Exercises — Work done by variable force — integration

Pehchano: $F$ ek fixed number hai — yeh $x$ par depend nahi karta. Yeh constant-force special case hai, isliye koi integral zaruri nahi (though ${\int }_0^{5}12dx=12\cdot 5$ same result deta hai). $W=Fd=12\times 5=60\text{J}$

• (a) $F$ constant, motion seedhi → $W=Fd$ theek hai.
• (b) $F$, $x$ par depend karta hai → integrate karna padega.
• (c) $F$ ki magnitude constant hai, lekin bent path ka matlab hai ki $\vec{F}$ aur motion ke beech ka angle badlega, isliye phir bhi $W=\int \vec{F}\cdot d\vec{r}$ chahiye — integrate karo (dekho Dot product and components of vectors).
• (d) $F$, $x$ par depend karta hai → integrate karna padega.

Work = curve ke neeche area = rectangle ka area. $W=10\times 6=60\text{J}$ Rectangle sirf constant-force case hai jo area ke roop mein draw kiya gaya hai — yeh reassuring hai ki dono pictures agree karte hain.

Integrate kyun? $F$, $x$ ke saath badhta hai, isliye koi ek value kaam nahi karti. $W={\int }_0^{3}4xdx=4{\left[\frac{x^2}{2}\right]}_0^3=2\left(3^2-0\right)=18\text{J}$ Geometry se check karo: $F=4x$ ka graph $(0,0)$ se $(3,12)$ tak seedhi line hai; base $3$, height $12$ wala triangle, area $\frac{1}{2}(3)(12)=18\text{J}$. ✓

$W={\int }_1^2(6x^2-2)dx=[2x^3-2x{]}_1^2$ Humne kya kiya: $6x^2$ ka antiderivative $2x^3$ hai (power ek badhao, nayi power se divide karo), aur $-2$ ka $-2x$ hai. $=(2\cdot 8-4)-(2-2)=(16-4)-0=12\text{J}$

$W={\int }_0^4(-3x)dx=-3{\left[\frac{x^2}{2}\right]}_0^4=-3\left(\frac{16}{2}\right)=-24\text{J}$ Sign ka matlab: har sliver $Fdx$ negative hai kyunki force displacement ka oppose karta hai — body se energy nikaali ja rahi hai (jaise ek brake). Area $x$-axis ke neeche hai.

Area ko ek triangle aur ek rectangle mein split karo.

• Triangle $(0\to 3)$: $\frac{1}{2}(3)(30)=45\text{J}$
• Rectangle $(3\to 7)$: $(7-3)(30)=120\text{J}$ $W=45+120=165\text{J}$

Signed area — axis ke neeche wala part subtract karta hai.

• Axis ke upar $(0\to 2)$: $+10\times 2=+20\text{J}$
• Axis ke neeche $(2\to 5)$: $-10\times 3=-30\text{J}$ $W_{\text{net}}=20+(-30)=-10\text{J}$ Kyun overall negative ho sakta hai: opposing stretch ke dauran jitni energy nikali gayi, usse zyada energy daali nahi gayi.

$W={\int }_0^4(x-2)dx={\left[\frac{x^2}{2}-2x\right]}_0^4=\left(8-8\right)-0=0\text{J}$ Zero kyun? Line $F=x-2$, $x<2$ ke liye negative hai (force motion ko oppose karta hai) aur $x>2$ ke liye positive. Axis ke neeche wala triangle $(0\to 2)$ area $\frac{1}{2}(2)(2)=2\text{J}$ below hai, aur upar wala triangle $(2\to 4)$ area $\frac{1}{2}(2)(2)=2\text{J}$ above hai. Yeh exactly cancel ho jaate hain.

Applied force $F_{\text{app}}=kx$ hai; spring force $F_{\text{spring}}=-kx$ hai (dekho Hooke's law and spring potential energy). (a) Tumhara work: $W_{\text{you}}={\int }_0^{0.1}kxdx=\frac{1}{2}k{x}_0^2=\frac{1}{2}(200)(0.1{)}^2=1\text{J}$ (b) Spring dwara kiya gaya work uska negative hai (opposite sign force, stretch ki same magnitude): $W_{\text{spring}}=-\frac{1}{2}k{x}_0^2=-1\text{J}$

Step 1 — work: $W={\int }_0^{2}6xdx=3{\left[x^2\right]}_0^2=3(4)=12\text{J}$ Step 2 — work–energy theorem kehta hai $W=\Delta KE=\frac{1}{2}m{v}^2-0$: $12=\frac{1}{2}(2)v^2=v^2\Rightarrow v=\sqrt{12}\approx 3.46\text{m/s}$

(b) Reversal point: force zero hoti hai jab $12-4x=0\Rightarrow x=3\text{m}$. Toh $0\to 3$ par force positive rehti hai (sirf end mein zero touch karti hai) — interval ke andar yeh actually reverse nahi karti. (a) Net work: $W={\int }_0^3(12-4x)dx={\left[12x-2x^2\right]}_0^3=(36-18)-0=18\text{J}$

(a) $W={\int }_0^2(4-x^2)dx={\left[4x-\frac{x^3}{3}\right]}_0^2=\left(8-\frac{8}{3}\right)-0=\frac{16}{3}\approx 5.33\text{J}$ (b) Work–energy theorem $v_i=0$ ke saath: $\frac{16}{3}=\frac{1}{2}(1)v^2\Rightarrow v^2=\frac{32}{3}\Rightarrow v=\sqrt{\frac{32}{3}}\approx 3.27\text{m/s}$ (c) $x>2$ ke liye, $x^2>4$, toh $F=4-x^2<0$ — force ab backward point karta hai, particle ko decelerate karta hai. Exactly $x=2$ par force zero hai (push ka turning-over point).

$W={\int }_2^{2}5xdx=0\text{J}$ Equal limits ⇒ zero-width interval ⇒ zero area ⇒ zero work. Rule: work ke liye displacement chahiye; koi net movement nahi (straight-line, single-valued force mein), toh slivers ka sum empty hai. (Back-and-forth paths ke liye integral ko legs mein split karo aur signed pieces add karo — lekin yahan position kabhi badi hi nahi.)

$F=6x^{-2}$ likhte hain. Uska antiderivative: power ko $-1$ kar do, $-1$ se divide karo: $6\cdot \frac{x^{-1}}{-1}=-\frac{6}{x}$. $W={\int }_1^{3}6x^{-2}dx={\left[-\frac{6}{x}\right]}_1^3=\left(-\frac{6}{3}\right)-\left(-\frac{6}{1}\right)=-2+6=4\text{J}$ Limiting behaviour: agar body bilkul $x\to \infty$ tak jaaye, $W={\left[-\frac{6}{x}\right]}_1^{\infty }=0-(-6)=6\text{J}$ Work finite rehta hai chahe infinite distance ho, kyunki force $1/x^2$ ki tarah fade karti hai — tail area itna chhota hai ki ek finite number mein sum ho jaata hai. Yeh bilkul wohi reasoning hai jo escape-energy calculations mein use hoti hai.