Exercises — Spring potential energy — derivation

Kya: seedha $U=\frac{1}{2}k{x}^2$ mein plug karo. Kyun: spring ideal hai, $x$ already relaxed length se measure ki gayi hai, toh koi adjustment nahi chahiye. $U=\frac{1}{2}(300)(0.05{)}^2=\frac{1}{2}(300)(0.0025)=0.375\text{J}$ Answer: $U=0.375\text{J}$.

Kya: $x=+0.05$ aur $x=-0.05$ ke liye $U$ compare karo. Kyun: $x$ $U=\frac{1}{2}k{x}^2$ mein square hota hai, toh sign khatam ho jaata hai. Dono mein $(0.05{)}^2=0.0025$ aata hai. $U=\frac{1}{2}(300)(-0.05{)}^2=0.375\text{J}$ Answer: bilkul same, $0.375\text{J}$. Barabar size ka compression aur stretch barabar energy store karte hain.

Kya: $U=\frac{1}{2}k{x}^2$ ko $k$ ke liye rearrange karo. Kyun: humein $U$ aur $x$ pata hai, $k$ chahiye — use isolate karo. $k=\frac{2U}{x^2}=\frac{2(2)}{(0.20{)}^2}=\frac{4}{0.04}=100\text{N/m}$ Answer: $k=100\text{N/m}$.

Kya: dono energies compute karo aur unka ratio lo. Kyun: kyunki $U\propto x^2$, $x$ ko teen guna karne par $U$ $3^2$ se scale honi chahiye. Verify karte hain. $U_1=\frac{1}{2}(250)(0.10{)}^2=1.25\text{J}$ $U_2=\frac{1}{2}(250)(0.30{)}^2=11.25\text{J}$ $\frac{U_2}{U_1}=\frac{11.25}{1.25}=9$ Answer: energy 9× badhti hai (3× nahi), kyunki $U\propto x^2$.

Kya: extra work stored energies ka difference hai. Kyun: tumhara kiya gaya work conservative spring ke against = stored PE mein change. Q1 se dono values hain. $W=U_2-U_1=11.25-1.25=10\text{J}$ Answer: $W=10\text{J}$. Kaisa dikhta hai: force–$x$ graph par yeh $x=0.10$ aur $x=0.30$ ke beech ka trapezoid strip hai — bada triangle minus chota triangle.

Kya: stored spring energy ko block ki kinetic energy ke barabar set karo. Kyun: Conservation of mechanical energy — koi friction nahi, toh saari elastic PE motion energy $\frac{1}{2}m{v}^2$ ban jaati hai. $U=\frac{1}{2}(800)(0.05{)}^2=\frac{1}{2}(800)(0.0025)=1\text{J}$ $\frac{1}{2}m{v}^2=1\Rightarrow \frac{1}{2}(0.4)v^2=1\Rightarrow v^2=\frac{1}{0.2}=5$ $v=\sqrt{5}\approx 2.236\text{m/s}$ Answer: $v\approx 2.24\text{m/s}$.

Kya (part 1): rest par spring force gravity balance karti hai: $kx=mg$. Kyun: "rests" ka matlab zero net force hai, toh upar ki spring pull neeche ke weight ke barabar hai. $x=\frac{mg}{k}=\frac{(0.5)(10)}{200}=\frac{5}{200}=0.025\text{m}$ Kya (part 2): us stretch par stored spring energy: $U=\frac{1}{2}k{x}^2=\frac{1}{2}(200)(0.025{)}^2=\frac{1}{2}(200)(0.000625)=0.0625\text{J}$ Answer: stretch $=0.025\text{m}$, stored energy $U=0.0625\text{J}$. Note: Gravitational potential energy $U=mgh$ (constant force) se contrast karo — yahan spring force change hoti hai, isliye triangle formula chahiye.

Kya: triangle area = $\frac{1}{2}\times \text{base}\times \text{height}$ compute karo, base $x$ aur height $F=kx$ ke saath. Kyun: force–displacement graph ke neeche ka area hi kiya gaya work (stored energy) hai.

• base $=x=0.025\text{m}$
• height $=kx=200(0.025)=5\text{N}$ $\text{area}=\frac{1}{2}(0.025)(5)=0.0625\text{J}$ Answer: Q1 se exactly match karta hai — $0.0625\text{J}$. Neeche ki figure dekho: height triangle ki tip hai, base stretch hai.

Kya: released energy $=U_{\text{start}}-U_{\text{end}}$. Kyun: spring partially relax hone par stored energy ka difference deti hai. $U_{\text{start}}=\frac{1}{2}(1000)(0.10{)}^2=5\text{J}$ $U_{\text{end}}=\frac{1}{2}(1000)(0.04{)}^2=0.8\text{J}$ $\Delta U=5-0.8=4.2\text{J}$ Answer: $4.2\text{J}$ release hua.

Kya: saari spring PE highest point par gravitational PE mein convert ho jaati hai (jahan speed $=0$). Kyun: Conservation of mechanical energy — frictionless hai, toh spring energy → gravitational energy $mgh$. $\frac{1}{2}k{x}^2=mgh$ $\frac{1}{2}(400)(0.20{)}^2=(0.5)(10)h$ $\frac{1}{2}(400)(0.04)=5h\Rightarrow 8=5h\Rightarrow h=1.6\text{m}$ Answer: $h=1.6\text{m}$ (vertical rise ke roop mein measured, ramp angle se independent).

Kya: spring energy = kinetic energy gain + friction mein gayi energy. Kyun: energy bookkeeping — friction $f\times d$ energy heat ke roop mein remove karta hai; baki motion ban jaati hai. $U=\frac{1}{2}(400)(0.20{)}^2=8\text{J}$ $E_{\text{friction}}=fd=2\times 1.5=3\text{J}$ $\frac{1}{2}m{v}^2=U-E_{\text{friction}}=8-3=5\text{J}$ $\frac{1}{2}(0.5)v^2=5\Rightarrow v^2=\frac{5}{0.25}=20\Rightarrow v=\sqrt{20}\approx 4.472\text{m/s}$ Answer: $v\approx 4.47\text{m/s}$.

Kya: woh point nikalo jahan block par net force zero ho. Kyun: speed tab tak badhti rehti hai jab tak spring push karte rehti hai ($F>0$) aur push khatam hote hi badhna band ho jaati hai. Maximum speed exactly wahan hoti hai jahan spring force zero reach kare — natural length, $x=0$. $x=0$ par saari stored energy $U=\frac{1}{2}(800)(0.05{)}^2=1\text{J}$ kinetic ban chuki hai: $\frac{1}{2}(0.4)v_{\max }^2=1\Rightarrow v_{\max }^2=5\Rightarrow v_{\max }=\sqrt{5}\approx 2.236\text{m/s}$ Answer: maximum speed relaxed length par hoti hai ($x=0$), $v_{\max }\approx 2.24\text{m/s}$.

Kya: applied force spring ko balance karti hai: $F_{\text{applied}}=k{x}^{\prime }+\beta x^{\prime 3}$. Ise $0$ se $x$ tak integrate karo. Kyun: parent note ki derivation logic general hai — Work done by a variable force ka matlab hai $U={\int }_0^x{F}_{\text{applied}}d{x}^{\prime }$ chahe koi bhi force law ho, jab tak woh conservative ho. $U={\int }_0^x(k{x}^{\prime }+\beta x^{\prime 3})d{x}^{\prime }=k\frac{x^2}{2}+\beta \frac{x^4}{4}=\frac{1}{2}k{x}^2+\frac{1}{4}\beta x^4$ Ab evaluate karo: $U=\frac{1}{2}(100)(0.10{)}^2+\frac{1}{4}(5000)(0.10{)}^4$ $=\frac{1}{2}(100)(0.01)+\frac{1}{4}(5000)(0.0001)$ $=0.5+0.125=0.625\text{J}$ Answer: $U(x)=\frac{1}{2}k{x}^2+\frac{1}{4}\beta x^4$; numerically $0.625\text{J}$. Insight: linear part hamara familiar triangle hai; cubic ek curved sliver upar add karta hai kyunki force straight line se zyada tezi se curve karta hai.

Kya (step 1): effective spring constant nikalo. Series mein, $\frac{1}{k_{\text{eff}}}=\frac{1}{k_1}+\frac{1}{k_2}$. Kyun: same tension $F$ har ek ko stretch karta hai ($F=k_1{x}_1=k_2{x}_2$), aur total extension $x=x_1+x_2$. Substitute karne par reciprocal rule milta hai. $\frac{1}{k_{\text{eff}}}=\frac{1}{300}+\frac{1}{600}=\frac{2}{600}+\frac{1}{600}=\frac{3}{600}=\frac{1}{200}$ $k_{\text{eff}}=200\text{N/m}$ Kya (step 2): total energy = $\frac{1}{2}{k}_{\text{eff}}{x}^2$. Kyun: series pair ek spring ki tarah behave karta hai stiffness $k_{\text{eff}}$ ke saath, toh triangle formula uس par apply hota hai. $U=\frac{1}{2}(200)(0.09{)}^2=\frac{1}{2}(200)(0.0081)=0.81\text{J}$ Answer: $U=0.81\text{J}$.

Kya (a): oscillator ki total mechanical energy maximum stretch, $x=A$ par spring energy ke barabar hai — wahan block momentarily still hai, toh saari energy spring mein PE ke roop mein baith jaati hai. Kyun (a): Simple Harmonic Motion mein energy spring PE aur kinetic energy ke beech aage-peeche slosh karti hai, lekin total constant rehta hai; turning point par evaluate karna (jahan kinetic $=0$ hai) poori amount padhne ki sabse aasaan jagah hai. $E=\frac{1}{2}k{A}^2=\frac{1}{2}(50)(0.12{)}^2=\frac{1}{2}(50)(0.0144)=0.36\text{J}$ Toh (a) $E=0.36\text{J}$. Kya (b): $x=0.06\text{m}$ par spring PE nikalo, phir total se subtract karo kinetic part pane ke liye, phir divide karo. Kyun (b): kisi bhi position par energy spring PE $=\frac{1}{2}k{x}^2$ aur kinetic $=E-U$ mein split hoti hai; fraction kinetic = kinetic energy divided by total. $U=\frac{1}{2}(50)(0.06{)}^2=\frac{1}{2}(50)(0.0036)=0.09\text{J}$ $\text{kinetic}=E-U=0.36-0.09=0.27\text{J}$ $\text{fraction kinetic}=\frac{0.27}{0.36}=0.75=75\%$ Toh (b) $75\%$ energy kinetic hai ($0.27\text{J}$ kinetic, $0.09\text{J}$ spring PE). Neat pattern: half amplitude par, PE ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$ total ka hai, toh kinetic $\frac{3}{4}$ hai. Wahi $x^2$ dependence phir!