$F_s = -kx$

where:

• $F_s$ = spring force (N)
• $k$ = spring constant (N/m) - measure of spring's stiffness
• $x$ = displacement from equilibrium position (m)
• Negative sign means force opposes displacement (restoring force)

💡 Hooke's Law: Named after Robert Hooke, this is valid for springs in their elastic region (not stretched/compressed too far).

Understanding the Spring Force

Equilibrium Position

The natural length of the spring - no force exerted ($x = 0$)

Compression ($x < 0$)

• Spring is squeezed
• Force pushes outward (away from compression)

Extension ($x > 0$)

• Spring is stretched
• Force pulls inward (back toward equilibrium)

Spring Constant $k$

• Large $k$: Stiff spring (hard to stretch/compress)
• Small $k$: Soft spring (easy to stretch/compress)
• Units: N/m (newtons per meter)

Elastic Potential Energy

Energy stored in a compressed or stretched spring:

$PE_{elastic} = \frac{1}{2}kx^2$

where:

• $PE_{elastic}$ (or $U_s$) = elastic potential energy (J)
• $k$ = spring constant (N/m)
• $x$ = displacement from equilibrium (m)

Key Properties

1. Always positive - $x^2$ is always positive (compressed or stretched)
2. Quadratic relationship - Double displacement = 4× energy
3. Minimum at equilibrium - PE = 0 when $x = 0$
4. Symmetric - Same energy whether compressed or stretched by $|x|$

Work Done by/on a Spring

Work done BY spring (spring releases)

$W_{by\,spring} = -\Delta PE_{elastic} = -\left(\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2\right)$

Work done ON spring (you compress/stretch it)

$W_{on\,spring} = \Delta PE_{elastic} = \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2$

Special case: From equilibrium to $x$

$W = \int_0^x F\,dx = \int_0^x kx'\,dx' = \frac{1}{2}kx^2$

This equals the elastic PE stored.

Why Is PE = ½kx², Not kx?

The spring force is variable: $F = kx$ (increases with displacement).

Average force from 0 to $x$: $F_{avg} = \frac{0 + kx}{2} = \frac{kx}{2}$

Work = Average force × distance: $W = F_{avg} \cdot x = \frac{kx}{2} \cdot x = \frac{1}{2}kx^2$

Alternatively, work is the area under the $F$ vs. $x$ graph:

• Triangle with base $x$ and height $kx$
• Area = $\frac{1}{2} \cdot base \cdot height = \frac{1}{2}kx^2$

Conservation of Energy with Springs

For a mass on a spring (horizontal, no friction):

$E_{total} = KE + PE_{elastic} = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = constant$

• At maximum compression/extension: $v = 0$, all energy is elastic PE
• At equilibrium: $x = 0$, all energy is kinetic

Spring Combinations

Springs in Series (end-to-end)

Effective spring constant is smaller:

$\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$

Easier to stretch than individual springs.

Springs in Parallel (side-by-side)

Effective spring constant is larger:

$k_{eff} = k_1 + k_2$

Harder to stretch than individual springs.

⚠️ Common Mistakes

Mistake 1: Forgetting the ½

❌ Wrong: $PE = kx^2$ ✅ Right: $PE = \frac{1}{2}kx^2$

Mistake 2: Using Wrong x

$x$ is displacement from equilibrium, not from one end of possible motion!

Mistake 3: Thinking PE Can Be Negative

Elastic PE is always ≥ 0 (because of $x^2$)

Mistake 4: Sign in Hooke's Law

$F = -kx$ has a negative sign (restoring force), but many problems just use magnitude: $|F| = k|x|$

Problem-Solving Strategy

1. Identify equilibrium position where $x = 0$
2. Measure displacement $x$ from equilibrium
3. For force: Use $F = kx$ (magnitude) with direction toward equilibrium
4. For energy: Use $PE = \frac{1}{2}kx^2$
5. Apply conservation of energy if appropriate

Applications

Vehicle Suspension

Springs absorb bumps, converting KE to elastic PE and back.

Pogo Sticks

Compression of spring stores PE, which launches you upward (converts to gravitational PE and KE).

Trampolines

Springs and elastic mat store PE when compressed, releasing it to bounce you up.

Molecular Bonds

Atoms in molecules act like masses connected by springs (vibrating).

Comparison: Elastic vs. Gravitational PE

Key Formulas Summary

Find: Maximum speed $v_{max}$

Analysis:

Maximum speed occurs at equilibrium position ($x = 0$) where all elastic PE converts to KE.

Step 1: Calculate initial energy (at maximum compression)

At $x = 0.2$ m:

• $KE_i = 0$ (released from rest)
• $PE_i = \frac{1}{2}kx_i^2 = \frac{1}{2}(100)(0.2)^2 = 50(0.04) = 2$ J

Total initial energy: $E_i = 2$ J

Step 2: Calculate final energy (at equilibrium)

At $x = 0$:

• $KE_f = \frac{1}{2}mv_{max}^2$ (unknown)
• $PE_f = \frac{1}{2}k(0)^2 = 0$ J

Total final energy: $E_f = \frac{1}{2}mv_{max}^2$

Step 3: Apply conservation of energy

$E_i = E_f$

$2 = \frac{1}{2}(0.5)v_{max}^2$

$2 = 0.25v_{max}^2$

$v_{max}^2 = 8$

$v_{max} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \text{ m/s}$

Alternative formula:

For a spring-mass system released from rest at maximum displacement:

$v_{max} = x_{max}\sqrt{\frac{k}{m}} = 0.2\sqrt{\frac{100}{0.5}} = 0.2\sqrt{200} = 0.2(10\sqrt{2}) = 2\sqrt{2}$ m/s ✓

Answer: The maximum speed is $2\sqrt{2}$ m/s or approximately 2.83 m/s.

(a) Find total mechanical energy

Step 1: Choose reference for gravitational PE

Let equilibrium position be $h = 0$ for gravitational PE.

Step 2: Calculate energy at release point

At release (0.1 m below equilibrium):

• $KE_i = 0$ (released from rest)
• $PE_{elastic,i} = \frac{1}{2}kx_i^2 = \frac{1}{2}(500)(0.1)^2 = 250(0.01) = 2.5$ J
• $PE_{grav,i} = mgh_i = (2)(9.8)(-0.1) = -1.96$ J (negative because below equilibrium)

Total energy: $E = 0 + 2.5 + (-1.96) = 0.54 \text{ J}$

(b) Find maximum height above release point

Step 3: At maximum height

At highest point, block momentarily stops ($v = 0$) and spring returns through equilibrium and compresses.

Let's say spring compresses by distance $d$ above equilibrium.

At this point:

• $KE = 0$
• $PE_{elastic} = \frac{1}{2}kd^2$
• $PE_{grav} = mg(d)$ (height $d$ above equilibrium, which is $d + 0.1$ above release)

Step 4: Apply conservation of energy

$E_{initial} = E_{final}$

$0.54 = 0 + \frac{1}{2}kd^2 + mgd$

$0.54 = 250d^2 + 19.6d$

$250d^2 + 19.6d - 0.54 = 0$

Step 5: Solve quadratic

Using quadratic formula: $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$d = \frac{-19.6 \pm \sqrt{(19.6)^2 - 4(250)(-0.54)}}{2(250)}$

$d = \frac{-19.6 \pm \sqrt{384.16 + 540}}{500}$

$d = \frac{-19.6 \pm \sqrt{924.16}}{500}$

$d = \frac{-19.6 \pm 30.4}{500}$

Taking positive solution: $d = \frac{10.8}{500} = 0.0216 \text{ m} = 2.16 \text{ cm}$

Height above release point: $h = d + 0.1 = 0.0216 + 0.1 = 0.1216$ m

Simplified approach (if we ignore spring compression at top):

If spring just returns to equilibrium ($d = 0$):

$0.54 = 0 + 0 + mg(h)$ $h = \frac{0.54}{19.6} \approx 0.0276$ m

But spring actually compresses slightly, giving total height ≈ 0.122 m.

Answers:

• (a) Total mechanical energy: 0.54 J
• (b) Height above release point: approximately 0.122 m or 12.2 cm

Note: This problem is complex because both gravitational and elastic PE change. The spring compresses slightly above equilibrium before the block stops.