Work-energy theorem — derivation from Newton's second law: Question bank

1.3.3 · D5 · Physics › Work, Energy & Power › Work-energy theorem — derivation from Newton's second law

Is page par har jagah yeh do ideas probe ki ja rahi hain:

Work $W = \int F \cdot d r$ — actual path ke saath force ka sum, sign ke saath.
Kinetic energy $K = \frac{1}{2} m v^{2}$ — speed squared se bana ek scalar, kabhi negative nahi, kabhi directional nahi.

True ya false — justify karo

Neeche har claim fact ki tarah stated hai. Decide karo true ya false aur ek-line ka reason do.

"Agar kisi object par net work zero hai, toh object move nahi kar raha."

False — zero net work ka matlab hai

Δ K = 0

, yaani speed unchanged hai, zero nahi; ek block jo constant velocity se khicha ja raha ho poore time move karta hai aur uska

W_{n e t} = 0

hota hai.

"Ek force kisi object par bade distance tak act kar sakti hai aur phir bhi zero work kar sakti hai."

True — agar force hamesha displacement ke perpendicular ho (jaise circular motion mein tension, slope par normal force), toh har step par

F \cdot d r = 0

hoga, isliye integral zero hoga.

"Kinetic energy negative ho sakti hai agar object backwards move kare."

False —

K = \frac{1}{2} m v^{2}

mein speed squared use hoti hai, isliye yeh direction ki parwah kiye bina

\geq 0

hoti hai; sirf work (jo sign carry karta hai) negative ho sakta hai.

"Kisi object ki speed double karne se uski kinetic energy double ho jaati hai."

False —

K \propto v^{2}

, isliye

v

double karne se

K

chaar guna ho jaati hai; isliye speed ke saath stopping distances itni tezi se badhte hain.

"Work–energy theorem sirf constant forces ke liye kaam karta hai."

False — derivation

\int F d x

integrate karta hai, isliye position par kisi bhi known dependence wali force ke liye kaam karta hai; constant-force case

W = F d

bas sabse aasan special case hai.

"Do identical cars, ek north jaati hui aur ek south jaati hui same speed par, inki kinetic energies opposite hoti hain."

False —

K

speed squared par depend karta hai, jo ek scalar hai; dono ki same positive

K

hoti hai, direction irrelevant hai.

"Agar tum ek box ko push karo aur woh move na kare, toh tumne uske par work kiya hai."

False — displacement zero hai, isliye

\int F \cdot d r = 0

; tumhari muscles thak jaati hain (biochemical cost) lekin box par physics work zero hai.

"Work–energy theorem ek nayi physical law hai jo Newton's laws se independent hai."

False — yeh

F = ma

ko distance par integrate karna hai; isme koi nayi physics nahi hai, bas time se position tak variable ka change hai.

"Friction hamesha negative work karta hai."

False — sliding object par kinetic friction negative work karta hai, lekin friction positive work bhi kar sakta hai: jaise conveyor belt wala friction jo ek box ko speed up karta hai, ya woh friction jo car ke tyres ko forward accelerate karta hai.

Error dhundo

Har item mein reasoning mein ek subtle mistake hai. Use dhundo aur fix batao.

"Final speed nikalne ke liye, sirf applied force dwara kiya gaya work compute karo, phir use

Δ K

ke barabar set karo."

Error: theorem net work use karta hai, applied work nahi; tumhe friction aur kisi bhi opposing forces ke work ko subtract karna hoga — warna

Δ K

zyada estimate ho jaata hai.

"Kyunki

W = F d

, motion ke saath

6 0^{\circ}

par ek force

4 m

mein

F = 10 N

ke saath

40 J

work karta hai."

Error:

W = F d

mein force parallel to motion chahiye; yahan

W = F d cos 6 0^{\circ} = 10 \times 4 \times 0.5 = 20 J

hai,

40 J

nahi.

"Derivation likhti hai

\frac{d v}{d t} d x = v d v

; yeh bas

d t

d x

ke saath fractions ki tarah cancel karna hai."

Error (justification mein): yeh naive cancellation nahi hai — hum

\frac{d x}{d t} = v

substitute karte hain, yaani

\frac{d v}{d t} d x = \frac{d x}{d t} d v = v d v

d t

chain rule ke zariye regroup hota hai, jo ek real theorem hai, fraction magic nahi.

"Seedha upar phenka hua ball slow down hota hai, isliye upar jaate waqt gravity zero work karta hai."

Error: gravity upar jaate waqt negative work karta hai (force neeche, displacement upar), isliye exactly

K

decrease hoti hai; "zero" ka matlab hota speed change nahi.

"Kyunki kinetic energy conserved hai, final speed initial speed ke barabar hogi."

Error: kinetic energy generally conserved nahi hoti; sirf tab jab net work zero ho. Theorem kehta hai

Δ K = W_{n e t}

, aur

W_{n e t}

usually nonzero hota hai.

"Variable force

F (x) = 6 x

ke liye, work hai

F \times d = 6 (2) \times 2 = 24 J

Error: varying force ke liye

F \times d

use nahi kar sakte; integrate karna padega:

\int_{0}^{2} 6 x d x = 12 J

. Final force value use karna overcount karta hai.

"Theorem bahut saare particles ke system par bhi apply hoti hai — bas total mass aur centre-of-mass displacement use karo."

Error: deformable/multi-particle systems ke liye,

W_{n e t} = Δ K

strictly ek single particle ke liye hold karta hai; system ke liye internal work aur energy account karni padti hai, isliye naive extension fail ho sakti hai.

Why wale questions

"Why" ka jawab actual mechanism se do, koi restatement nahi.

"Hum Newton's law ko

d x

se kyun multiply karte hain

d t

se nahi?"

Kyunki hum work

\int F d x

chahte hain (force over distance);

d t

se multiply karne par impulse

\int F d t = Δ p

milta hai, jo Impulse–Momentum Theorem hai — ek alag, time-based sibling.

v d v

trick 'time eliminate' kyun karti hai?"

Substitution

\frac{d x}{d t} = v

time derivative ko velocity mein absorb kar leta hai,

m v d v

chhod deta hai jisme koi

t

nahi hota; time drop out ho jaata hai kyunki hum position par integrate karte hain time ki jagah.

"Theorem

F = ma

se zyada convenient kyun hai jab force position par depend kare?"

F = ma

solve karne ke liye

a (t)

nikalna padta hai phir twice integrate karna; theorem

\int F d x

ko seedha speed change se link karta hai, messy time-dependence skip karte hue.

"Dot product

F \cdot d r

direction ko 'automatically' kyun handle kar leta hai?"

Yeh

F d r cos θ

return karta hai, isliye yeh positive hota hai jab force motion aid kare, negative jab oppose kare, aur zero jab perpendicular ho — work ka sign geometry se automatically aa jaata hai, koi extra rules nahi.

Δ K

ek scalar kyun hai jabki force aur displacement vectors hain?"

Dot product do vectors ko ek single number mein badal deta hai, aur

\frac{1}{2} m v^{2}

sirf speed par depend karta hai; direction information dot product mein 'use up' ho jaati hai aur discard ho jaati hai.

"Braking negative work kyun correspond karta hai?"

Friction/braking force displacement ke opposite point karta hai, isliye

F \cdot d r < 0

; yeh kinetic energy remove karta hai, jo exactly slow down hone ka matlab hai.

Edge cases

Boundary aur degenerate scenarios jinhe theorem survive karna chahiye.

"Perfectly constant velocity se move karte object ke liye

W_{n e t}

kya hai?"

Exactly zero —

Δ K = 0

kyunki speed unchanged hai, isliye saari forces ke work ka net cancel ho jaata hai (koi bhi applied work opposing work se balance hoti hai).

"Theorem kya kehti hai agar object same speed se start aur end kare lekin curved path le?"

W_{n e t} = Δ K = 0

, chahe individual forces ne raaste mein bade positive aur negative works kiye hon — sirf path par net matter karta hai.

"Ek object apne rise ke top par momentarily rest par hai. Kya uski kinetic energy zero hai? Kya net force zero hai?"

K = 0

(speed zero hai) lekin net force zero nahi hai — gravity abhi bhi act kar rahi hai, isliye yeh turant phir move karna shuru kar deta hai; zero

K

kabhi zero force imply nahi karta.

"Ek force apply ki gayi hai lekin displacement zero hai (wall ko push karna). Work kya hai?"

Zero, kyunki

\int F \cdot d r = 0

jab

d r = 0

; koi distance nahi matlab koi work nahi chahe force kitni bhi badi ho.

"Agar path ke har jagah net force zero hai, toh

Δ K

kya hai?"

Zero —

W_{n e t} = \int F_{n e t} d x = 0

, isliye kinetic energy unchanged hai; object constant speed par drift karta hai (Newton's first law, special case ke roop mein recover hota hai).

"Ek object speed up hota hai, phir wapas apni original speed tak slow ho jaata hai. Total net work kya hai?"

Overall zero, kyunki

Δ K = K_{f} - K_{i} = 0

; speed-up ke dauran positive work exactly slow-down ke dauran negative work se undo ho jaata hai.

Connections

Newton's Second Law — woh single input jis par har trap trace back karta hai.
Kinetic Energy — kyun $K \geq 0$ aur $v^{2}$ ke saath scale karta hai, kai traps ka source.
Work done by a variable force — integral vs. $F d$ confusion yahan rehti hai.
Conservation of Mechanical Energy — "K is conserved" trap actually kab hota hai iske baare mein hai.
Power — work ki rate, work se alag.
Impulse–Momentum Theorem — $d t$ -sibling jo hamare $d x$ -theorem se contrast hota hai.