Intuition The one-line ideaWhen you push something, your push does work , and that work doesn't vanish — it shows up as a change in how fast the object moves, i.e. its kinetic energy . The Work–Energy Theorem is just Newton's F = m a F=ma F = ma integrated over distance instead of time .
Definition Work–Energy TheoremThe net work done by all forces on a particle equals the change in its kinetic energy :
W n e t = Δ K = K f − K i = 1 2 m v f 2 − 1 2 m v i 2 W_{net} = \Delta K = K_f - K_i = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 W n e t = Δ K = K f − K i = 2 1 m v f 2 − 2 1 m v i 2 K = 1 2 m v 2 K = \tfrac{1}{2}mv^2 K = 2 1 m v 2
WHAT it relates: force-over-distance (W W W Δ K \Delta K Δ K WHY it's useful: it skips time and acceleration. If you know force vs. position, you get the final speed directly — no need to solve for a ( t ) a(t) a ( t )
Newton's law tells us how force changes velocity in time : F = m d v d t F = m\,\dfrac{dv}{dt} F = m d t d v
But we want to know how force changes velocity over distance . So the whole game is: convert the time derivative into a position derivative. That single trick produces the entire theorem.
v d v v\,dv v d v d v d t ⋅ d x = d x d t ⋅ d v = v d v \frac{dv}{dt}\cdot dx = \frac{dx}{dt}\cdot dv = v\,dv d t d v ⋅ d x = d t d x ⋅ d v = v d v "how does speed change each second?" we ask "how does speed change each metre?" — and metres are exactly what work cares about.
For motion in 3-D with F ⃗ = m d v ⃗ d t \vec{F}=m\dfrac{d\vec v}{dt} F = m d t d v F ⃗ ⋅ d r ⃗ = m d v ⃗ d t ⋅ v ⃗ d t = m v ⃗ ⋅ d v ⃗ = d ( 1 2 m v ⃗ ⋅ v ⃗ ) = d ( 1 2 m v 2 ) \vec F\cdot d\vec r = m\frac{d\vec v}{dt}\cdot \vec v\,dt = m\,\vec v\cdot d\vec v = d\!\left(\tfrac12 m\,\vec v\cdot\vec v\right)=d\!\left(\tfrac12 m v^2\right) F ⋅ d r = m d t d v ⋅ v d t = m v ⋅ d v = d ( 2 1 m v ⋅ v ) = d ( 2 1 m v 2 ) ∫ F ⃗ ⋅ d r ⃗ = Δ K \displaystyle \int \vec F\cdot d\vec r = \Delta K ∫ F ⋅ d r = Δ K
Worked example 1 — Block pushed across a floorA 2 kg 2\,\text{kg} 2 kg 10 N 10\,\text{N} 10 N 5 m 5\,\text{m} 5 m
W n e t = F d = 10 × 5 = 50 J W_{net} = F\,d = 10\times5 = 50\,\text{J} W n e t = F d = 10 × 5 = 50 J Why? Constant force ⇒ ∫ F d x = F d \int F\,dx = Fd ∫ F d x = F d Δ K = 50 J \Delta K = 50\,\text{J} Δ K = 50 J K i = 0 K_i=0 K i = 0 K f = 50 J K_f = 50\,\text{J} K f = 50 J Why? Theorem: net work = change in K K K 1 2 ( 2 ) v f 2 = 50 ⇒ v f 2 = 50 ⇒ v f = 50 ≈ 7.07 m/s \tfrac12(2)v_f^2 = 50 \Rightarrow v_f^2 = 50 \Rightarrow v_f = \sqrt{50}\approx 7.07\,\text{m/s} 2 1 ( 2 ) v f 2 = 50 ⇒ v f 2 = 50 ⇒ v f = 50 ≈ 7.07 m/s Why? Solve K = 1 2 m v 2 K=\tfrac12 mv^2 K = 2 1 m v 2 v v v
Worked example 2 — Variable (spring-like) forceA force F ( x ) = 6 x N F(x) = 6x\ \text{N} F ( x ) = 6 x N 3 kg 3\,\text{kg} 3 kg x = 0 x=0 x = 0 x = 2 m x=2\,\text{m} x = 2 m
W = ∫ 0 2 6 x d x = [ 3 x 2 ] 0 2 = 12 J W = \int_0^2 6x\,dx = [3x^2]_0^2 = 12\,\text{J} W = ∫ 0 2 6 x d x = [ 3 x 2 ] 0 2 = 12 J Why? Force varies, so we must integrate, not multiply.1 2 ( 3 ) v f 2 = 12 ⇒ v f 2 = 8 ⇒ v f = 2 2 ≈ 2.83 m/s \tfrac12(3)v_f^2 = 12 \Rightarrow v_f^2 = 8 \Rightarrow v_f = 2\sqrt2 \approx 2.83\,\text{m/s} 2 1 ( 3 ) v f 2 = 12 ⇒ v f 2 = 8 ⇒ v f = 2 2 ≈ 2.83 m/s Why? This is the payoff of the theorem — we never needed a ( t ) a(t) a ( t )
Worked example 3 — Friction slowing a car (negative work)A 1000 kg 1000\,\text{kg} 1000 kg 20 m/s 20\,\text{m/s} 20 m/s
K i = 1 2 ( 1000 ) ( 20 ) 2 = 200,000 J K_i = \tfrac12(1000)(20)^2 = 200{,}000\,\text{J} K i = 2 1 ( 1000 ) ( 20 ) 2 = 200 , 000 J K f = 0 K_f=0 K f = 0 W n e t = Δ K = 0 − 200000 = − 200,000 J W_{net} = \Delta K = 0 - 200000 = -200{,}000\,\text{J} W n e t = Δ K = 0 − 200000 = − 200 , 000 J Why? Friction opposes motion, so it removes kinetic energy — work is negative.
Common mistake "Work = Force × distance, always."Why it feels right: the formula W = F d W=Fd W = F d The fix: that's only true if F F F constant and parallel to motion. In general W = ∫ F ⃗ ⋅ d r ⃗ W=\int \vec F\cdot d\vec r W = ∫ F ⋅ d r
applied force in the theorem."
Why it feels right: you control the force you apply, so it seems central.
The fix: the theorem uses net work (all forces — friction, gravity, normal). A block dragged at constant velocity has W n e t = 0 W_{net}=0 W n e t = 0
Common mistake "Negative work is impossible / means no work."Why it feels right: "work" sounds like effort, which feels positive.
The fix: sign comes from F ⃗ ⋅ d r ⃗ \vec F\cdot d\vec r F ⋅ d r decreases . That's exactly braking.
Common mistake "Kinetic energy depends on direction of velocity."Why it feels right: velocity is a vector.
The fix: K = 1 2 m v 2 K=\tfrac12 m v^2 K = 2 1 m v 2 speed squared ; it's a scalar, always ≥ 0 \ge 0 ≥ 0 5 m/s 5\,\text{m/s} 5 m/s K K K
What time-to-position trick converts F = m a F=ma F = ma
Which "force" appears in W n e t = Δ K W_{net}=\Delta K W n e t = Δ K
Why is friction's work negative?
(Answers: chain rule d v d t d x = v d v \frac{dv}{dt}dx = v\,dv d t d v d x = v d v net force; force opposes displacement so F ⃗ ⋅ d r ⃗ < 0 \vec F\cdot d\vec r<0 F ⋅ d r < 0
Recall Feynman: explain to a 12-year-oldImagine pushing a toy car. The harder and longer you push (that's "work"), the faster it ends up going (that's "energy of motion"). If you grab it and push backwards , you slow it down — you're doing "negative work" and stealing its speed. The theorem just says: all the pushing you do, added up over the whole distance, exactly equals how much the car's go-fast energy changed. Nothing is lost; it's just bookkeeping for motion.
"Net work nets you new speed." And for the derivation: "Multiply by d x dx d x v d v v\,dv v d v
State the Work–Energy Theorem in words. The net work done on a particle equals the change in its kinetic energy:
W n e t = Δ K W_{net}=\Delta K W n e t = Δ K .
What single calculus trick converts F = m a F=ma F = ma Writing
d v d t d x = d x d t d v = v d v \frac{dv}{dt}\,dx = \frac{dx}{dt}\,dv = v\,dv d t d v d x = d t d x d v = v d v , eliminating time.
What is kinetic energy and why is it a scalar? K = 1 2 m v 2 K=\tfrac12 mv^2 K = 2 1 m v 2 ; it depends on speed squared, not on velocity's direction.
In W n e t = Δ K W_{net}=\Delta K W n e t = Δ K ALL forces — the net force (applied, friction, gravity, normal, etc.).
When is W = F d W=Fd W = F d W = ∫ F ⃗ ⋅ d r ⃗ W=\int \vec F\cdot d\vec r W = ∫ F ⋅ d r Only when force is constant and parallel to displacement.
A block is dragged at constant velocity. What is W n e t W_{net} W n e t Zero, since
Δ K = 0 \Delta K=0 Δ K = 0 (your work is cancelled by friction's negative work).
Why can work be negative? When force has a component opposite to displacement,
F ⃗ ⋅ d r ⃗ < 0 \vec F\cdot d\vec r<0 F ⋅ d r < 0 , removing kinetic energy.
Force F ( x ) = 6 x F(x)=6x F ( x ) = 6 x ∫ 0 2 6 x d x = 12 J \int_0^2 6x\,dx = 12\,\text{J} ∫ 0 2 6 x d x = 12 J .
Why is the theorem more convenient than F = m a F=ma F = ma It links force-over-distance directly to speed, skipping acceleration and time.
generalizes via dot product
Time derivative of velocity
Change in kinetic energy ΔK
Force vs position gives final speed
Intuition Hinglish mein samjhoDekho, Work–Energy Theorem ek bahut hi sundar shortcut hai. Newton ka F = m a F=ma F = ma time ke saath kaise badalti hai. Lekin kai baar humein time pata hi nahi hota — humein bas itna jaanna hai ki ek object kitni doori tak force lagne ke baad kitna fast ho jayega. Bas yahi theorem karta hai: net work (W = ∫ F d x W=\int F\,dx W = ∫ F d x W n e t = 1 2 m v f 2 − 1 2 m v i 2 W_{net}=\tfrac12 mv_f^2-\tfrac12 mv_i^2 W n e t = 2 1 m v f 2 − 2 1 m v i 2
Derivation ka asli jaadu ek chhoti si chain-rule trick hai. F = m d v d t F=m\frac{dv}{dt} F = m d t d v d x dx d x d v d t d x = d x d t d v = v d v \frac{dv}{dt}\,dx = \frac{dx}{dt}\,dv = v\,dv d t d v d x = d t d x d v = v d v Δ K \Delta K Δ K d x dx d x v d v v\,dv v d v
Do important baatein jo students bhool jaate hain: pehla, theorem mein net force ka work chahiye — sirf jo tum laga rahe ho woh nahi, balki friction aur gravity sab. Isliye constant velocity par block khinchne par W n e t = 0 W_{net}=0 W n e t = 0 negative bhi ho sakta hai — jab force motion ke opposite ho (jaise brake lagana), tab kinetic energy kam hoti hai.
Kyun important hai? Kyunki variable force (spring, gravity) ke saath a ( t ) a(t) a ( t ) F F F x x x
