Exercises — Work — definition, dot product F·d, sign convention

Friction motion ke opposite point karta hai, isliye force aur displacement ke beech angle $\theta =180°$ hai. Kyunki $\cos 180°=-1$, work negative hai. $W=Fd\cos 180°=-Fd<0$ Yeh kaisa dikhta hai: figure s01 ke left panel mein (jiska title hai "A: friction vs motion") coral arrow (friction) aur mint arrow (motion) opposite directions mein point karte hain.

Perfect circle ke liye centre ki taraf pull har instant par velocity ke perpendicular hota hai, isliye $\theta =90°$ aur $\cos 90°=0$. $W=Fd\cos 90°=0$ Force sirf steer karta hai; yeh kabhi speed up ya slow down nahi karta. Zero work. (Figure s01 ka middle panel, "B: gravity in orbit".)

Tension upar point karta hai, displacement upar hai, isliye $\theta =0°$ aur $\cos 0°=+1$. $W=Fd\cos 0°=+Fd>0$ Positive work — rope crate mein energy daal rahi hai. (Figure s01 ka right panel, "C: tension lifting".)

Displacement horizontal hai, aur $\theta =60°$ rope aur us horizontal direction ke beech ka angle hai, isliye hum force ka horizontal share use karte hain $=F\cos \theta$. $W=Fd\cos \theta =50\times 8\times \cos 60°=400\times 0.5=\boxed{200\text{J}}$ $\cos$ kyun, $\sin$ kyun nahi? Humein force ka woh part chahiye jo us road ke along ho jis par sled chalti hai; woh hamesha cosine part hota hai.

Koi angle nahi diya gaya, isliye component form use karo. Yahan $\hat{ı}$ aur $\hat{ȷ}$ unit arrows hain jo $x$ aur $y$ ke along point karte hain; aage ke numbers $F_x,F_y$ aur $d_x,d_y$ hain. $W=F_x{d}_x+F_y{d}_y=(6)(3)+(-2)(5)=18-10=\boxed{8\text{J}}$

Gravity ki magnitude $F=mg=4\times 9.8=39.2\text{N}$ hai, neeche point karti hai. Displacement bhi neeche hai, isliye $\theta =0°$. Hum standard $W=Fd\cos \theta$ use karte hain $F=mg$ aur $d=3\text{m}$ ke saath: $W=Fd\cos 0°=(mg)d\cos 0°=39.2\times 3\times 1=\boxed{117.6\text{J}}$ (Yahi familiar shortcut "$mgh$" ki origin hai: seedhe vertical drop ke liye displacement magnitude $d$ wohi hoti hai jitna height drop hua, aur $\cos 0°=1$.)

(a) Applied force motion ke along, $\theta =0°$: $W_a=40\times 6\times \cos 0°=+240\text{J}$ (b) Friction motion ko oppose karta hai, $\theta =180°$: $W_f=15\times 6\times \cos 180°=-90\text{J}$ (c) Gravity ki magnitude $mg=10\times 9.8=98\text{N}$ hai, vertical hai, motion horizontal hai, $\theta =90°$: $W_g=mg\times 6\times \cos 90°=98\times 6\times 0=0\text{J}$ (d) Normal force vertical hai, motion horizontal hai, $\theta =90°$: $W_N=0\text{J}$. (e) Net work = sab works ka sum: $W_{\text{net}}=240-90+0+0=\boxed{150\text{J}}$ Inhe kyun add karte hain? Work ek scalar hai, isliye total energy delivered sirf plain sum hoti hai — koi vector bookkeeping nahi. Yeh net work woh hai jo crate ki kinetic energy change karega.

Gravity seedhi neeche point karti hai magnitude $mg=2\times 9.8=19.6\text{N}$ ke saath. Displacement $5\text{m}$ ramp ke along hai. Seedhe-neeche aur ramp ki downhill direction ke beech angle $\theta =90°-30°=60°$ hai. $W_g=mgd\cos 60°=19.6\times 5\times 0.5=\boxed{49\text{J}}$ Height se check karo: block vertical height $h=d\sin 30°=5\times 0.5=2.5\text{m}$ drop karta hai, aur $W_g=mgh=19.6\times 2.5=49\text{J}$. Same answer — kyunki gravity ke work ke liye sirf vertical drop matter karta hai. Figure s02 mein mint arrow slanted path dikhata hai aur dashed lines vertical drop $h$ dikhati hain; dono routes same work dete hain.

Puck rest se start karta hai, isliye initial kinetic energy zero hai ($v_0=0$). Theorem kehta hai net work kinetic energy ki change ke barabar hai: $W_{\text{net}}=\frac{1}{2}m{v}^2-0\Rightarrow 24=\frac{1}{2}(3)v^2=1.5v^2$ $v^2=\frac{24}{1.5}=16\Rightarrow v=\boxed{4\text{m/s}}$

Kyunki crate steady speed par chalti hai, friction ($15\text{N}$) applied force ke barabar hona chahiye — lekin ruko, Exercise 3.1 mein numbers ($F_a=40\text{N}$, $f=15\text{N}$) equal nahi hain, toh woh scenario accelerating tha. Is power question ke liye hum constant-speed case lete hain aur sirf applied force ke kiye gaye work ke baare mein poochte hain, jo is baat par depend nahi karta ki doosre forces balance karte hain ya nahi. Exercise 3.1 se, applied force ne $W_a=F_{a}d\cos 0°=40\times 6=240\text{J}$ kiya. Power work per unit time hai: $P_{\text{avg}}=\frac{W_a}{t}=\frac{240}{3}=\boxed{80\text{W}}$ $P=Fv$ se cross-check: crate ki speed $v=\frac{6\text{m}}{3\text{s}}=2\text{m/s}$ hai, aur $P=F_{a}v\cos 0°=40\times 2=80\text{W}$. ✓ Key distinction: power yahan ek specific force ka work ($W_a$) use karta hai, net work nahi. Net work kinetic energy ki change se relate karta hai (Exercise 4.1); ek individual force ki power $W_{\text{that force}}/t$ hoti hai.

Upar jaate waqt: displacement $50\text{m}$ upar hai, gravity neeche hai, $\theta =180°$: $W_{\text{up}}=mg(50)\cos 180°=-12\times 9.8\times 50=-5880\text{J}$ Neeche aate waqt: displacement $50\text{m}$ neeche hai, gravity neeche hai, $\theta =0°$: $W_{\text{down}}=mg(50)\cos 0°=+5880\text{J}$ Total: $W_{\text{total}}=-5880+5880=\boxed{0\text{J}}$ Zero kyun? Round trip ka net displacement zero hota hai, aur gravity ek conservative force hai — iska work sirf start aur end heights par depend karta hai, jo yahan same hain.

(a) $\vec{F}$ ka work component form se (sirf $x$-parts survive karte hain, kyunki $d_y=0$): $W_F=F_x{d}_x+F_y{d}_y=(30)(4)+(-40)(0)=\boxed{120\text{J}}$ (b) Push ka ek downward component $F_y=-40\text{N}$ hai, yaani woh $40\text{N}$ se neeche press karta hai. Floor ko weight aur is extra downward push dono support karne padte hain, isliye normal force hai $N=mg+|F_y|=5\times 9.8+40=49+40=89\text{N}$ Friction magnitude: $f=\mu N=0.2\times 89=17.8\text{N}$ (c) Friction motion ko oppose karta hai, $\theta =180°$: $W_f=fd\cos 180°=17.8\times 4\times (-1)=\boxed{-71.2\text{J}}$ (d) Net work (gravity aur normal vertical hain, horizontal motion ke $\perp$, isliye woh zero work karte hain): $W_{\text{net}}=120-71.2+0+0=\boxed{48.8\text{J}}$

Leg 1: $W_1=(2)(4)+(3)(0)=8\text{J}$. Leg 2: $W_2=(2)(0)+(3)(3)=9\text{J}$. Total: $W=8+9=\boxed{17\text{J}}$. Straight-line check: overall displacement ${\vec{d}}_1+{\vec{d}}_2=(4\hat{ı}+3\hat{ȷ})\text{m}$ hai, isliye $W=(2)(4)+(3)(3)=8+9=17\text{J}.✓$ Yeh match kyun karte hain: $\vec{F}$ constant hai, aur dot product vector addition par distribute karta hai: $\vec{F}\cdot ({\vec{d}}_1+{\vec{d}}_2)=\vec{F}\cdot {\vec{d}}_1+\vec{F}\cdot {\vec{d}}_2$. Constant force ke liye, sirf net displacement matter karta hai — dekho Dot Product (Scalar Product).

Ramp ki downhill direction ko "along motion" set karo. Gravity ka along-ramp component (jo use neeche drive karta hai): $W_g=mg\sin 30°\times d=4\times 9.8\times 0.5\times 10=+196\text{J}$ Normal force: motion ke perpendicular, $\theta =90°$, isliye $W_N=0$. Friction: pehle normal force $N=mg\cos 30°=4\times 9.8\times 0.866=33.95\text{N}$ hai, isliye $f=\mu N=0.25\times 33.95=8.49\text{N}$, motion ko oppose karta hai ($\theta =180°$): $W_f=-fd=-8.49\times 10=-84.87\text{J}$ Net work: $W_{\text{net}}=196+0-84.87=\boxed{111.13\text{J}}$ Har force account kiya gaya, har sign check kiya gaya: gravity energy feed karti hai, friction use drain karta hai, normal force kuch nahi karta.