Classical-Mechanics · Unit 14 · Video 2 · Interactive Practice

Kinetic Energy: Why v Squared, and the Work Equation Hidden in Kinematics

IKey Formulas

Formula	Name	What you need
$K = \tfrac{1}{2}mv^2$	Kinetic energy	Mass and speed; a scalar with $K \ge 0$
$\Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$	Change in kinetic energy	Initial and final speeds
$a_x\,\Delta x = \tfrac{1}{2}v_{x,f}^2 - \tfrac{1}{2}v_{x,i}^2$	Kinematic identity	Constant $a_x$ in one dimension
$F_x\,\Delta x = K_f - K_i$	Constant-force result	Newton's second law, $F_x = ma_x$

Key Insight: $a_x\Delta x = \Delta\!\left(\tfrac{1}{2}v^2\right)$ is pure kinematics; multiplying both sides by the mass $m$ is the single step that turns it into energy physics.

IIVisualization 1 — Why the Square of Speed

$K$ is proportional to the area of a square of side $v$ , so each $+10$ mph adds a wider L.

Car B initial speed 50 mph

IIIVisualization 2 — The Same Cars, a Moving Observer

Every velocity drops by the observer's speed, and both energy changes — and their ratio — move with it.

Observer speed

u

0 mph

IVVisualization 3 — Acceleration Times Displacement

Graph $\tfrac{1}{2}v^2$ against position: constant acceleration makes it a straight line of slope $a_x$ .

Acceleration

a_x

(m/s²) 0.6 Displacement Δx (m) 6.0

💡 Multiply this identity by $m$ and use $F_x = ma_x$ : it becomes $F_x\,\Delta x = K_f - K_i$ , and for a varying force $\int_{x_i}^{x_f} F_x\,dx = K_f - K_i$ — the quantity about to be named work.

VQuiz Questions

Problem 1 · Computing $\Delta K$

Given: A car of mass $m = 1200\ \text{kg}$ speeds up in a straight line from $v_i = 8.0\ \text{m/s}$ to $v_f = 12.0\ \text{m/s}$ — find its change in kinetic energy.

✅ Correct!

\Delta K = \tfrac{1}{2}(1200)(144 - 64) = 600 \times 80 = 4.8 \times 10^4\ \text{J}

❌ Not quite. That is

\tfrac{1}{2}m(\Delta v)^2 = 600(4.0)^2

. The definition squares each speed before subtracting:

v_f^2 - v_i^2 \ne (v_f - v_i)^2

❌ That is $K_f$ , not $\Delta K$ . The car already carried

\tfrac{1}{2}(1200)(8.0)^2 = 38.4\ \text{kJ}

before it sped up; subtract it.

❌ Check the factor of $\tfrac{1}{2}$ . You computed

m(v_f^2 - v_i^2)

; kinetic energy carries a factor of one half.

❌ Not quite. Use

\Delta K = \tfrac{1}{2}m\left(v_f^2 - v_i^2\right)

with

v_f^2 = 144

and

v_i^2 = 64

Show solution

Square each speed first, then subtract:

\Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = \tfrac{1}{2}m\left(v_f^2 - v_i^2\right)

\Delta K = \tfrac{1}{2}(1200\ \text{kg})\left[(12.0)^2 - (8.0)^2\right]\ \text{m}^2/\text{s}^2

\Delta K = 600(144 - 64) = 600(80) = 48\,000\ \text{J} = 48\ \text{kJ}

Units check: $\text{kg}\cdot\text{m}^2/\text{s}^2 = \text{J}$ , the joule.

Problem 2 · Equal Speed Gains, Unequal Energy

Given: Two identical trucks each gain $10\ \text{m/s}$ . Truck 1 goes from $10$ to $20\ \text{m/s}$ ; truck 2 goes from $30$ to $40\ \text{m/s}$ — find the ratio $\Delta K_2 / \Delta K_1$ .

✅ Correct! The common factor

\tfrac{1}{2}m

cancels, leaving

(1600 - 900)/(400 - 100) = 700/300 = 7/3

❌ This is the trap. Equal speed gains are not equal energy gains:

K

is quadratic in

v

, so the faster truck gains far more.

❌ That is $K_{2f}/K_{2i} = 40^2/30^2$ . You compared truck 2's final kinetic energy with its own initial energy; the question asks for truck 2's change divided by truck 1's change.

❌ Not quite.

4 = (40/20)^2

compares final speeds squared, not

v_f^2 - v_i^2

for each truck.

❌ Not quite. Write the ratio in full: the factors

\tfrac{1}{2}m

cancel, leaving

\left(v_{2f}^2 - v_{2i}^2\right)/\left(v_{1f}^2 - v_{1i}^2\right)

Show solution

Identical masses, so every $\tfrac{1}{2}m$ cancels:

\frac{\Delta K_2}{\Delta K_1} = \frac{\tfrac{1}{2}m\left(v_{2f}^2 - v_{2i}^2\right)}{\tfrac{1}{2}m\left(v_{1f}^2 - v_{1i}^2\right)} = \frac{40^2 - 30^2}{20^2 - 10^2}

= \frac{1600 - 900}{400 - 100} = \frac{700}{300} = \frac{7}{3} \approx 2.33

In general $v_f^2 - v_i^2 = (v_f - v_i)(v_f + v_i)$ , so for a fixed gain $v_f - v_i = 10$ the energy change grows in proportion to $v_f + v_i$ — the faster pair of speeds always wins.

Problem 3 · From Force and Displacement to Speed

Given: A $2.0\ \text{kg}$ block slides on a frictionless surface at $v_i = 3.0\ \text{m/s}$ when a constant force $F_x = 4.0\ \text{N}$ , pointing along the motion, acts over a displacement $\Delta x = 4.0\ \text{m}$ — find the final speed.

✅ Correct!

F_x\Delta x = 16\ \text{J} = \Delta K

, so

v_f^2 = 9 + 16 = 25

and

v_f = 5.0\ \text{m/s}

❌ You dropped $K_i$ .

F_x\Delta x

equals

K_f - K_i

, not

K_f

; the block already had

\tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}

❌ Check the acceleration.

a_x = F_x/m = 4.0/2.0 = 2.0\ \text{m/s}^2

, not

4.0\ \text{m/s}^2

— the mass must divide the force.

❌ $\Delta x$ is a displacement, not a time.

v_f = v_i + a_x t

needs a time; over a displacement use

a_x\Delta x = \tfrac{1}{2}v_f^2 - \tfrac{1}{2}v_i^2

❌ Not quite. Apply

F_x\Delta x = K_f - K_i

with

K_i = \tfrac{1}{2}(2.0)(3.0)^2

Show solution

Step 1 — Energy delivered along the displacement:

F_x\,\Delta x = (4.0\ \text{N})(4.0\ \text{m}) = 16\ \text{J}

Step 2 — Set it equal to the change in kinetic energy:

16\ \text{J} = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = (1.0)\left(v_f^2 - 9.0\right)

v_f^2 = 9.0 + 16 = 25 \quad\Rightarrow\quad v_f = 5.0\ \text{m/s}

Check with kinematics: $a_x = F_x/m = 2.0\ \text{m/s}^2$ , and

a_x\Delta x = (2.0)(4.0) = 8.0 = \tfrac{1}{2}(5.0)^2 - \tfrac{1}{2}(3.0)^2 = 12.5 - 4.5\ \checkmark

Problem 4 · Whose Energy Change? (Video Example)

Given: Car A speeds up from $10$ to $20$ mph and car B from $50$ to $60$ mph, both of mass $m$ , along the same straight road (the video's Example 13.1). Now measure both from a reference frame moving along the road in the same direction at speed $u$ .

With $u = 15$ mph, what does that observer measure for car A?

With $u = 25$ mph, car A's change in kinetic energy is:

✅ Excellent!

\Delta K_A \propto (v_f - v_i)(v_f + v_i - 2u)

vanishes at

u = 15

mph and turns negative beyond it.

❌ Check the velocities in that frame. Car A runs from

10 - 15 = -5

mph to

20 - 15 = +5

mph: two different velocities, but the same speed.

❌ Check the speeds in that frame. Car A runs from

-15

mph to

-5

mph, so it slows down as that observer sees it.

Show solution

In a frame moving at $u$ , every velocity drops by $u$ , so

\Delta K_A = \tfrac{1}{2}m\left[(v_f - u)^2 - (v_i - u)^2\right] = \tfrac{1}{2}m\,(v_f - v_i)\left(v_f + v_i - 2u\right)

At $u = 15$ mph: $10 \to -5$ mph and $20 \to +5$ mph. Both speeds are $5$ mph, so $K_f = K_i$ and

\Delta K_A = \tfrac{1}{2}m(10)(30 - 30) = 0

At $u = 25$ mph: $10 \to -15$ mph and $20 \to -5$ mph, so the speed falls from $15$ to $5$ mph:

\Delta K_A = \tfrac{1}{2}m\left(5^2 - 15^2\right) = \tfrac{1}{2}m(-200) < 0

Kinetic energy is built from speed, and observers in different frames measure different speeds — so $K$ and $\Delta K$ are frame dependent. (Car B is safe here: $\Delta K_B \propto 1100 - 20u$ stays positive until $u = 55$ mph.)

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