Work and Potential Energy — Energy Stored in Position 🏔️

A roller coaster climbs slowly to the top of a tall hill, pauses for a beat, then plunges downward at terrifying speed 🎢. At the top, velocity was zero — so kinetic energy was zero. But it carried hidden energy, ready to be released. That’s potential energy, and it’s captured by simple formulas for gravity and for springs.

The core idea in one paragraph 📌

Potential energy ($U$) is stored energy that depends on an object’s position or configuration, not on its motion. The two main kinds in grade 10: gravitational $U_g = mgh$ (from height in a gravitational field) and elastic $U_e = \tfrac{1}{2}kx^2$ (from a compressed or stretched spring). Subtle point: the absolute value of $U$ doesn’t matter — only $\Delta U$ (the change) does, because you always pick an arbitrary reference.

Why “work” is in the title 🔗

If you lift an object against gravity with force $F$, you do work. That work isn’t lost — it’s stored as potential energy in the object. The golden relation:

$$\boxed{W_\text{done on object} = \Delta U}$$

So potential energy is essentially “banked” work, waiting to be released later.

Type 1 — gravitational potential energy 🏔️

$$\boxed{U_g = m g h}$$

Numerical example

Lift a 2-kg object to height 3 m:
$$U_g = 2 \times 9.8 \times 3 = 58.8\ \text{J}$$

If you drop it and let it hit the ground, that 58.8 J becomes kinetic energy — impact speed:
$$v = \sqrt{\frac{2 U_g}{m}} = \sqrt{58.8} \approx 7.67\ \text{m/s}$$

Where’s the reference?

A subtlety: the absolute value of $U_g$ depends on your choice of reference (floor? ground? sea level?). But $\Delta U$ (the difference between two positions) is independent of that choice. In a problem, pick one reference and stick with it 🎯.

Type 2 — elastic potential energy 🏹

For a spring with spring constant $k$ deformed by $x$ from its natural length:

$$\boxed{U_e = \tfrac{1}{2} k x^2}$$

Hooke’s law

The spring’s restoring force: $F = -kx$ (negative because it always pushes back toward equilibrium).

Starting from this force, $U_e$ emerges as the area under the $F$–$x$ curve — a triangle of height $kx$ and base $x$, so area = $\tfrac{1}{2} kx \cdot x = \tfrac{1}{2}kx^2$. A beautiful preview of integration.

Example — a sport bow

A bow with $k = 900$ N/m drawn by $x = 0.5$ m:
$$U_e = \tfrac{1}{2}(900)(0.5)^2 = 112.5\ \text{J}$$

If a 20-g arrow (0.02 kg) captures all this energy, its exit speed:
$$v = \sqrt{\frac{2 U_e}{m}} = \sqrt{\frac{225}{0.02}} \approx 106\ \text{m/s}$$

Over 380 km/h — exactly what professional compound bows actually achieve 🏹.

Comparing the two types 📊

Property Gravitational $U_g = mgh$ Elastic $U_e = \tfrac{1}{2}kx^2$
Source Gravity Spring force (Hooke’s law)
Linear/quadratic Linear in $h$ Quadratic in $x$
Sign Positive if $h>0$ Always positive ($x^2$)
Reference Arbitrary Equilibrium (natural length)

Python analysis 🐍

1) $U_g \to K$ conversion in free fall

import numpy as np
import matplotlib.pyplot as plt

m, g = 2.0, 9.8
h_top = 5.0
h = np.linspace(h_top, 0, 100)   # top to bottom
U = m*g*h                         # potential
K = m*g*(h_top - h)               # kinetic (no friction: lower h → more K)
E_total = U + K                   # total, constant

plt.plot(h, U, label='U (potential)')
plt.plot(h, K, label='K (kinetic)')
plt.plot(h, E_total, '--', label='E_total', color='black')
plt.xlabel('Height (m)'); plt.ylabel('Energy (J)')
plt.legend(); plt.grid(); plt.gca().invert_xaxis()
plt.title('Energy conversion in free fall')
plt.savefig('free_fall_energy.png', dpi=120)

for hh in [5, 3, 1, 0]:
    U = m*g*hh
    K = m*g*(h_top - hh)
    print(f"h={hh}m → U={U:.1f}J, K={K:.1f}J, total={U+K:.1f}J")
# At every height total = 98 J

2) Hooke’s law and $U_e$ via integration

import numpy as np

k = 900     # N/m
x_max = 0.5 # m

# Analytic
U_analytic = 0.5 * k * x_max**2

# Numeric (Riemann integral of F = kx)
x = np.linspace(0, x_max, 10_000)
F = k * x
U_numeric = np.trapz(F, x)   # area under the F–x curve

print(f"Analytic: U = {U_analytic:.4f} J")
print(f"Numeric:  U = {U_numeric:.4f} J")
# Both 112.5 J

3) Compound bow — efficiency and arrow speed

import numpy as np

k = 900       # N/m
x = 0.5       # m draw
m_arrow = 0.02  # kg
efficiency = 0.85   # only ~85% is transferred to the arrow

U_stored = 0.5 * k * x**2
K_arrow = efficiency * U_stored
v_arrow = np.sqrt(2 * K_arrow / m_arrow)

print(f"Stored energy: {U_stored:.1f} J")
print(f"To the arrow:  {K_arrow:.1f} J")
print(f"Arrow speed:   {v_arrow:.1f} m/s ≈ {v_arrow*3.6:.0f} km/h")
# Speed ~98 m/s ≈ 353 km/h

Take-home summary 🎁

$U_g = mgh$ (gravitational, linear), $U_e = \tfrac{1}{2}kx^2$ (elastic, quadratic). $W_\text{done} = \Delta U$. Reference is arbitrary — only $\Delta U$ matters. $U_e$ is the area under the spring-force graph — a preview of integration. Python shows how $U$ falls and $K$ rises during free fall, with the sum constant — a foreshadowing of energy conservation 😎.

“Nice to know” box: the gravitational battery 💡

The largest “batteries” in the world actually work via gravitational potential energy. The technique is pumped hydroelectric storage:
– Nights (cheap or excess electricity): pump water into an upper reservoir → $U_g$ is stored
– Days (peak demand): release the water through turbines → generate electricity

Round-trip efficiency ~80%. Largest example: Bath County, Virginia, capacity 3,003 MW — enough to power 750,000 homes. A mountain lake IS a giant battery ⚡.

Test yourself 📝

References and further exploration 📚

Articles and reference

Videos (YouTube)

External simulators

On this site 🔗

In the next section we hit one of the most beautiful laws in physics: conservation of mechanical energy ⚖️ — energy swaps between $U$ and $K$ but the total is invariant. See you there 👋

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