarily at rest, so ½kx² = 12500.
- 3.x = √(25000/5250) ≈ 2.18 m. The model ignores spring mass, friction, and permanent deformation.
- 4.The supplied source prints 2.00 m, but that does not satisfy its stated spring constant: ½×5250×2² = 10500 J. Use 2.18 m for these given values.
Check your understanding
Quiz
A spring compressed to x = −0.2 m has U
Explanation: Its elastic energy is ½kx².
The spring force at positive x points
Explanation: It is restoring: Fs = −kx.
Slow stretching from 0 to A requires external work
Explanation: It is the triangular force-position area.
At maximum deformation, the speed is
Explanation: The mass turns back there.
Doubling spring deformation multiplies stored energy by
Explanation: Elastic energy depends on x².
Key Takeaways
• Hooke’s law has a restoring minus sign. • Integrate variable force to obtain ½kx². • Recompute numerical answers and verify the assumptions.