High School Physics · Grades 10–12

18Mechanical Energy and Its Conservation

A roller coaster has no engine past the first hill: every loop, every burst of speed afterwards is spent from the account opened on the way up. This chapter turns the bookkeeping into two numbers — kinetic and potential energy — and one rule: while only weight works, their sum does not change. Friction is the fee; we will learn to read it off the books.

18.1 Kinetic energy

Definition 18.1 (Kinetic energy)

A body of mass $m$ ( $\mathrm{kg}$ ) moving at speed $v$ ( $\mathrm{m}/\mathrm{s}$ ) carries the kinetic energy

E_k = \tfrac12\, m v^2 .

Units check: $\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}^{2} = \mathrm{N}\,\mathrm{m} = \mathrm{J}$ — joules, like every energy (Chapter 9).

Example 18.2 (Orders of magnitude)

a pedestrian, $70\,\mathrm{kg}$ at $1.4\,\mathrm{m}/\mathrm{s}$ : $E_k = \tfrac12 \times 70 \times 1.4^2 \approx 69\,\mathrm{J}$ ;
a car, $1300\,\mathrm{kg}$ at $130\,\mathrm{km}/\mathrm{h}$ $= 36.1\,\mathrm{m}/\mathrm{s}$ : $E_k \approx 8.5 \times 10^{5}\,\mathrm{J}$ ;
a high-speed train, $m = 3.85 \times 10^{5}\,\mathrm{kg}$ , at $320\,\mathrm{km}/\mathrm{h}$ $= 88.9\,\mathrm{m}/\mathrm{s}$ : $E_k \approx 1.5 \times 10^{9}\,\mathrm{J}$ ;
a rifle bullet, $8.0\,\mathrm{g}$ at $800\,\mathrm{m}/\mathrm{s}$ : $E_k \approx 2.6\,\mathrm{kJ}$ .

Seven orders of magnitude separate the stroll from the train. And $E_k$ grows like the square of the speed: at $130\,\mathrm{km}/\mathrm{h}$ the car carries $4$ times the energy it has at $65\,\mathrm{km}/\mathrm{h}$ — four times what its brakes must remove.

18.2 The kinetic-energy theorem

Theorem 18.3 (Kinetic-energy theorem)

For a body of mass $m$ moving along a straight line from $A$ to $B$ , the change of kinetic energy equals the total work (Chapter 17) received:

\Delta E_k = E_k(B) - E_k(A) = \sum W_{AB}(\vect F),

the sum running over all forces acting on the body. It extends to motion along any path, as we also admit here.

Proof. Admitted at this level. ∎

Example 18.4 (Free fall, checked frame by frame)

A $0.200\,\mathrm{kg}$ ball is dropped and filmed; the frames give:

$t$ ( $\mathrm{s}$ )	0.10	0.20	0.30	0.40
fallen height $h$ ( $\mathrm{m}$ )	0.049	0.196	0.441	0.785
speed $v$ ( $\mathrm{m}/\mathrm{s}$ )	0.98	1.96	2.94	3.92
$E_k = \tfrac12 mv^2$ ( $\mathrm{J}$ )	0.096	0.384	0.864	1.54
$W = mgh$ ( $\mathrm{J}$ )	0.096	0.385	0.865	1.54

Starting from $E_k = 0$ , at every frame $E_k$ matches the work $mgh$ of the weight: $\Delta E_k = W(\vect P)$ , i.e. $v^2 = 2gh$ — the theorem, verified to the precision of the readings.

Remark 18.5 (Where the proof lives)

The general statement — valid along any path, not only a straight line — follows from Newton’s second law and a little calculus, both later in this book (Chapters 25 and 29); until then the free-fall check is our warrant, and we admit the curved-path version and use it freely.

18.3 Gravitational potential energy

Definition 18.6 (Gravitational potential energy)

A body of mass $m$ at altitude $z$ , measured upward from a chosen reference level where $z = 0$ , stores the gravitational potential energy

E_p = m g z, \qquad g = 9.81\,\mathrm{N}/\mathrm{kg}.

Remark 18.7 (Only differences matter)

Moving from $A$ to $B$ changes $E_p$ by $mg(z_B - z_A) = -W_{AB}(\vect P)$ : the opposite of the work of the weight (Chapter 17). The reference level is a free choice — floor, sea level, table top — shifting $E_p$ everywhere by a constant that drops out of every difference. Put it where the numbers are simplest, and do not move it mid-problem.

Example 18.8 (A book on a shelf)

A $1.2\,\mathrm{kg}$ book sits $1.8\,\mathrm{m}$ above the floor, itself $9.0\,\mathrm{m}$ above the street. Reference at the floor: $E_p = 1.2 \times 9.81 \times 1.8 \approx 21\,\mathrm{J}$ ; in the street: $E_p \approx 127\,\mathrm{J}$ . Falling to the floor releases $21\,\mathrm{J}$ in both accounts.

18.4 Mechanical energy and its conservation

Definition 18.9 (Mechanical energy)

The mechanical energy of a body is

E_m = E_k + E_p = \tfrac12 m v^2 + m g z .

Theorem 18.10 (Conservation of mechanical energy)

If the only force working on a body is its weight (no friction; other forces, like the reaction of a frictionless track, perpendicular to the motion), $E_m$ is conserved: for any two points $A$ , $B$ of the motion,

\tfrac12 m v_A^2 + m g z_A = \tfrac12 m v_B^2 + m g z_B .

Proof. By the kinetic-energy theorem, $E_k(B) - E_k(A) = W_{AB}(\vect P) = mg(z_A - z_B) = E_p(A) - E_p(B)$ : what $E_k$ gains, $E_p$ loses. ∎

Example 18.11 (Water slide)

A child starts at rest atop a frictionless slide of height $h = 3.2\,\mathrm{m}$ . Reference at the bottom: $mgh = \tfrac12 mv^2$ , so $v = \sqrt{2gh} \approx 7.9\,\mathrm{m}/\mathrm{s}$ ( $29\,\mathrm{km}/\mathrm{h}$ ) — whatever the mass and whatever the shape of the slide: straight, curved or spiral, only the height drop enters.

Example 18.12 (Pendulum)

A pendulum bob is pulled aside until it rises $h = 12\,\mathrm{cm}$ above its lowest point, then released. At the extremes $v = 0$ ; at the bottom the whole $mgh$ is kinetic, $v = \sqrt{2gh} \approx 1.5\,\mathrm{m}/\mathrm{s}$ ; on the far side the bob climbs back to exactly $12\,\mathrm{cm}$ .

The pendulum trades E_p for E_k and back: all potential at the extremes, all kinetic at the bottom, the same rise h twice. — The pendulum trades $E_p$ for $E_k$ and back: all potential at the extremes, all kinetic at the bottom, the same rise $h$ twice.

Energy bar chart of a frictionless coaster: at every position the E_p and E_k bars stack to the same total E_m (dashed line). — Energy bar chart of a frictionless coaster: at every position the $E_p$ and $E_k$ bars stack to the same total $E_m$ (dashed line).

Example 18.13 (Ski jump landing speed)

A ski jumper leaves the takeoff at $v_A = 25\,\mathrm{m}/\mathrm{s}$ and lands $\Delta z = 40\,\mathrm{m}$ lower. Neglecting air resistance, $v_B = \sqrt{v_A^2 + 2g\,\Delta z} = \sqrt{625 + 785} \approx 38\,\mathrm{m}/\mathrm{s}$ — no knowledge of the flight path needed.

Ski jump: whatever the curve flown from A to B, only the drop z enters the energy balance. — Ski jump: whatever the curve flown from $A$ to $B$ , only the drop $\Delta z$ enters the energy balance.

18.5 Dissipation by friction

Definition 18.14 (Dissipation)

Energy leaving the mechanical account is said to be dissipated: it reappears as thermal energy in the rubbing surfaces and the air. Total energy is conserved (Chapter 9); mechanical energy alone is not.

Proposition 18.15 (The deficit measures friction)

If a friction force $\vect f$ also works along the motion,

\Delta E_m = E_m(B) - E_m(A) = W_{AB}(\vect f) < 0:

the loss of mechanical energy equals the work of friction.

Proof. The kinetic-energy theorem now reads $\Delta E_k = W_{AB}(\vect P) + W_{AB}(\vect f) = -\Delta E_p + W_{AB}(\vect f)$ ; move $\Delta E_p$ across. Friction opposes the motion, so its work is negative. ∎

Example 18.16 (Toboggan audit)

A child and sled ( $m = 40\,\mathrm{kg}$ ) start at rest, drop $h = 5.0\,\mathrm{m}$ along $12\,\mathrm{m}$ of snow, and arrive at $7.0\,\mathrm{m}/\mathrm{s}$ instead of the frictionless $\sqrt{2gh} \approx 9.9\,\mathrm{m}/\mathrm{s}$ . Reference at the bottom: $\Delta E_m = \tfrac12 \times 40 \times 7.0^2 - 40 \times 9.81 \times 5.0 = 980 - 1962 = -982\,\mathrm{J}$ . So $W(\vect f) = -982\,\mathrm{J}$ over $12\,\mathrm{m}$ : an average friction force $f = 982/12 \approx 82\,\mathrm{N}$ , and $982\,\mathrm{J}$ of warmth in the runners and the snow.

E_m along a motion: level without friction, decreasing with it — the drop at each instant is the heat produced so far. — $E_m$ along a motion: level without friction, decreasing with it — the drop at each instant is the heat produced so far.

Method 18.17 (Energy bookkeeping)

Choose a reference level and keep it; name $A$ (data), $B$ (question).
Write $E_m = \tfrac12 mv^2 + mgz$ at $A$ and at $B$ .
No friction: $E_m(A) = E_m(B)$ ; solve — typically $v_B = \sqrt{v_A^2 + 2g(z_A - z_B)}$ .
Friction: $E_m(B) - E_m(A) = -f\ell$ ( $\ell$ = path length); solve for the unknown.
Check: the $E_k$ / $E_p$ bars stack to $E_m$ , level unless friction.

18.6 Exercises

Exercise 18.1 ★

Compute the kinetic energy of (a) a $90\,\mathrm{kg}$ scooter and rider at $25\,\mathrm{km}/\mathrm{h}$ ; (b) a $1200\,\mathrm{kg}$ car at $130\,\mathrm{km}/\mathrm{h}$ . (c) By what factor does $E_k$ change when the speed doubles? Triples?

Solution

Solution of Exercise 18.1.

(a) $v = 6.94\,\mathrm{m}/\mathrm{s}$ : $E_k = \tfrac12 \times 90 \times 6.94^2 \approx 2.2 \times 10^{3}\,\mathrm{J}$ . (b) $v = 36.1\,\mathrm{m}/\mathrm{s}$ : $E_k = \tfrac12 \times 1200 \times 36.1^2 \approx 7.8 \times 10^{5}\,\mathrm{J}$ . (c) $\times 4$ ; $\times 9$ — the square.

Exercise 18.2 ★

A $75\,\mathrm{kg}$ climber gains $850\,\mathrm{m}$ of altitude. Compute the gain of potential energy. Does it depend on the reference level? On the route?

Solution

Solution of Exercise 18.2.

$\Delta E_p = 75 \times 9.81 \times 850 \approx 6.3 \times 10^{5}\,\mathrm{J}$ . No: the reference shifts both values equally. No: only the height difference enters.

Exercise 18.3 ★

A flowerpot falls $20\,\mathrm{m}$ from a balcony. Neglecting air resistance, find its ground speed in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ . Where did the mass go?

Solution

Solution of Exercise 18.3.

$v = \sqrt{2 \times 9.81 \times 20} = 19.8\,\mathrm{m}/\mathrm{s} \approx 71\,\mathrm{km}/\mathrm{h}$ . $mgh = \tfrac12 mv^2$ : the mass cancels.

Exercise 18.4 ★

A $1200\,\mathrm{kg}$ car brakes from $25\,\mathrm{m}/\mathrm{s}$ to rest in $50\,\mathrm{m}$ . Use the kinetic-energy theorem to find the work of the brakes, then the average braking force.

Solution

Solution of Exercise 18.4.

$W = \Delta E_k = 0 - \tfrac12 \times 1200 \times 25^2 = -3.75 \times 10^{5}\,\mathrm{J}$ ; $F = 3.75 \times 10^{5}/50 = 7.5 \times 10^{3}\,\mathrm{N}$ .

Exercise 18.5 ★

A pendulum bob is released $8.0\,\mathrm{cm}$ above its lowest point. Find its speed at the bottom and the height reached on the other side.

Solution

Solution of Exercise 18.5.

$v = \sqrt{2 \times 9.81 \times 0.080} \approx 1.3\,\mathrm{m}/\mathrm{s}$ ; it rises back to $8.0\,\mathrm{cm}$ (conservation).

Exercise 18.6 ★★

Using Example 18.2: by what factor does the train’s kinetic energy exceed the bullet’s? To what height could its own $E_k$ lift the train ( $h = v^2/2g$ )?

Solution

Solution of Exercise 18.6.

$1.5 \times 10^{9}/2.6 \times 10^{3} \approx 5.9 \times 10^{5}$ : about six hundred thousand bullets. $h = 88.9^2/(2 \times 9.81) \approx 4.0 \times 10^{2}\,\mathrm{m}$ — its own speed could lift the train $400$ meters.

Exercise 18.7 ★★

Two frictionless slides drop from the same $3.0\,\mathrm{m}$ platform, one straight and steep, one long and winding. Compare the arrival speeds (compute them) and the arrival times (no computation).

Solution

Solution of Exercise 18.7.

Same speed $v = \sqrt{2 \times 9.81 \times 3.0} \approx 7.7\,\mathrm{m}/\mathrm{s}$ (only the drop enters). The steep slide wins on time: it reaches high speed sooner and its path is shorter.

Exercise 18.8 ★★

A ball is thrown straight up at $12\,\mathrm{m}/\mathrm{s}$ . Find the maximum height reached, then the speed at $4.0\,\mathrm{m}$ — on the way up and on the way down.

Solution

Solution of Exercise 18.8.

$h = v^2/2g = 144/19.62 \approx 7.3\,\mathrm{m}$ . At $4.0\,\mathrm{m}$ : $v = \sqrt{144 - 2 \times 9.81 \times 4.0} \approx 8.1\,\mathrm{m}/\mathrm{s}$ — the same both ways: $E_m$ depends on height, not on direction of travel.

Exercise 18.9 ★★

A ski jumper leaves the takeoff at $23\,\mathrm{m}/\mathrm{s}$ and lands $35\,\mathrm{m}$ lower. Predict the landing speed without air resistance; is the real value larger or smaller, and why?

Solution

Solution of Exercise 18.9.

$v = \sqrt{23^2 + 2 \times 9.81 \times 35} = \sqrt{1216} \approx 35\,\mathrm{m}/\mathrm{s}$ . Smaller in reality: air resistance dissipates part of $E_m$ .

Exercise 18.10 ★★

A $45\,\mathrm{kg}$ sled starts at rest, drops $6.0\,\mathrm{m}$ along $15\,\mathrm{m}$ of slope, arriving at $8.0\,\mathrm{m}/\mathrm{s}$ . Find the mechanical energy lost, then the average friction force. Where did the energy go?

Solution

Solution of Exercise 18.10.

$\Delta E_m = \tfrac12 \times 45 \times 8.0^2 - 45 \times 9.81 \times 6.0 = 1440 - 2649 \approx -1.2 \times 10^{3}\,\mathrm{J}$ ; $f = 1209/15 \approx 81\,\mathrm{N}$ . Into heat, in the runners and the snow.

Exercise 18.11 ★★

Galileo’s peg: a pendulum released $15\,\mathrm{cm}$ above its lowest point meets a peg blocking the upper half of the string at the vertical. Find the speed at the bottom and the rise beyond the peg; what does the peg change, and what not?

Solution

Solution of Exercise 18.11.

$v = \sqrt{2 \times 9.81 \times 0.15} \approx 1.7\,\mathrm{m}/\mathrm{s}$ ; it rises to $15\,\mathrm{cm}$ again. The peg changes the path (a tighter circle), not the energy: heights and speeds are untouched.

Exercise 18.12 ★★★

A skateboarder drops into a half-pipe from $2.5\,\mathrm{m}$ ; each traverse (side to side) dissipates $10\%$ of the mechanical energy. Find the speed at the bottom of the first descent, then — computing the successive rise heights — the first traverse ending below $1.0\,\mathrm{m}$ .

Solution

Solution of Exercise 18.12.

$v = \sqrt{2 \times 9.81 \times 2.5} = 7.0\,\mathrm{m}/\mathrm{s}$ . Rise heights scale with $E_m$ , so $\times 0.9$ each traverse: $2.25$ , $2.03$ , $1.82$ , $1.64$ , $1.48$ , $1.33$ , $1.20$ , $1.08$ , $0.97$ — first below $1.0\,\mathrm{m}$ on the $9$ th traverse.

Exercise 18.13 ★★★

From a $45\,\mathrm{m}$ cliff top, a stone is thrown at $15\,\mathrm{m}/\mathrm{s}$ — upward, horizontally, or at any angle. Show that the landing speed is the same in all cases and compute it. What does depend on the angle?

Solution

Solution of Exercise 18.13.

Same initial $E_m = \tfrac12 m \times 15^2 + m \times 9.81 \times 45$ whatever the angle, so the ground speed is $v = \sqrt{225 + 2 \times 9.81 \times 45} = \sqrt{1108} \approx 33\,\mathrm{m}/\mathrm{s}$ . The angle fixes the direction of the landing velocity, the flight time and the range — not the landing speed.

Exercise 18.14 ★★★

An $80\,\mathrm{kg}$ downhill skier starts at rest and descends $120\,\mathrm{m}$ of drop along an $800\,\mathrm{m}$ run, against a constant $65\,\mathrm{N}$ friction force. Find the arrival speed and the fraction of $E_m$ dissipated.

Solution

Solution of Exercise 18.14.

$\tfrac12 mv^2 = mgh - f\ell = 80 \times 9.81 \times 120 - 65 \times 800 = 94\,176 - 52\,000 = 4.22 \times 10^{4}\,\mathrm{J}$ , so $v = \sqrt{2 \times 42\,176/80} \approx 32\,\mathrm{m}/\mathrm{s}$ ( $117\,\mathrm{km}/\mathrm{h}$ ). Fraction dissipated: $52000/94176 \approx 55\%$ .

Exercise 18.15 ★★★

A pumped-storage plant lifts $V = 2.0 \times 10^{6}\,\mathrm{m}^{3}$ of water ( $1000\,\mathrm{kg}$ per $\mathrm{m}^{3}$ ) through $300\,\mathrm{m}$ . Compute the stored energy in joules and in $\mathrm{kW}\,\mathrm{h}$ ; turbined back at $85\%$ efficiency (Chapter 9), how many homes drawing $10\,\mathrm{kW}\,\mathrm{h}$ per day does it feed for one day?

Solution

Solution of Exercise 18.15.

$m = 2.0 \times 10^{9}\,\mathrm{kg}$ : $E_p = 2.0 \times 10^{9} \times 9.81 \times 300 \approx 5.9 \times 10^{12}\,\mathrm{J} = 5.9 \times 10^{12}/3.6 \times 10^{6} \approx 1.6 \times 10^{6}\,\mathrm{kW}\,\mathrm{h}$ . Recovered: $0.85 \times 1.64 \times 10^{6} \approx 1.4 \times 10^{6}\,\mathrm{kW}\,\mathrm{h}$ — about $1.4 \times 10^{5}$ homes for a day.

18.7 Problem: Designing the loop

Problem 18.1

Weekend problem — an energy audit of a roller coaster: the launch height a loop dictates, the speeds it promises, the friction confessed by two sensors, the brakes that close the account — the whole ride hanging from one number

A train of mass $m = 2.0 \times 10^{3}\,\mathrm{kg}$ is released at rest from a launch hill of height $H$ and coasts, engineless, through a vertical circular loop of radius $R = 8.0\,\mathrm{m}$ whose top sits at height $2R$ . Circular-motion analysis (next year) supplies the one datum we borrow: at the loop top the train presses on the track only if $v_{\text{top}}^2 \geq gR$ . Parts I and II ignore friction.

Part I — The hill the loop dictates.

Compute the minimum safe speed at the top of the loop.
Ground as reference level: compute the train’s mechanical energy at the loop top at that minimum speed.
Deduce $H_{\min}$ and show $H_{\min} = 5R/2$ — no mass, no $g$ .
The team takes $H = 24\,\mathrm{m}$ : compute the actual top-of-loop speed and the ratio $v_{\text{top}}^2/(gR)$ .
Where did $m$ go? Why does the design fit full and empty trains alike?

Part II — What the height promises ( $H = 24\,\mathrm{m}$ ).

Speed at the bottom of the loop ( $z = 0$ ), in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ ?
Speed at the side of the loop ( $z = R$ )?
Compute $E_p$ and $E_k$ at launch, bottom, side and top of the loop; check that each pair sums to $mgH$ .
A $12\,\mathrm{m}$ camel-back bump follows the loop: speed at its crest?
Give the general formula for $v(z)$ ; where is the ride fastest, and slowest?

Part III — The friction audit. Two sensors at the same height $z = 2.0\,\mathrm{m}$ , $60\,\mathrm{m}$ of track apart, read $v_1 = 20.4\,\mathrm{m}/\mathrm{s}$ then $v_2 = 19.2\,\mathrm{m}/\mathrm{s}$ .

Compute the change of mechanical energy between the sensors. What made the same-height placement clever?
Deduce the work of friction and the average friction force.
Where has the missing energy gone? Name the form and the places.
What fraction of the train’s $E_m$ is lost per $60\,\mathrm{m}$ ? Why would the bare $H_{\min}$ be unsafe in the real, rubbing world?
About $90\,\mathrm{m}$ of track separate launch and loop top: estimate $E_m$ there; does the margin of question 4 still save the loop?

Part IV — Closing the account. The last sensor, at the entry of the straight horizontal braking run ( $z = 0$ ), reads $18.0\,\mathrm{m}/\mathrm{s}$ .

Compute the train’s kinetic energy there.
The brakes must stop it in $d = 45\,\mathrm{m}$ : compute the required (constant) braking force.
Comfort check: compute the deceleration $a = F/m$ ; compare to $g$ .
How much energy did track friction dissipate over the whole ride? Verify that friction $+$ brakes $= mgH$ , to the joule.
Punchline, one sentence: which single number fixed the loop’s safety, every speed, the friction bill and the brake force — and what was the price of ignoring friction?

Solution

Solution of Problem 18.1.

1. $v_{\text{top}} = \sqrt{gR} = \sqrt{9.81 \times 8.0} \approx 8.9\,\mathrm{m}/\mathrm{s}$ .

2. $E_m = mg\,(2R) + \tfrac12 m gR = 3.14 \times 10^{5} + 7.8 \times 10^{4} \approx 3.92 \times 10^{5}\,\mathrm{J}$ .

3. $mgH_{\min} = E_m$ gives $H_{\min} = 20\,\mathrm{m}$ ; algebraically $H_{\min} = 2R + \frac{gR}{2g} = \frac{5R}{2}$ .

4. $v_{\text{top}} = \sqrt{2g(H - 2R)} = \sqrt{2 \times 9.81 \times 8.0} \approx 12.5\,\mathrm{m}/\mathrm{s}$ ; $v_{\text{top}}^2/(gR) = 2.0$ — twice the minimum.

5. Every energy is proportional to $m$ , so $m$ cancels: same speeds and same safety, full or empty.

6. $v = \sqrt{2 \times 9.81 \times 24} \approx 21.7\,\mathrm{m}/\mathrm{s} \approx 78\,\mathrm{km}/\mathrm{h}$ .

7. $v = \sqrt{2 \times 9.81 \times (24 - 8.0)} \approx 17.7\,\mathrm{m}/\mathrm{s}$ .

8. $mgH = 4.71 \times 10^{5}\,\mathrm{J}$ . $(E_p, E_k)$ : launch $(4.71 \times 10^{5}, 0)$ ; bottom $(0, 4.71 \times 10^{5})$ ; side $(1.57 \times 10^{5}, 3.14 \times 10^{5})$ ; top $(3.14 \times 10^{5}, 1.57 \times 10^{5})$ — each pair sums to $4.71 \times 10^{5}\,\mathrm{J}$ .

9. $v = \sqrt{2 \times 9.81 \times (24 - 12)} \approx 15.3\,\mathrm{m}/\mathrm{s}$ .

10. $v(z) = \sqrt{2g(H - z)}$ : fastest at the lowest point of the track, slowest at the top of the loop.

11. Same height, so $\Delta E_p = 0$ and $\Delta E_m = \Delta E_k = \tfrac12 \times 2000 \times (19.2^2 - 20.4^2) \approx -4.75 \times 10^{4}\,\mathrm{J}$ : the placement makes the deficit pure kinetic, read off two speedometers.

12. $W(\vect f) = -4.75 \times 10^{4}\,\mathrm{J}$ ; $f = 4.75 \times 10^{4}/60 \approx 7.9 \times 10^{2}\,\mathrm{N}$ .

13. Thermal energy, in the wheels, the rails and the air.

14. $E_m$ at the first sensor: $4.16 \times 10^{5} + 3.9 \times 10^{4} = 4.55 \times 10^{5}\,\mathrm{J}$ ; loss $4.75 \times 10^{4}/4.55 \times 10^{5} \approx 10\%$ per $60\,\mathrm{m}$ . At $H_{\min}$ the top of the loop is reached exactly at $\sqrt{gR}$ ; any loss drops it below the safety speed.

15. Losing $\approx 10\%$ per $60\,\mathrm{m}$ , about $15\%$ of $mgH$ ( $\approx 7.1 \times 10^{4}\,\mathrm{J}$ ) is gone: $E_m \approx 4.00 \times 10^{5}\,\mathrm{J}$ , still above the required $3.92 \times 10^{5}\,\mathrm{J}$ — safe, with only $2\%$ to spare. The margin was no luxury.

16. $E_k = \tfrac12 \times 2000 \times 18.0^2 = 3.24 \times 10^{5}\,\mathrm{J}$ .

17. $F = 3.24 \times 10^{5}/45 = 7.2 \times 10^{3}\,\mathrm{N}$ .

18. $a = 7200/2000 = 3.6\,\mathrm{m}/\mathrm{s}^{2} \approx 0.37\,g$ : firm but comfortable.

19. Track friction: $mgH - 3.24 \times 10^{5} = 470\,880 - 324\,000 = 1.47 \times 10^{5}\,\mathrm{J}$ . Audit: $1.47 \times 10^{5} + 3.24 \times 10^{5} = 4.71 \times 10^{5} = mgH$ — every joule accounted for.

20. The launch height $H = 24\,\mathrm{m}$ fixed the loop’s safety, every speed $\sqrt{2g(H - z)}$ , the friction bill and the brake force; the price of ignoring friction is the margin — the $4\,\mathrm{m}$ of hill that the rubbing world quietly consumes.