Intuition The big pictureImagine a ball falling. As it drops, it speeds up (gains kinetic energy) but loses height (loses potential energy). The magic is: the amount of KE gained is exactly the amount of PE lost. Nothing is created or destroyed — energy just changes costume . The total stays frozen. That frozen total is what we call mechanical energy , and proving it stays constant is this whole note.
Definition Mechanical energyThe mechanical energy E E E E = K + U E = K + U E = K + U K = 1 2 m v 2 K = \tfrac{1}{2}mv^2 K = 2 1 m v 2 U U U U = m g h U=mgh U = m g h
Definition Conservative forceA force is conservative if the work it does on an object moving between two points is independent of the path taken (equivalently, work over any closed loop is zero). Gravity and spring forces are conservative; friction is not .
The claim: If only conservative forces do work, then E = K + U E = K + U E = K + U constant in time.
Two facts do all the heavy lifting:
Work–energy theorem (from Newton's 2nd law): net work = change in kinetic energy.Definition of potential energy : the work done by a conservative force is minus the change in its potential energy.
Put those side by side and the result falls out. Let's derive each piece from scratch .
Start from Newton's second law in one dimension, F n e t = m a = m d v d t F_{net} = ma = m\dfrac{dv}{dt} F n e t = ma = m d t d v
The net work done as the object moves a small displacement d x dx d x d W n e t = F n e t d x = m d v d t d x dW_{net} = F_{net}\,dx = m\frac{dv}{dt}\,dx d W n e t = F n e t d x = m d t d v d x
Worked example Why this step?We use a trick of the chain rule to swap variables. Since d x = v d t dx = v\,dt d x = v d t d v d t d x = d v d t ( v d t ) = v d v \frac{dv}{dt}\,dx = \frac{dv}{dt}\,(v\,dt) = v\,dv d t d v d x = d t d v ( v d t ) = v d v v v v t t t
So:
d W n e t = m v d v dW_{net} = m\,v\,dv d W n e t = m v d v
Integrate from state 1 (v 1 v_1 v 1 v 2 v_2 v 2 W n e t = ∫ v 1 v 2 m v d v = 1 2 m v 2 2 − 1 2 m v 1 2 = Δ K W_{net} = \int_{v_1}^{v_2} m\,v\,dv = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2 = \Delta K W n e t = ∫ v 1 v 2 m v d v = 2 1 m v 2 2 − 2 1 m v 1 2 = Δ K
For a conservative force, we define potential energy so that the work it does equals the drop in potential energy:
W c o n s = − Δ U = − ( U 2 − U 1 ) W_{cons} = -\Delta U = -(U_2 - U_1) W co n s = − Δ U = − ( U 2 − U 1 )
Intuition Why the minus sign?Lift a ball up against gravity: gravity does negative work, yet PE goes up . To make "work by gravity" and "change in PE" line up with opposite signs, we bake the minus into the definition. Check: ball falls ⇒ \Rightarrow ⇒ ⇒ U \Rightarrow U ⇒ U
For gravity near Earth, F = − m g F = -mg F = − m g h 1 h_1 h 1 h 2 h_2 h 2 W g r a v = ∫ h 1 h 2 ( − m g ) d h = − m g ( h 2 − h 1 ) = − ( m g h 2 − m g h 1 ) W_{grav} = \int_{h_1}^{h_2}(-mg)\,dh = -mg(h_2 - h_1) = -(mgh_2 - mgh_1) W g r a v = ∫ h 1 h 2 ( − m g ) d h = − m g ( h 2 − h 1 ) = − ( m g h 2 − m g h 1 ) W c o n s = − Δ U W_{cons} = -\Delta U W co n s = − Δ U U = m g h U = mgh U = m g h We derived it, not assumed it.
If the only force doing work is conservative, then W n e t = W c o n s W_{net} = W_{cons} W n e t = W co n s Δ K = W n e t = W c o n s = − Δ U \Delta K = W_{net} = W_{cons} = -\Delta U Δ K = W n e t = W co n s = − Δ U
Δ K + Δ U = 0 ⟹ Δ ( K + U ) = 0 \boxed{\Delta K + \Delta U = 0 \quad\Longrightarrow\quad \Delta(K+U)=0} Δ K + Δ U = 0 ⟹ Δ ( K + U ) = 0
Recall Forecast before reading the answerA ball is dropped from rest at height h h h Predict its speed at the ground using energy conservation, then verify with kinematics.
Energy: K 1 + U 1 = K 2 + U 2 ⇒ 0 + m g h = 1 2 m v 2 + 0 ⇒ v = 2 g h K_1+U_1 = K_2+U_2 \Rightarrow 0 + mgh = \tfrac12 mv^2 + 0 \Rightarrow v=\sqrt{2gh} K 1 + U 1 = K 2 + U 2 ⇒ 0 + m g h = 2 1 m v 2 + 0 ⇒ v = 2 g h Kinematics: v 2 = u 2 + 2 g h = 0 + 2 g h ⇒ v = 2 g h v^2 = u^2 + 2gh = 0 + 2gh \Rightarrow v=\sqrt{2gh} v 2 = u 2 + 2 g h = 0 + 2 g h ⇒ v = 2 g h Same answer — energy method skipped time entirely.
Worked example Example 1 — Pendulum bobA bob of mass m m m h = L h=L h = L
Setup: Tension is always perpendicular to motion ⇒ does zero work ⇒ only gravity (conservative) works ⇒ energy conserved.
Why this step? If a non-working force were present, we'd be forced to use the work–energy theorem with extra terms; checking tension does no work is what licenses the conservation law.
m g L = 1 2 m v 2 ⇒ v = 2 g L mgL = \tfrac12 mv^2 \Rightarrow v=\sqrt{2gL} m g L = 2 1 m v 2 ⇒ v = 2 g L
Worked example Example 2 — Block on a springA block (m m m k k k x x x
Spring PE is U s p r i n g = 1 2 k x 2 U_{spring}=\tfrac12 kx^2 U s p r in g = 2 1 k x 2 W s p r i n g = ∫ 0 x ( − k x ′ ) d x ′ = − 1 2 k x 2 = − Δ U W_{spring}=\int_0^x(-kx')dx' = -\tfrac12kx^2 = -\Delta U W s p r in g = ∫ 0 x ( − k x ′ ) d x ′ = − 2 1 k x 2 = − Δ U Why this step? We must build the correct U U U this conservative force before plugging in.
1 2 k x 2 = 1 2 m v 2 ⇒ v = x k m \tfrac12 kx^2 = \tfrac12 mv^2 \Rightarrow v = x\sqrt{\tfrac{k}{m}} 2 1 k x 2 = 2 1 m v 2 ⇒ v = x m k
Worked example Example 3 — Mixing heights and springsA ball (m m m h h h x x x k k k Why this step? At maximum compression the ball is momentarily at rest (K = 0 K=0 K = 0 m g h = 1 2 k x 2 ⇒ k = 2 m g h x 2 mgh = \tfrac12 kx^2 \Rightarrow k = \frac{2mgh}{x^2} m g h = 2 1 k x 2 ⇒ k = x 2 2 m g h
Common mistake "Energy is always conserved, so I can always writeK 1 + U 1 = K 2 + U 2 K_1+U_1=K_2+U_2 K 1 + U 1 = K 2 + U 2 Why it feels right: the total energy of the universe really is conserved, so the phrase sounds bulletproof.
The fix: Mechanical energy is conserved only when non-conservative forces do no work . Friction turns KE into heat; that heat is energy, but it's no longer in K + U K+U K + U K 2 + U 2 = K 1 + U 1 + W f r i c t i o n K_2+U_2 = K_1+U_1 + W_{friction} K 2 + U 2 = K 1 + U 1 + W f r i c t i o n W f r i c t i o n < 0 W_{friction}<0 W f r i c t i o n < 0
Common mistake "I'll include the work done by gravityand the gravitational PE."
Why it feels right: gravity does work, so surely we add its work term.
The fix: That's double-counting . Potential energy is the bookkeeping for the conservative force's work. Either track gravity as a force (work–energy theorem) or as U = m g h U=mgh U = m g h
Common mistake "Tension / normal force must do work because they're big forces."Why it feels right: large magnitude feels energetic.
The fix: Work = F ⃗ ⋅ d ⃗ = \vec F\cdot\vec d = F ⋅ d perpendicular to motion (tension on a swinging string, normal force on a slope) do zero work regardless of size.
Recall Feynman: explain to a 12-year-oldThink of energy like pocket money you can keep in two pockets: a "moving" pocket (kinetic) and a "stored-up" pocket (potential — like being high up or a squished spring). When a ball falls, money slides from the "high-up" pocket into the "moving" pocket — but the total money in your two pockets never changes . The only way to lose money is if a sticky thief called friction grabs some and turns it into heat. No friction, no thief → your total stays exactly the same forever.
"K Up, U Down — Total's Frozen."
And for when it works: "No friction, no problem" (only conservative forces ⇒ energy conserved).
What is mechanical energy? The sum of kinetic and potential energy,
E = K + U E=K+U E = K + U .
State the work–energy theorem. The net work done on an object equals its change in kinetic energy,
W n e t = Δ K W_{net}=\Delta K W n e t = Δ K .
How is potential energy defined from a conservative force? As the negative of the work done by that force:
W c o n s = − Δ U W_{cons}=-\Delta U W co n s = − Δ U .
Under what condition is mechanical energy conserved? When the only forces doing work are conservative (no net work by non-conservative forces like friction).
Derive v v v h h h m g h = 1 2 m v 2 ⇒ v = 2 g h mgh=\tfrac12mv^2 \Rightarrow v=\sqrt{2gh} m g h = 2 1 m v 2 ⇒ v = 2 g h .
Why does the minus sign appear in W c o n s = − Δ U W_{cons}=-\Delta U W co n s = − Δ U So that work done by the force and the change in stored PE have opposite signs (force does +work ⇒ PE drops).
Why is energy conservation easier than kinematics here? It eliminates time and force direction; you only compare two states.
What replaces Δ E = 0 \Delta E=0 Δ E = 0 Δ ( K + U ) = W n c \Delta(K+U)=W_{nc} Δ ( K + U ) = W n c , the work done by non-conservative forces (negative for friction).
Why does string tension do no work on a pendulum? It is always perpendicular to the bob's velocity, so
F ⃗ ⋅ d ⃗ = 0 \vec F\cdot\vec d=0 F ⋅ d = 0 .
Spring potential energy formula and its derivation source? U = 1 2 k x 2 U=\tfrac12kx^2 U = 2 1 k x 2 , from
− ∫ 0 x ( − k x ′ ) d x ′ -\int_0^x(-kx')dx' − ∫ 0 x ( − k x ′ ) d x ′ .
Mechanical energy conserved
Friction non-conservative
Intuition Hinglish mein samjhoSocho ek ball ko tum upar se neeche girate ho. Jaise-jaise woh girti hai, uski speed badhti jaati hai — yaani kinetic energy badh rahi hai. Par saath hi uski height kam ho rahi hai — yaani potential energy (m g h mgh m g h bilkul utni hi KE badh jaati hai. Energy gayab nahi hoti, sirf ek roop se doosre roop mein badalti hai. Dono ka total — jise hum mechanical energy kehte hain — same rehta hai. Bas yahi conservation of mechanical energy hai.
Iska proof do simple cheezon se aata hai. Pehla: work-energy theorem , jo Newton ke second law se aata hai — net work = change in kinetic energy (W n e t = Δ K W_{net}=\Delta K W n e t = Δ K W c o n s = − Δ U W_{cons}=-\Delta U W co n s = − Δ U W n e t = W c o n s W_{net}=W_{cons} W n e t = W co n s Δ K = − Δ U \Delta K = -\Delta U Δ K = − Δ U Δ ( K + U ) = 0 \Delta(K+U)=0 Δ ( K + U ) = 0
Sabse important baat — ye law tabhi chalta hai jab friction jaisi non-conservative force kaam na kare. Friction energy ko heat mein badal deta hai, jo K + U K+U K + U K 1 + U 1 = K 2 + U 2 K_1+U_1=K_2+U_2 K 1 + U 1 = K 2 + U 2
Yeh method kyu useful hai? Kyunki ismein na time chahiye, na force ki direction ka tension. Sirf do states — start aur end — compare karo, aur answer mil jaata hai. Pendulum, spring, ramp — sab jagah ek hi trick.
