1.3.5Work, Energy & Power

Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²)

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WHAT is potential energy?

WHY the minus sign? If a force does positive work on an object (helps it move), it spends stored energy, so UU goes down. Conversely, if you push against the force, you store energy and UU goes up. The minus sign encodes "stored energy = work done against the force."

The inverse relation (force from energy): Fx=dUdxF=UF_x = -\frac{dU}{dx} \qquad \vec F = -\nabla U WHY: force points "downhill" in the energy landscape, toward lower PE — like a ball rolling into a valley.


HOW to derive each case (from scratch)

1. Gravitational PE near Earth's surface — mghmgh

Derivation. Near the surface the gravitational force is constant: F=mgy^\vec F = -mg\,\hat y (downward). Lift an object from y=0y=0 to y=hy=h. The force does work: Wgrav=0h(mg)dy=mghW_{\text{grav}} = \int_0^h (-mg)\,dy = -mgh Then ΔU=Wgrav=(mgh)=mgh.  \Delta U = -W_{\text{grav}} = -(-mgh) = mgh.\;\checkmark Why this step? The displacement is upward (+dy+dy) while gravity points down, so Fdr<0\vec F\cdot d\vec r<0 — gravity does negative work, and we store +mgh+mgh.

2. General gravitational PE — GMmr-\dfrac{GMm}{r}

Derivation. Now the force varies: F=GMmr2r^\vec F = -\dfrac{GMm}{r^2}\hat r (attractive, inward). Move a mass from infinity to distance rr: W=r(GMmr2)dr=GMm[1r]r=GMm(1r+0)=GMmrW = \int_{\infty}^{r}\left(-\frac{GMm}{r'^2}\right)dr' = -GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = -GMm\left(-\frac1r + 0\right)=\frac{GMm}{r} U(r)=W=GMmr.  U(r) = -W = -\frac{GMm}{r}.\;\checkmark Why negative? At infinity U=0U=0. As the mass falls inward, gravity does positive work, so UU drops below zero. The object is in a "well." You'd need to add energy to climb back out to infinity.

3. Elastic (spring) PE — 12kx2\tfrac12 kx^2

Derivation. Hooke's law: the spring force is F=kxF = -kx (restoring, opposes stretch). Stretch slowly from 00 to xx: Wspring=0x(kx)dx=12kx2W_{\text{spring}} = \int_0^x (-kx')\,dx' = -\tfrac12 kx^2 U=Wspring=12kx2.  U = -W_{\text{spring}} = \tfrac12 kx^2.\;\checkmark Why x2x^2 (not xx)? Because force grows with stretch — stretching the last bit costs more than the first bit. Integrating a linear force gives a quadratic energy. Note U(+x)=U(x)U(+x)=U(-x): compressing stores the same as stretching.

Figure — Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²)

Worked Examples


Common Mistakes (Steel-manned)


Flashcards

Define potential energy in one sentence.
The negative of work done by a conservative force moving the system from a reference configuration to the current one; ΔU=Wcons\Delta U=-W_{cons}.
Why does ΔU=W\Delta U=-W carry a minus sign?
Positive work by the force spends stored energy, so UU decreases; doing work against the force stores energy.
Why do only conservative forces have a PE?
Their work is path-independent, so UU can be a single-valued function of position; friction's work depends on path.
Derive mghmgh.
ΔU=0h(mg)dy=mgh\Delta U=-\int_0^h(-mg)dy=mgh (constant downward force, upward displacement).
Derive GMm/r-GMm/r and state the reference.
U=r(GMm/r2)dr=GMm/rU=-\int_\infty^r(-GMm/r'^2)dr'=-GMm/r; reference U=0U=0 at r=r=\infty.
Why is bound gravitational PE negative?
Reference is at infinity; falling inward gravity does positive work, dropping UU below zero into a "well."
Show GMm/r-GMm/r reduces to mghmgh.
ΔUGMmh/R2=mgh\Delta U\approx GMm\,h/R^2=mgh using g=GM/R2g=GM/R^2 for hRh\ll R.
Derive elastic PE.
U=0x(kx)dx=12kx2U=-\int_0^x(-kx')dx'=\tfrac12kx^2.
Why is spring PE quadratic, not linear?
Force kxkx grows with stretch; integrating a linear force gives a quadratic energy.
How do you get force from PE?
F=dU/dxF=-dU/dx (force points toward lower PE).
Energy to escape Earth from surface?
GMm/RGMm/R (full depth of the well, since UUR=GMm/RU_\infty-U_R=GMm/R).
Whom does PE belong to?
The whole system (two interacting objects), not a single object.

Recall Feynman: explain to a 12-year-old

Imagine a slide at a playground. When you climb up, you're working to lift yourself — you store that effort. At the top you have lots of "stored go." Slide down and the stored energy turns into speed. That stored climbing effort is potential energy. A stretched rubber band is the same: you store your pull, and it snaps back. Gravity stores it when you go up (mghmgh); a spring stores it when you squish or stretch it (12kx2\tfrac12kx^2). The deep-space version is a "hole in space" around a planet — the closer you fall in, the deeper (more negative) you are, and you'd need energy to climb out.

Connections

Concept Map

stored as

defined by

path independent

path dependent, so

invert to get

force points downhill

constant near surface

inverse square law

spring force -kx

approximate for small h

reference U=0 at h=0

reference U=0 at infinity

Work against conservative force

Potential energy U

Delta U = -W_cons

Conservative force

Friction

No PE defined

F = -dU/dx

U = mgh

U = -GMm/r

U = half k x squared

Surface reference

Infinite reference

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Potential energy ka matlab hai stored work — jab aap kisi conservative force ke against kaam karte ho (book uthana, spring kheenchna), to woh energy gayab nahi hoti, configuration mein "bank" ho jaati hai. Definition yaad rakho: ΔU=Wcons\Delta U = -W_{cons}. Minus sign isliye hai kyunki agar force khud positive work karta hai, to stored energy kharch hoti hai, yaani UU kam hota hai.

Teen cases hain. Ground ke paas gravity constant hai, to integrate karke U=mghU=mgh milta hai — simple linear. Space mein gravity badalti hai (1/r21/r^2), to integrate karke U=GMm/rU=-GMm/r aata hai, jisme reference infinity par U=0U=0 rakha jaata hai — isliye bound energy negative hoti hai (ek "well" ki tarah). Aur dono agree karte hain: GMm/r-GMm/r ko chhoti height ke liye expand karo to wahi mghmgh ban jaata hai. Spring ke liye force kxkx hai (badhti hai stretch ke saath), to integrate karke U=12kx2U=\tfrac12 kx^2 — yeh ½ integration se aata hai, bhoolna mat!

Sabse important baat: PE kisi system ki hoti hai (object + Earth, ya ball + spring), akele object ki nahi. Aur sign reference point pe depend karta hai — sirf difference ΔU\Delta U physical hai. Force wapas nikalne ke liye rule hai F=dU/dxF=-dU/dx, matlab force hamesha lower PE ki taraf point karta hai, jaise ball valley mein lurhakta hai. Exams mein yeh energy conservation (K+U=K+U= constant) ke saath hamesha use hota hai, isliye derivation samajhna 80/20 ka asli faayda deta hai.

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Connections