Topic 3.3: Spring (Elastic) Potential Energy

Name: _______________________________ Date: _______________ Period: _______

Before We Start: Yesterday's Recap

Last class: for two masses, $U_g=-Gm_1m_2/r$ increases as they move apart. For a spring, is it always true that stretching or compressing farther increases stored energy — and what's the reference position where the stored energy is zero?

Today's New Concepts

AP CED Alignment: Unit 3, Topic 3.3

Spring Potential Energy: $U_s=\tfrac12k(\Delta x)^2$

• $\Delta x$ is measured from the spring's relaxed length; $U_s=0$ only there
• $U_s$ is positive for stretch OR compression (the displacement is squared)
• $U_s\propto x^2$: double the stretch, get 4× the energy; the graph of $U_s$ vs. $x$ is a parabola with its minimum at $x=0$

Worked Example: A block on a horizontal spring ($k=50\text{ N/m}$) is moved slowly from $x=-0.20\text{ m}$ through $0$ to $+0.20\text{ m}$. $U_s=1.0\text{ J}\to0\to1.0\text{ J}$: it decreases, then increases; net change zero.

Equal Steps, Unequal Energy

• Compress a spring by $d$ ($\Delta U_1$), then an additional $d$ ($\Delta U_2$): $\Delta U_2>\Delta U_1$
• For any starting compression $x_0$: $\Delta U_2-\Delta U_1=kd^2>0$

Worked Example: $k=200\text{ N/m}$: $U_s(0.10)=1.0\text{ J}$, $U_s(0.20)=4.0\text{ J}$. $\Delta U_1=1.0\text{ J}$, $\Delta U_2=3.0\text{ J}$, and $\Delta U_2-\Delta U_1=2.0\text{ J}=kd^2$.

Where the Formula Comes From, and Springs Plus Gravity

• $F=kx$ is a straight line from the origin; the area under it from $0$ to $x$ is a triangle: $\tfrac12(x)(kx)=\tfrac12kx^2=U_s$
• On a vertical spring at equilibrium, $kx=mg$; lowering an object onto it changes BOTH $U_g$ (decreases) and $U_s$ (increases)

Worked Example: A $0.10\text{ kg}$ egg is lowered slowly onto a vertical spring ($k=20\text{ N/m}$). $x=mg/k=0.050\text{ m}$; $\Delta U_g=-0.050\text{ J}$; $\Delta U_s=+0.025\text{ J}$.