Topic 2.8 Spring Forces: apply Hooke's law to relate the force from an ideal spring to its displacement, and use it in equilibrium and dynamics problems.

What this topic is asking

The College Board (Topic 2.8) wants you to apply Hooke's law to find the force from an ideal spring, to interpret the spring constant, and to use the spring force in equilibrium and Newton's-second-law problems. The defining feature of a spring force is that it is a restoring force: it always points back toward the spring's natural length, which is what makes springs oscillate.

Hooke's law

An "ideal" spring obeys this linear law exactly, no matter the direction of displacement: stretch it or compress it by the same distance and you get the same magnitude of force. Real springs follow Hooke's law well within their elastic limit.

The spring constant and the restoring force

This restoring behavior is the physical reason springs oscillate. Displace a mass on a spring and the spring force pushes it back toward equilibrium; it overshoots, the spring force reverses, and the mass swings back and forth. The further you displace it, the larger the restoring force, which is the signature of Hooke's law.

Using spring forces in problems

Spring forces enter free-body diagrams and Newton's laws like any other force:

• In equilibrium, the spring force balances the others. A hanging mass settles where $kx = mg$, so the equilibrium extension is $x = \dfrac{mg}{k}$.
• Away from equilibrium, the spring force contributes to the net force, and $F_{net} = ma$ gives the acceleration at that instant.

The procedure is unchanged from earlier topics: draw the diagram, include the spring force (magnitude $kx$, directed toward the natural length), and apply equilibrium or the second law.

Why the restoring force leads to oscillation

The reason this topic sits at the end of Unit 2 is that the spring force ties together everything before it and points ahead to simple harmonic motion. Because the restoring force grows linearly with displacement and always points back toward equilibrium, a mass on a spring does not just return to equilibrium and stop; it accelerates most strongly when furthest out, races through the equilibrium point at top speed, and decelerates as the spring on the other side resists, producing a smooth back-and-forth oscillation. The same restoring-force idea explains why a vertical spring with a hanging mass oscillates about its stretched equilibrium position rather than its natural length: gravity simply shifts where equilibrium sits, and the spring force still provides a linear restoring force about that new point. Being comfortable that "spring force equals $kx$ toward equilibrium" lets you handle both the static question (where does it hang?) and the dynamic question (what is the net force when displaced?), which are the two ways the exam probes this topic. The energy stored in a spring, $\tfrac{1}{2}kx^2$, builds on this same displacement in Unit 3.

Try this

Q1. A spring of constant $150$ N/m is stretched $0.20$ m. Calculate the spring force. [2 points]

• Cue. $F = kx = (150)(0.20) = 30$ N, directed back toward the natural length.

Q2. A $1.0$ kg mass hangs in equilibrium from a spring of constant $100$ N/m. Calculate the extension ($g = 9.8$ m/s squared). [2 points]

• Cue. $x = \dfrac{mg}{k} = \dfrac{(1.0)(9.8)}{100} = 0.098$ m.