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

WHAT to notice: the lift height $0.80\text{m}$ is microscopic compared to Earth's radius $6.371\times 10^6\text{m}$. Over that tiny range gravity is essentially constant, so we use the flat-Earth formula $U=mgh$, not $-GMm/r$. WHY $mgh$: the general well $-GMm/r$ curves, but zoom in on a few metres and the curve looks straight — a constant downward pull $mg$. That is exactly the assumption behind $mgh$. $\Delta U=mgh=3\times 9.8\times 0.80=23.52\text{J}$ Positive because we lifted against gravity, banking energy.

WHAT: $G,M,m,r$ are all positive, and there is a leading minus sign, so $U(r)<0$ — negative. WHY negative: the reference $U=0$ is chosen at $r=\infty$ (infinitely far, no interaction). Any real orbit is closer than infinity, sitting inside the gravitational "well," so it is below zero. $U=0$ would mean the satellite has just barely escaped to infinite distance with nothing left over. Answer: negative; $U=0$ corresponds to being infinitely far away.

WHAT: $U=\frac{1}{2}k{x}^2$ depends on $x^2$, and $(-0.05{)}^2=(+0.05{)}^2$. WHY: squaring erases the sign, so stretch and compression by equal amounts store the same energy. The spring does not care about direction, only how far from natural length. Answer: $0.4\text{J}$ — identical. The implied $k$: solve $0.4=\frac{1}{2}k(0.05{)}^2=\frac{1}{2}k(0.0025)$, giving $k=0.4/(0.00125)=320\text{N/m}$.

Step 1 — store: $U=\frac{1}{2}k{x}^2=\frac{1}{2}(500)(0.08{)}^2=\frac{1}{2}(500)(0.0064)=1.6\text{J}$. Step 2 — convert: frictionless, so all PE becomes kinetic energy: $\frac{1}{2}m{v}^2=1.6$. WHY conservation applies: no non-conservative force removes energy (see Conservation of mechanical energy). $v=\sqrt{\frac{2\times 1.6}{0.20}}=\sqrt{16}=4\text{m/s}$ Answer: $U=1.6\text{J}$, $v=4\text{m/s}$.

WHAT: the whole well depth must be paid: $\Delta U=U_{\infty }-U_R=0-\left(-\frac{GMm}{R}\right)=\frac{GMm}{R}$. $\Delta U=\frac{(6.674\times 10^{-11})(5.972\times 10^{24})(1200)}{6.371\times 10^6}\approx 7.506\times 10^{10}\text{J}$ WHY this is the full well: at infinity the interaction vanishes ($U=0$); on the surface $U$ is at its most negative, so climbing out costs exactly $GMm/R$. This is why Escape velocity satisfies $\frac{1}{2}m{v}_e^2=GMm/R$.

WHAT: use the master rule $F=-\frac{dU}{dx}$ from Force from potential — F = -dU/dx. $\frac{dU}{dx}=8x-3\Rightarrow F=-(8x-3)=3-8x$ At $x=2$: $F=3-16=-13\text{N}$. WHY the minus: force points downhill in the energy landscape, toward lower $U$. Negative $F$ means it pushes toward $-x$.

(a) Flat-Earth: $\Delta U=mgh=10\times 9.8\times 200=19600\text{J}$. (b) Exact: $\Delta U=-GMm\left(\frac{1}{R+h}-\frac{1}{R}\right)$ with $R=6.371\times 10^6$, $h=200$: $\Delta U\approx 1.9636\times 10^4\text{J}\approx 19636\text{J}$ WHAT this shows: the two differ by about $0.18\%$ — negligible for a $200\text{m}$ lift. WHY they agree: for $h\ll R$, Taylor-expanding gives $\Delta U\approx GMmh/R^2=(GM/R^2)mh=mgh$, since $g=GM/R^2$. $mgh$ is literally the first term of the exact expression.

What the figure below shows: the blue curve is the full well $U(r)=-GMm/r$, which bends upward toward $U=0$ (the dotted yellow line) as $r\to \infty$. The pink straight line is the tangent to that curve at the surface $r=R$ — its slope is the local $mgh$ rule. Over a tiny height like $200\text{m}$ the blue curve and the pink tangent are visually indistinguishable, which is why $mgh$ works near the ground: you are riding the straight-line approximation of a gently curving well.

WHAT: force is zero where $\frac{dU}{dx}=0$ (flat spot in the landscape). $\frac{dU}{dx}=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)=0\Rightarrow x=1,3$ Classify with the second derivative $\frac{d^{2}U}{d{x}^2}=6x-12$:

• At $x=1$: $6(1)-12=-6<0$ → a maximum of $U$ → unstable (ball rolls away).
• At $x=3$: $6(3)-12=+6>0$ → a minimum of $U$ → stable (ball settles back). WHY: a valley bottom ($U$ min) pushes a nudged bead back; a hilltop ($U$ max) pushes it further away.

What the figure below shows: the blue curve is $U(x)$ plotted across $x$. The pink dot at $x=1$ sits on a local hilltop — nudge the bead either way and $U$ drops, so it accelerates away: unstable. The yellow dot at $x=3$ sits in a local valley — nudge it and $U$ rises on both sides, so the force pushes it back: stable. The picture makes concrete why the sign of the second derivative (curving down vs curving up) is exactly the stability test.

Stored PE: $U=\frac{1}{2}(300)(0.10{)}^2=1.5\text{J}$. Momentum: starts at rest, so total momentum stays zero: $m_1{v}_1=m_2{v}_2$, i.e. $0.30v_1=0.60v_2\Rightarrow v_1=2v_2$. Energy: $\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=1.5$. Substitute $v_1=2v_2$: $\frac{1}{2}(0.30)(2v_2{)}^2+\frac{1}{2}(0.60)v_2^2=0.60v_2^2+0.30v_2^2=0.90v_2^2=1.5$ $v_2=\sqrt{1.5/0.90}=\sqrt{5/3}\approx 1.291\text{m/s},v_1=2v_2\approx 2.582\text{m/s}$ WHY both laws: energy alone gives one equation for two unknowns; momentum conservation supplies the second. The lighter block flies off faster.

WHAT: at maximum compression the ball is momentarily at rest, so all gravitational PE lost equals spring PE stored. The ball falls a total height $h+x$ (the drop plus the compression). Energy balance (with $U=0$ at the spring's top): $mg(h+x)=\frac{1}{2}k{x}^2$ $0.50\times 9.8\times (1.2+x)=\frac{1}{2}(800)x^2$ $4.9(1.2+x)=400x^2\Rightarrow 400x^2-4.9x-5.88=0$ Solve the quadratic: $x=\frac{4.9\pm \sqrt{4.9^2+4(400)(5.88)}}{2(400)}=\frac{4.9\pm \sqrt{9432.01}}{800}$ $x\approx \frac{4.9\pm 97.12}{800}\Rightarrow x\approx +0.1275\text{m}\text{ or }x\approx -0.1152\text{m}$ WHY discard the negative root: $x$ is a compression distance, a physical length the spring squashes down — it must be positive. The negative root would mean the spring stretches upward, but a dropped ball only pushes the spring down, so that root is unphysical and we keep $x\approx +0.128\text{m}$.

Escape speed check: $v_e=\sqrt{2GM/R}=\sqrt{2(6.674\times 10^{-11})(5.972\times 10^{24})/6.371\times 10^6}\approx 11186\text{m/s}$. Since $8000<11186$, it does not escape — good, a finite $r_{\max }$ exists. Energy conservation (KE + PE conserved; at $r_{\max }$ speed = 0): $\frac{1}{2}m{v}_0^2-\frac{GMm}{R}=0-\frac{GMm}{r_{\max }}$ Cancel $m$, solve for $r_{\max }$: $\frac{GM}{r_{\max }}=\frac{GM}{R}-\frac{1}{2}{v}_0^2\Rightarrow r_{\max }=\frac{GM}{\frac{GM}{R}-\frac{1}{2}{v}_0^2}$ Numbers: $GM=3.986\times 10^{14}$, $GM/R=6.256\times 10^7$, $\frac{1}{2}{v}_0^2=3.2\times 10^7$. $r_{\max }=\frac{3.986\times 10^{14}}{6.256\times 10^7-3.2\times 10^7}=\frac{3.986\times 10^{14}}{3.056\times 10^7}\approx 1.304\times 10^7\text{m}$ WHY use $-GMm/r$, not $mgh$: the rocket climbs thousands of km, where gravity weakens noticeably — $mgh$'s constant-$g$ assumption breaks. Answer: $r_{\max }\approx 1.30\times 10^7\text{m}$ (about $2.05R$).

WHAT "leaving" means: the ball leaves when the contact force from the spring drops to zero. The spring can only push (once it reaches natural length it can't pull the ball). So contact is lost exactly at the natural length, $x_{\text{leave}}=0$. WHY not at the equilibrium point: a common guess is the point where spring force balances gravity ($kx=mg\Rightarrow x=mg/k=4.9/800\approx 0.006\text{m}$). But there the spring is still compressed and pushing; the ball only separates when the spring can no longer push, which is at natural length. Beyond that the spring would have to pull — impossible for a non-attached ball. Answer: $x_{\text{leave}}=0$ (natural length). The equilibrium compression $mg/k\approx 0.0061\text{m}$ is where net force is zero, but the ball keeps going and leaves at $x=0$.

$r\to \infty$: $U\to 0^{-}$. The two masses are infinitely far apart, effectively non-interacting. This is the chosen zero reference — clean and physical. $r\to 0^{+}$: $U\to -\infty$. The formula predicts an infinitely deep well. WHERE it breaks: real objects have finite size. You cannot bring two centres to $r=0$ because you hit the surface first: for a planet of radius $R_p$, the point-mass law $-GMm/r$ only holds outside the body ($r\ge R_p$). Once you burrow inside, the mass "below" you no longer all pulls inward and Newton's shell theorem changes the law entirely — $U$ actually stays finite and smooth at the true centre, it does not blow up. On top of that, at extreme density Newtonian gravity itself gives way to general relativity. So $r\to 0$ is a purely mathematical limit of an equation used past its domain, never a state reachable by ordinary bodies. WHY track limits: they reveal that the "bottomless well" is an artefact of pretending the mass is a point. The real, physical minimum of $U$ occurs at the surface $r=R$, giving $U=-GMm/R$ — the finite depth we actually pay to escape (as in L2.2).

WHAT the sign means:

• $E<0$ → bound: KE cannot pay the full well depth, so at some $r_{\max }$ the speed hits zero and it falls back.
• $E=0$ → marginal: just reaches $r=\infty$ with zero speed (exactly escape).
• $E>0$ → unbound: leftover KE at infinity, escapes and keeps going. Test: $\frac{1}{2}(1)(11186{)}^2=6.256\times 10^7\text{J}$; $\frac{GMm}{R}=\frac{3.986\times 10^{14}}{6.371\times 10^6}=6.256\times 10^7\text{J}$. $E=6.256\times 10^7-6.256\times 10^7\approx 0$ WHY: $v=11186\text{m/s}$ is the escape speed, so $E=0$ — the boundary case. Any faster ⇒ $E>0$ ⇒ escapes; any slower ⇒ $E<0$ ⇒ bound. This single sign is the whole story of orbits.

$mgh$ is valid only for $h\ll R$ ::: over large climbs use $-GMm/r$. Spring energy squares the displacement ::: $\frac{1}{2}k{x}^2$, never $\frac{1}{2}kx$. Equal spring force ⇒ equal momentum, not equal speed ::: lighter block goes faster. Force is zero at flat spots of $U$; stable = minimum, unstable = maximum ::: use $d^{2}U/d{x}^2$ sign. Total energy sign decides fate ::: $E<0$ bound, $E=0$ escape, $E>0$ unbound. A ball leaves a spring at natural length ::: not at the force-balance point.