Intuition The big pictureA moving object can do work — a hammer drives a nail, wind spins a turbine. The energy it carries because of its motion is called kinetic energy . The deep question is: how much work was needed to get it moving, and how much can it pay back? We will derive this amount from Newton's laws, not just quote it.
Definition Kinetic energyThe kinetic energy of a body of mass m m m v v v work done to accelerate it from rest to that speed . Symbol K K K joule (J) , a scalar.
WHY a scalar? Because it is built from work (W = F ⃗ ⋅ d ⃗ W=\vec F\cdot\vec d W = F ⋅ d v 2 v^2 v 2
We start from the definition of work and Newton's second law . No memorised formula allowed.
This is the Work–Energy Theorem : ==net work done = change in kinetic energy==,
W n e t = K f − K i = 1 2 m v f 2 − 1 2 m v i 2 . W_{net} = K_f - K_i = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2. W n e t = K f − K i = 2 1 m v f 2 − 2 1 m v i 2 .
Worked example 1 — Straight plug-inA 2 kg 2\,\text{kg} 2 kg 3 m/s 3\,\text{m/s} 3 m/s K K K K = 1 2 m v 2 = 1 2 ( 2 ) ( 3 ) 2 = 9 J K=\tfrac12 mv^2=\tfrac12(2)(3)^2 = 9\,\text{J} K = 2 1 m v 2 = 2 1 ( 2 ) ( 3 ) 2 = 9 J Why this step? Direct definition; speed squared, then half of mass × that.
Worked example 2 — Doubling speed (forecast first!)Forecast: if speed doubles, does K K K 1000 kg 1000\,\text{kg} 1000 kg 10 m/s 10\,\text{m/s} 10 m/s 20 m/s 20\,\text{m/s} 20 m/s K 1 = 1 2 ( 1000 ) ( 10 ) 2 = 50,000 J , K 2 = 1 2 ( 1000 ) ( 20 ) 2 = 200,000 J K_1=\tfrac12(1000)(10)^2=50{,}000\,\text{J},\quad K_2=\tfrac12(1000)(20)^2=200{,}000\,\text{J} K 1 = 2 1 ( 1000 ) ( 10 ) 2 = 50 , 000 J , K 2 = 2 1 ( 1000 ) ( 20 ) 2 = 200 , 000 J Verify: K 2 / K 1 = 4 K_2/K_1 = 4 K 2 / K 1 = 4 Because K ∝ v 2 K\propto v^2 K ∝ v 2 Double the speed → quadruple the energy (and stopping distance!).
Worked example 3 — Work–energy theoremA 0.5 kg 0.5\,\text{kg} 0.5 kg 4 m/s 4\,\text{m/s} 4 m/s 1 m/s 1\,\text{m/s} 1 m/s W = 1 2 ( 0.5 ) ( 1 2 − 4 2 ) = 1 2 ( 0.5 ) ( 1 − 16 ) = − 3.75 J W=\tfrac12(0.5)(1^2-4^2)=\tfrac12(0.5)(1-16)=-3.75\,\text{J} W = 2 1 ( 0.5 ) ( 1 2 − 4 2 ) = 2 1 ( 0.5 ) ( 1 − 16 ) = − 3.75 J Why negative? Friction removes energy, so it does negative work — sign carries physical meaning.
K = 1 2 m v K = \tfrac12 mv K = 2 1 m v Why it feels right: momentum is p = m v p=mv p = m v The fix: Kinetic energy comes from a s = 1 2 v 2 as=\tfrac12 v^2 a s = 2 1 v 2 v 2 v^2 v 2 v v v
Common mistake "Doubling speed doubles energy"Why it feels right: linear thinking is the default.
The fix: K ∝ v 2 K\propto v^2 K ∝ v 2
Common mistake "KE can be negative"Why it feels right: velocity can be negative.
The fix: v 2 ≥ 0 v^2 \ge 0 v 2 ≥ 0 K ≥ 0 K\ge 0 K ≥ 0 changes in K K K
Recall Feynman: explain to a 12-year-oldImagine pushing a toy car. The harder and longer you push, the faster it goes — you're loading "go-energy" into it. If you push twice as fast a result, the car doesn't just hit twice as hard, it hits four times as hard, because the energy follows the square of the speed. That stored "go-energy" is kinetic energy: half of the weight-ish number times speed times speed.
"Half-em-vee-squared" → 1 2 m v 2 \tfrac12 m v^2 2 1 m v 2 "Speed gets SQUARED, danger gets quadrupled."
What is kinetic energy in words? The work done to bring a body from rest to its current speed.
Formula for kinetic energy? K = 1 2 m v 2 K=\tfrac12 mv^2 K = 2 1 m v 2 Which kinematic equation is used in the derivation? v 2 = u 2 + 2 a s v^2=u^2+2as v 2 = u 2 + 2 a s , giving
a s = 1 2 v 2 as=\tfrac12 v^2 a s = 2 1 v 2 from rest.
State the work–energy theorem. Net work = change in kinetic energy,
W n e t = K f − K i W_{net}=K_f-K_i W n e t = K f − K i .
If speed triples, kinetic energy multiplies by? 9 (since
K ∝ v 2 K\propto v^2 K ∝ v 2 ).
Why is kinetic energy a scalar? It's built from a dot product (work) and
v 2 v^2 v 2 , both directionless.
Can kinetic energy be negative? No, because
v 2 ≥ 0 v^2\ge0 v 2 ≥ 0 ; only its change can be negative.
Unit of kinetic energy? Joule (J) = kg·m²/s².
In the calculus derivation, what substitution converts ∫ F d s \int F\,ds ∫ F d s d s d t = v \frac{ds}{dt}=v d t d s = v , giving
∫ m v d v \int mv\,dv ∫ m v d v .
Intuition Hinglish mein samjhoDekho, kinetic energy ka matlab hai — kisi cheez me jo "chalne wali energy" stored hai uske motion ki wajah se. Jab tum kisi object ko rest se push karke speed dete ho, to jitna kaam (work) tum lagate ho, wahi energy uske andar jama ho jaati hai. Isliye hum bolte hain: KE = work done to bring it from rest to speed v v v
Derivation simple hai. Work ka definition hai W = F s W=Fs W = F s F = m a F=ma F = ma W = m a s W=mas W = ma s v 2 = u 2 + 2 a s v^2=u^2+2as v 2 = u 2 + 2 a s u = 0 u=0 u = 0 a s = 1 2 v 2 as=\frac12 v^2 a s = 2 1 v 2 K = 1 2 m v 2 K=\frac12 mv^2 K = 2 1 m v 2
Sabse important baat: K K K v v v square . Matlab speed double karo to energy 4 guna ho jaati hai, 9 guna agar speed triple. Isi liye fast accidents itne dangerous hote hain — chhoti si extra speed bohot zyada energy carry karti hai. Aur ek galti se bacho: K = 1 2 m v K=\frac12 mv K = 2 1 m v
