Kinetic energy — derivation: Exercises — Physics deep dive

This page leans on three tools you already own:

$K = \frac{1}{2} m v^{2}$ — the kinetic energy of mass $m$ at speed $v$ .
$W_{n e t} = K_{f} - K_{i}$ — the Work–Energy Theorem (net work equals change in kinetic energy).
$W = F \cdot d$ — the definition of work (force times the along-motion part of displacement).

Level 1 — Recognition

Can you plug numbers into $K = \frac{1}{2} m v^{2}$ and read the formula backwards?

Recall Solution L1·1

WHAT: apply the definition directly. WHY: we know $m$ and $v$ , nothing else needed. $K = \frac{1}{2} m v^{2} = \frac{1}{2} (4) (5)^{2} = \frac{1}{2} (4) (25) = 50 J$ Answer: $== 50 J ==$ .

Recall Solution L1·2

WHAT: solve $K = \frac{1}{2} m v^{2}$ for $v$ . WHY: $v$ is the unknown, so isolate it. $v = \frac{2 K}{m} = \frac{2 ( 64 )}{8} = 16 = 4 m/s$ Answer: $== 4 m/s ==$ .

Recall Solution L1·3

WHAT: carry units through the formula. WHY: confirms the formula is dimensionally honest. $[K] = kg \times (m/s)^{2} = kg \cdot \frac{m ^{2}}{s ^{2}} = J .$ A joule is literally "kilogram-metre-squared per second-squared."

Level 2 — Application

Now feed kinetic energy into the Work–Energy Theorem.

Recall Solution L2·1

WHAT: use $W_{n e t} = K_{f} - K_{i}$ . WHY: work asked, speeds given — the theorem links them. $W_{n e t} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} (1200) (2 0^{2} - 8^{2})$ $= \frac{1}{2} (1200) (400 - 64) = 600 \times 336 = 201600 J$ Answer: $== 201600 J \approx 2.02 \times 1 0^{5} J ==$ (positive — energy added).

Recall Solution L2·2

WHAT: friction is the only horizontal force, so its work $W_{friction}$ equals the net work, hence $Δ K$ . WHY: work–energy theorem again, but the final speed is smaller. $W_{friction} = W_{n e t} = \frac{1}{2} (0.2) (2^{2} - 6^{2}) = \frac{1}{2} (0.2) (4 - 36) = 0.1 \times (- 32) = - 3.2 J$ Answer: $== - 3.2 J ==$ . The minus sign means energy left the puck — friction opposes motion ( $θ = 18 0^{\circ}$ , so $cos θ = - 1$ ), so its work is negative.

Recall Solution L2·3

WHAT: the box gains kinetic energy; that came from work. WHY: unknown force, known distance and speeds — combine work with $W_{n e t} = Δ K$ . Because the force points along the displacement here, the angle between them is $θ = 0^{\circ}$ , so $cos θ = 1$ and the general work formula collapses to a special case: $W_{n e t} = F s cos θ = F s cos 0^{\circ} = F s .$ (This is not a different formula — it is $W = F d cos θ$ with $θ = 0^{\circ}$ . If the force were tilted, you would keep the $cos θ$ , exactly as in L3·2.) Now: $W_{n e t} = \frac{1}{2} (5) (1 0^{2} - 0) = 250 J, F = \frac{W _{n e t}}{s} = \frac{250}{4} = 62.5 N$ Answer: $== 62.5 N ==$ .

Level 3 — Analysis

Reason about proportions, ratios, and geometry.

Recall Solution L3·1

WHAT: compare energies using $K \propto v^{2}$ . WHY: masses cancel, so only the speed ratio survives. $\frac{K _{B}}{K _{A}} = \frac{\frac{1}{2} m v _{B}^{2}}{\frac{1}{2} m v _{A}^{2}} = (\frac{v _{B}}{v _{A}})^{2} = 3^{2} = 9$ Answer: $== 9 ==$ . Since braking friction does a fixed force over the stopping distance $d$ (so $K = F d$ ), nine times the energy needs nine times the stopping distance. Look at figure s01: the energy bars grow as the square of speed.

Recall Solution L3·2

WHAT: only the along-motion part of the force does work, via $W = F d cos θ$ . WHY: the dot product keeps only the horizontal component; the vertical part is perpendicular to the motion, so it does zero work. See figure s02. $W_{rope} = F d cos θ = 20 \times 10 \times cos 6 0^{\circ} = 200 \times 0.5 = 100 J$ On frictionless ice the rope is the only force doing work, so $W_{n e t} = 100 J$ . That becomes kinetic energy: $100 = \frac{1}{2} (3) v^{2} \Rightarrow v^{2} = \frac{200}{3} = 66.67 \Rightarrow v = 8.165 m/s$ Answer: $== v \approx 8.16 m/s ==$ .

Recall Solution L3·3

WHAT: set the energies equal and solve for the speed ratio. WHY: equal $K$ links the two speeds through their masses. $\frac{1}{2} (2) v_{2}^{2} = \frac{1}{2} (8) v_{8}^{2} \Rightarrow 2 v_{2}^{2} = 8 v_{8}^{2} \Rightarrow \frac{v _{2}^{2}}{v _{8}^{2}} = 4 \Rightarrow \frac{v _{2}}{v _{8}} = 2$ Answer: $== 2 ==$ . The lighter mass must move twice as fast to carry the same energy.

Level 4 — Synthesis

Chain kinetic energy with other chapter tools.

Recall Solution L4·1

WHAT: the stored potential energy $U = m g h$ is converted into kinetic energy as the ball falls. WHY: this is energy conservation — lost height buys speed. Gravity is the only force acting, so all of $U$ becomes $K$ . $U = m g h = \frac{1}{2} m v^{2} \Rightarrow v = 2 g h = 2 (10) (5) = 100 = 10 m/s$ Answer: $== 10 m/s ==$ . Notice mass cancelled — all objects gain the same speed in free fall.

Recall Solution L4·2

WHAT: average power = energy delivered ÷ time. WHY: the engine's work all became kinetic energy. $K = \frac{1}{2} (1000) (3 0^{2}) = \frac{1}{2} (1000) (900) = 450000 J$ $P_{a v g} = \frac{K}{t} = \frac{450000}{12} = 37500 W = 37.5 kW$ Answer: $== 37500 W = 37.5 kW ==$ .

Recall Solution L4·3

WHAT: rewrite $K$ in terms of momentum. WHY: momentum $p = m v$ is given, not speed, so express $K$ using $v = p / m$ . $K = \frac{1}{2} m (\frac{p}{m})^{2} = \frac{p ^{2}}{2 m}$ $K_{X} = \frac{1 2 ^{2}}{2 ( 2 )} = \frac{144}{4} = 36 J, K_{Y} = \frac{144}{2 ( 6 )} = \frac{144}{12} = 12 J$ $\frac{K _{X}}{K _{Y}} = \frac{36}{12} = 3$ Answer: the lighter cart X has $== 3 ==$ times the energy. For equal momentum, $K \propto 1/ m$ : lighter means faster means more energy.

Level 5 — Mastery

Multi-step problems where you must decide which tool to reach for.

Recall Solution L5·1

WHAT: on the ramp the potential energy $U = m g h$ becomes kinetic energy; on the floor friction removes it all. WHY: energy accounting across the whole trip — final $K = 0$ . Kinetic energy at the bottom (equal to the potential energy given up on the frictionless ramp): $K_{bottom} = U = m g h = 2 \times 10 \times 3 = 60 J$ On the floor friction does negative work $W_{friction} = - f d$ until the block stops. The theorem gives $W_{friction} = K_{f} - K_{i} = 0 - 60$ : $0 - 60 = - f d \Rightarrow d = \frac{60}{f} = \frac{60}{4} = 15 m$ Answer: $== 15 m ==$ .

Recall Solution L5·2

WHAT: force varies with position, so work is an integral. WHY: $W_{n e t} = \int F d x$ handles any force history — exactly the variable-force branch of the parent derivation. $W_{n e t} = \int_{0}^{4} 6 x d x = [3 x^{2}]_{0}^{4} = 3 (16) = 48 J$ Then $W_{n e t} = \frac{1}{2} m v^{2}$ : $48 = \frac{1}{2} (3) v^{2} \Rightarrow v^{2} = 32 \Rightarrow v = 32 = 42 \approx 5.657 m/s$ Answer: $== v = 42 \approx 5.66 m/s ==$ . Figure s03 shows the work as the area under the force–position line.

Recall Solution L5·3

WHAT: total mechanical energy $K + U$ (kinetic plus potential) stays constant. WHY: only gravity acts, so Conservation of Mechanical Energy holds. Initial energy (all kinetic, since it starts at ground level where $U = 0$ ): $E = \frac{1}{2} (1) (8^{2}) = 32 J$ (a) At the top, all energy is potential ( $K = 0$ ): $U = m g h_{ma x} = 32 \Rightarrow h_{ma x} = \frac{32}{1 \times 10} = 3.2 m$ (b) At $h = 2 m$ : $U = m g h = 1 \times 10 \times 2 = 20 J$ , so $K = 32 - 20 = 12 J$ : $\frac{1}{2} (1) v^{2} = 12 \Rightarrow v^{2} = 24 \Rightarrow v = 24 = 26 \approx 4.899 m/s$ Answers: $== h_{ma x} = 3.2 m ==$ , $== v = 26 \approx 4.90 m/s ==$ .

Exercises — Kinetic energy — derivation