Intuition The big pictureWhen you stretch or compress a spring, you do work against the spring's restoring force. That work doesn't vanish — it gets stored inside the spring as elastic potential energy, ready to spring back. The tricky part: the force is not constant (the more you stretch, the harder it pulls), so we can't just write W = F × d W = F\times d W = F × d F F F add up tiny bits of work — that's an integral.
Definition Spring potential energyThe energy stored in an ideal spring stretched or compressed by a displacement x x x U = 1 2 k x 2 U = \tfrac{1}{2}kx^2 U = 2 1 k x 2 k k k spring constant (stiffness, units N/m \text{N/m} N/m x x x m \text{m} m
Intuition Hooke's law makes force position-dependentAn ideal spring obeys Hooke's law : the restoring force is
F spring = − k x F_{\text{spring}} = -kx F spring = − k x back toward equilibrium . At x = 0 x=0 x = 0 x = 0.1 m x = 0.1\,\text{m} x = 0.1 m F F F x x x F × d F\times d F × d triangle under the force graph .
We compute the work you do to stretch the spring slowly from 0 0 0 x x x F applied = + k x F_{\text{applied}} = +kx F applied = + k x
Step 1 — chop the motion into tiny pieces.
Over a tiny displacement d x ′ dx' d x ′ k x ′ kx' k x ′ effectively constant .
Why this step? Work-as-force-times-distance only works when force is constant. Making the slice infinitesimal makes that true.
Step 2 — work done on one slice.
d W = F applied d x ′ = k x ′ d x ′ dW = F_{\text{applied}}\,dx' = kx'\,dx' d W = F applied d x ′ = k x ′ d x ′
Why this step? This is just W = F d W=F\,d W = F d F F F
Step 3 — add all the slices (integrate) from 0 0 0 x x x
W = ∫ 0 x k x ′ d x ′ = k ∫ 0 x x ′ d x ′ = k [ x ′ 2 2 ] 0 x = 1 2 k x 2 W = \int_0^x kx'\,dx' = k\int_0^x x'\,dx' = k\left[\frac{x'^2}{2}\right]_0^x = \frac{1}{2}kx^2 W = ∫ 0 x k x ′ d x ′ = k ∫ 0 x x ′ d x ′ = k [ 2 x ′2 ] 0 x = 2 1 k x 2
Why this step? The total work is the sum of every infinitesimal d W dW d W
Step 4 — equate work done to stored energy.
Since you did this work against a conservative force and the spring now holds it,
U = 1 2 k x 2 \boxed{U = \tfrac{1}{2}kx^2} U = 2 1 k x 2
Why this step? For a conservative force, work done against it = potential energy gained. No energy is lost to heat in an ideal spring.
Worked example Predict before you computeA spring has k = 200 N/m k = 200\,\text{N/m} k = 200 N/m Forecast: if you double the stretch from 0.1 m 0.1\,\text{m} 0.1 m 0.2 m 0.2\,\text{m} 0.2 m
Guess first... Many say "doubles." Now verify:
U 1 = 1 2 ( 200 ) ( 0.1 ) 2 = 1 J U_1 = \tfrac12(200)(0.1)^2 = 1\,\text{J} U 1 = 2 1 ( 200 ) ( 0.1 ) 2 = 1 J U 2 = 1 2 ( 200 ) ( 0.2 ) 2 = 4 J U_2 = \tfrac12(200)(0.2)^2 = 4\,\text{J} U 2 = 2 1 ( 200 ) ( 0.2 ) 2 = 4 J
Energy went up 4× , not 2×. Why? Because U ∝ x 2 U\propto x^2 U ∝ x 2 x x x U U U
Worked example Worked example — compressionA spring (k = 500 N/m k=500\,\text{N/m} k = 500 N/m compressed 0.04 m 0.04\,\text{m} 0.04 m U = 1 2 ( 500 ) ( 0.04 ) 2 = 1 2 ( 500 ) ( 0.0016 ) = 0.4 J U = \tfrac12 (500)(0.04)^2 = \tfrac12(500)(0.0016) = 0.4\,\text{J} U = 2 1 ( 500 ) ( 0.04 ) 2 = 2 1 ( 500 ) ( 0.0016 ) = 0.4 J Why this step? We square x x x
Worked example Worked example — launching a blockThe compressed spring above releases a 0.2 kg 0.2\,\text{kg} 0.2 kg 1 2 k x 2 = 1 2 m v 2 ⇒ 0.4 = 1 2 ( 0.2 ) v 2 \tfrac12 kx^2 = \tfrac12 m v^2 \Rightarrow 0.4 = \tfrac12(0.2)v^2 2 1 k x 2 = 2 1 m v 2 ⇒ 0.4 = 2 1 ( 0.2 ) v 2 v 2 = 0.4 0.1 = 4 ⇒ v = 2 m/s v^2 = \frac{0.4}{0.1}=4 \Rightarrow v = 2\,\text{m/s} v 2 = 0.1 0.4 = 4 ⇒ v = 2 m/s Why this step? Energy conservation: stored elastic energy becomes motion energy with no losses.
Common mistake "Work is justF × x = k x ⋅ x = k x 2 F \times x = kx \cdot x = kx^2 F × x = k x ⋅ x = k x 2 Why it feels right: F = k x F=kx F = k x x x x Why it's wrong: k x kx k x maximum force, reached at the very end. At the start the force was zero. Using the final force everywhere over-counts by a factor of 2.
Fix: Use the average force 0 + k x 2 = k x 2 \frac{0+kx}{2}=\frac{kx}{2} 2 0 + k x = 2 k x W = F ˉ x = 1 2 k x 2 W=\bar{F}x=\tfrac12 kx^2 W = F ˉ x = 2 1 k x 2
Common mistake "Compression stores negative energy"Why it feels right: x x x Fix: U = 1 2 k x 2 U=\tfrac12 kx^2 U = 2 1 k x 2 x x x squared → always ≥ 0 \ge 0 ≥ 0
Common mistake Forgettingx x x Why it feels right: People plug in total length of spring.
Fix: x x x displacement from the natural (relaxed) length , not the absolute length.
Recall Feynman: explain to a 12-year-oldImagine pulling a rubber band. At first it's easy, but the more you pull, the harder it tugs back. So the "pulling work" you do is small at the start and big at the end. To find the total work, you can't use the hardest pull for the whole way — you'd be cheating. Instead use the average between the easy start (zero) and the hard end. Half of "stiffness times stretch times stretch" — that's all the energy the rubber band is now holding, waiting to snap back and fling something.
"Half-k-x-squared" — sing it like a chant. And: spring energy is a triangle , not a rectangle (area = ½·base·height). The "½" is the triangle.
Why can't we use W = F x W=Fx W = F x
What's the area under the F F F x x x
If x x x U U U
Why is spring force not constant? By Hooke's law
F = − k x F=-kx F = − k x , force grows linearly with displacement from equilibrium.
What is the spring potential energy formula? U = 1 2 k x 2 U=\tfrac12 kx^2 U = 2 1 k x 2 .
Derive U U U F = − k x F=-kx F = − k x W = ∫ 0 x k x ′ d x ′ = 1 2 k x 2 W=\int_0^x kx'\,dx'=\tfrac12kx^2 W = ∫ 0 x k x ′ d x ′ = 2 1 k x 2 ; this equals stored PE for a conservative force.
Why a factor of ½ in 1 2 k x 2 \tfrac12kx^2 2 1 k x 2 Average force is
k x 2 \tfrac{kx}{2} 2 k x (linear from 0 to
k x kx k x ); equivalently area of triangle under
F F F –
x x x .
Does compression store positive or negative energy? Positive —
x x x is squared, so
U ≥ 0 U\ge0 U ≥ 0 for both stretch and compression.
If stretch doubles, how does stored energy change? It quadruples, since
U ∝ x 2 U\propto x^2 U ∝ x 2 .
From where is x x x From the spring's natural (relaxed/equilibrium) length.
What does the area under the force-displacement graph represent? The work done = energy stored.
Triangle area under F-x line
Same energy stretch or compress
Intuition Hinglish mein samjhoDekho, spring ko stretch ya compress karte ho toh tum uske restoring force ke against kaam (work) karte ho, aur wo kaam spring ke andar elastic potential energy ban kar store ho jaata hai. Problem yeh hai ki spring ka force constant nahi hota — Hooke's law kehta hai F = − k x F = -kx F = − k x W = F × d W = F \times d W = F × d
Trick yeh hai ki hum motion ko chhote-chhote tukdon (slices) mein toad dete hain. Itne chhote ki har slice mein force almost constant lage. Ek slice ka kaam d W = k x ′ d x ′ dW = kx'\,dx' d W = k x ′ d x ′ 0 0 0 x x x U = 1 2 k x 2 U = \tfrac12 kx^2 U = 2 1 k x 2 F F F x x x k x 2 \tfrac{kx}{2} 2 k x
Important baat: x x x natural length se naapo, total length se nahi. Aur kyunki x x x x 2 x^2 x 2
