Exercises — Conservation of mechanical energy — derivation

Kinetic energy ::: $K=\frac{1}{2}m{v}^2$ Gravitational PE ::: $U=mgh$ (measured from a chosen zero height) Spring PE ::: $U=\frac{1}{2}k{x}^2$ (derived here) Conservation ::: $K_1+U_1=K_2+U_2$ when only conservative forces work; subscript 1 = start, 2 = end With friction ::: $K_2+U_2=K_1+U_1+W_{nc}$, where $W_{nc}<0$ (why)

As the block descends, $U=mgh$ falls (height drops) and $K=\frac{1}{2}m{v}^2$ rises (it speeds up). Neither alone is constant — this is exactly the pocket-swapping shown in the figure above. Because the ramp is frictionless and the normal force is perpendicular to motion (does no work), the only working force is gravity — conservative. So $E=K+U$ is constant.

Answer: mechanical energy $E$.

False. Gravity does negative work on the way up, but that lost kinetic energy is stored as gravitational PE — it is not removed from the mechanical total. Money moves from the "moving" pocket to the "stored-up" pocket. $E=K+U$ stays frozen (no friction, no air drag assumed).

Only gravity works ⇒ conservation. State 1 = at the top (at rest), state 2 = at the ground. Take the ground as $U=0$. $K_1+U_1=K_2+U_2\Rightarrow 0+mgh=\frac{1}{2}m{v}^2+0$ Why "mass cancels": every surviving term contains $m$, so the equation reads $mgh=\frac{1}{2}m{v}^2$. Divide both sides by $m$ — this is allowed because $m=2\text{kg}\ne 0$, so $m$ is a nonzero common factor: $\frac{mgh}{m}=\frac{\frac{1}{2}m{v}^2}{m}\Rightarrow gh=\frac{1}{2}{v}^2$ $v=\sqrt{2gh}=\sqrt{2(9.8)(20)}=\sqrt{392}\approx 19.8\text{m/s}$ Notice the mass disappeared: a heavy stone and a light one hit the ground at the same speed.

Stored spring PE (state 1) becomes kinetic energy (state 2). Floor is level so $U_{grav}$ doesn't change and drops out. $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2\Rightarrow v=x\sqrt{\frac{k}{m}}=0.10\sqrt{\frac{800}{0.50}}=0.10\sqrt{1600}=0.10(40)=4.0\text{m/s}$

Tension is perpendicular to the velocity everywhere ⇒ does zero work ⇒ only gravity works. State 1 = release (string horizontal, at rest), state 2 = lowest point.

Why the drop is exactly $h=L$: look at the figure below. With the pivot fixed, the release point is level with the pivot (string horizontal), and the lowest point hangs a full string-length straight down from the pivot. So the bob's vertical fall from state 1 to state 2 is the vertical distance from the pivot's level down to the bottom of the string — which is precisely $L$. The horizontal shift doesn't enter energy at all; only the vertical drop feeds $U=mgh$.

$mgL=\frac{1}{2}m{v}^2$ Divide both sides by $m\ne 0$: $gL=\frac{1}{2}{v}^2$, so $v=\sqrt{2gL}=\sqrt{2(9.8)(1.25)}=\sqrt{24.5}\approx 4.95\text{m/s}$

Friction is non-conservative, so mechanical energy is not conserved. Use the corrected law (state 1 = top, state 2 = bottom): $K_2+U_2=K_1+U_1+W_{friction}$ With $U_2=0$, $K_1=0$, $U_1=mgh$: $\frac{1}{2}m{v}^2=mgh+W_{friction}=(2.0)(9.8)(5.0)+(-30)=98-30=68\text{J}$ Here we do not cancel $m$, because the friction term $-30\text{J}$ has no factor of $m$ in it — the masses no longer sit in every term. Solve directly: $v=\sqrt{\frac{2(68)}{2.0}}=\sqrt{68}\approx 8.25\text{m/s}$

In the figure below, the loop's centre sits at height $R$ and its top at height $2R=6.0\text{m}$ (a full diameter above the track). Frictionless ⇒ conservation between start (state 1, at rest, height $h$) and loop-top (state 2, height $2R$): $mgh=\frac{1}{2}m{v}_{top}^2+mg(2R)$ Why we can divide by $m$: every term contains $m$ (the $mg(2R)$ term too), so $m\ne 0$ is a common factor of the whole equation. Dividing both sides by $m$: $gh=\frac{1}{2}{v}_{top}^2+g(2R)\Rightarrow \frac{1}{2}{v}_{top}^2=g(h-2R)=9.8(10-6)=39.2$ $v_{top}=\sqrt{78.4}\approx 8.85\text{m/s}$

Three energy states, but we only need the first (top, at rest — state 1) and last (max compression, momentarily at rest — state 2). At max compression $K=0$ because the ball has stopped. All gravitational PE has become spring PE: $mgh=\frac{1}{2}k{x}^2\Rightarrow k=\frac{2mgh}{x^2}=\frac{2(0.40)(9.8)(1.8)}{(0.12{)}^2}=\frac{14.112}{0.0144}=980\text{N/m}$

This problem has two different reference points and we must keep them straight — see the figure below.

• The spring measures its stored energy from its natural (uncompressed) length: $U_{spring}=\frac{1}{2}k{x}^2$ where $x$ is the compression. At release the spring is compressed by $x=0.20\text{m}$, so it stores $\frac{1}{2}k{x}^2$. Once the ball leaves the spring, the spring is back to natural length and stores nothing.
• Gravity measures its stored energy from a height-zero we choose. We put gravity's zero at the release point (the compressed position). So at release $U_{grav}=0$; at the highest point $U_{grav}=mgH$.

These two references are independent — one belongs to the spring, one to gravity — and that's fine, because each energy store is bookkept separately. The offset between "spring natural length" and "release point" is exactly the compression $x$, but it never appears explicitly: we only need the spring's stored energy at release (state 1) and the gravitational PE at the top (state 2). Both states have $K=0$ (ball momentarily at rest).

State 1 (release): $K=0$, spring stores $\frac{1}{2}k{x}^2$, $U_{grav}=0$. State 2 (top): $K=0$, spring stores $0$, $U_{grav}=mgH$. $\frac{1}{2}k{x}^2=mgH$ $H=\frac{k{x}^2}{2mg}=\frac{200(0.20{)}^2}{2(0.50)(9.8)}=\frac{200(0.04)}{9.8}=\frac{8}{9.8}\approx 0.816\text{m}$ ($H$ is measured from the release point, exactly as asked.)

"Just completes" means the minimum speed at the top where the track can still push zero — gravity provides all the centripetal force: $mg=\frac{m{v}_{top}^2}{R}$ Divide both sides by $m\ne 0$: $g=\frac{v_{top}^2}{R}$, so $v_{top}^2=gR$. Now conservation from start (rest at $h$, state 1) to loop-top (height $2R$, state 2): $mg{h}_{min}=\frac{1}{2}m{v}_{top}^2+mg(2R)=\frac{1}{2}m(gR)+2mgR=\frac{5}{2}mgR$ Divide by $m\ne 0$: $\boxed{h_{min}=\frac{5}{2}R=2.5R}$ Independent of mass and $g$ — a clean, famous result. For $R=3.0\text{m}$: $h_{min}=7.5\text{m}$.

Total energy equals the energy at the turning point ($x=A$, all spring PE, $v=0$): $E=\frac{1}{2}k{A}^2$. At a general $x$: $E=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$. Set equal: $\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2\Rightarrow m{v}^2=k(A^2-x^2)$ $v=\sqrt{\frac{k}{m}}\sqrt{A^2-x^2}$ Check $x=0$ (equilibrium): $v=\sqrt{k/m}A$ — this is the maximum speed, correct since all energy is kinetic there. Check $x=A$ (extreme): $v=0$ — correct, the mass is momentarily at rest at a turning point. This is the energy view of SHM.

Energy has no direction — only the magnitude of speed enters $K=\frac{1}{2}m{v}^2$. So whether $u$ points up or down gives the same $K_1$. $\frac{1}{2}m{u}^2+mgh=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{u^2+2gh}=\sqrt{25+2(9.8)(15)}=\sqrt{25+294}=\sqrt{319}\approx 17.9\text{m/s}$ Thrown up with the same speed $5.0\text{m/s}$: on the way back down it repasses $h=15\text{m}$ moving at exactly $5.0\text{m/s}$ again (energy is restored), then continues — same landing speed. Energy is blind to direction; that's its superpower.