Naloge iz energije
še neki sm ugotovu
\(F_1 = F_2 + 2mgsin \alpha\)
\(F_1 v_1 = F_2 v_2\)
\((F_2 + 2mgsin \alpha ) v_1 = F_2 v_2\)
\(F_2 v_1 + 2mgsin \alpha v_1 = F_2 v_2\)
\(2mgsin \alpha v_1 = F_2 v_2 - F_2 v_1\)
\(2mgsin \alpha v_1 = F_2 (v_2 - v_1)\)
\(F_2 = \frac{2mgsin \alpha v_1}{(v_2 - v_1)}\)
potem \(F_2 v_2 = F_3 v_3\)
\(\frac{2mgsin \alpha v_1}{(v_2 - v_1)} = mgk_t v_3\)
\(\frac{2sin \alpha v_1 v_2}{(v_2 - v_1)}= k_t v_3\)
\(k_t v_3= 11,57\)
sm že bližje
\(\frac{F_1 - mg sin \alpha}{mg cos \alpha} v_3 = 11,57\) namesto \(k_t\) vstavim \(k_t= \frac{F_1 - mg sin \alpha}{mg cos \alpha}\)
\(\frac{F_2 + 2mgsin \alpha - mg sin \alpha}{mg cos \alpha} v_3 = 11,57\) namest \(F_1\) vstavim \(F_1 = F_2 + 2mgsin \alpha\)
\(\frac{\frac{2mgsin \alpha v_1}{(v_2 - v_1)} + 2mgsin \alpha - mg sin \alpha}{mg cos \alpha} v_3 = 11,57\)
prava zmeda....sem vsaj na pravi poti?
\(F_1 = F_2 + 2mgsin \alpha\)
\(F_1 v_1 = F_2 v_2\)
\((F_2 + 2mgsin \alpha ) v_1 = F_2 v_2\)
\(F_2 v_1 + 2mgsin \alpha v_1 = F_2 v_2\)
\(2mgsin \alpha v_1 = F_2 v_2 - F_2 v_1\)
\(2mgsin \alpha v_1 = F_2 (v_2 - v_1)\)
\(F_2 = \frac{2mgsin \alpha v_1}{(v_2 - v_1)}\)
potem \(F_2 v_2 = F_3 v_3\)
\(\frac{2mgsin \alpha v_1}{(v_2 - v_1)} = mgk_t v_3\)
\(\frac{2sin \alpha v_1 v_2}{(v_2 - v_1)}= k_t v_3\)
\(k_t v_3= 11,57\)
sm že bližje
\(\frac{F_1 - mg sin \alpha}{mg cos \alpha} v_3 = 11,57\) namesto \(k_t\) vstavim \(k_t= \frac{F_1 - mg sin \alpha}{mg cos \alpha}\)
\(\frac{F_2 + 2mgsin \alpha - mg sin \alpha}{mg cos \alpha} v_3 = 11,57\) namest \(F_1\) vstavim \(F_1 = F_2 + 2mgsin \alpha\)
\(\frac{\frac{2mgsin \alpha v_1}{(v_2 - v_1)} + 2mgsin \alpha - mg sin \alpha}{mg cos \alpha} v_3 = 11,57\)
prava zmeda....sem vsaj na pravi poti?
\(P_1 = P_2\)
\(mg\left (sin\alpha + k_tcos\alpha\right ) v_1= mg\left (k_t cos\alpha - sin\alpha\right ) v_2\)
\(\left (sin\alpha + k_t cos\alpha\right ) v_1= \left (k_t cos\alpha - sin\alpha\right ) v_2\)
\(sin\alpha v_1 + k_t cos\alpha v_1 = k_tcos\alpha v_2 - sin\alpha v_2\)
\(k_t = \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)}\)
Pol daš pa ta \(k_t\) v enačbo \(P_2 = P_3\) iz katere \(v_3\) ven izračunaš. Upam, da se nism kje zmotu
\(mg\left (sin\alpha + k_tcos\alpha\right ) v_1= mg\left (k_t cos\alpha - sin\alpha\right ) v_2\)
\(\left (sin\alpha + k_t cos\alpha\right ) v_1= \left (k_t cos\alpha - sin\alpha\right ) v_2\)
\(sin\alpha v_1 + k_t cos\alpha v_1 = k_tcos\alpha v_2 - sin\alpha v_2\)
\(k_t = \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)}\)
Pol daš pa ta \(k_t\) v enačbo \(P_2 = P_3\) iz katere \(v_3\) ven izračunaš. Upam, da se nism kje zmotu
neki mi ne pride prav....noro
\(P_2 = P_3\)
\(F_2 v_2 = F_3 v_3\)
\(mg(k_t cos \alpha - sin \alpha) v_2 = mgk_t v_3\)
\(( \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)} cos \alpha - sin \alpha) v_2 = \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)} v_3\)
\(( \frac{-sin \alpha (v_1 + v_2)}{v_1 - v_2} - sin\alpha) v_2 = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(-sin\alpha \frac{v_1+v_2}{v_1-v_2}v_2 - sin\alpha v_2 = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(-sin \alpha \left v_2 \left (\frac{v_1+v_2}{v_1-v_2} + 1) = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(\frac{-sin \alpha \left v_2 \left (\frac{v_1+v_2}{v_1-v_2} + 1)}{-tg \frac{v_1 + v_2}{v_1-v_2}}=v_3\)
\(v_3=\frac{cos\alpha \left (\frac{v_1+v_2}{v_1-v_2} + 1)}{\frac{v_1 + v_2}{v_1-v_2}}\)
\(v_3=\frac{cos10° (\frac{20+50}{20-50} +1)}{\frac{20+50}{20-50}}\)
\(v_3=\frac{cos10° (-1,33)}{-2,33}\)
\(v_3=°\frac{1,33cos10}{2,33}=0,56 \frac{m}{s}\)
kaj je tuki narobe? moralo bi priti \(28,14 \frac{m}{s}\)
hvala
\(P_2 = P_3\)
\(F_2 v_2 = F_3 v_3\)
\(mg(k_t cos \alpha - sin \alpha) v_2 = mgk_t v_3\)
\(( \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)} cos \alpha - sin \alpha) v_2 = \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)} v_3\)
\(( \frac{-sin \alpha (v_1 + v_2)}{v_1 - v_2} - sin\alpha) v_2 = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(-sin\alpha \frac{v_1+v_2}{v_1-v_2}v_2 - sin\alpha v_2 = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(-sin \alpha \left v_2 \left (\frac{v_1+v_2}{v_1-v_2} + 1) = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(\frac{-sin \alpha \left v_2 \left (\frac{v_1+v_2}{v_1-v_2} + 1)}{-tg \frac{v_1 + v_2}{v_1-v_2}}=v_3\)
\(v_3=\frac{cos\alpha \left (\frac{v_1+v_2}{v_1-v_2} + 1)}{\frac{v_1 + v_2}{v_1-v_2}}\)
\(v_3=\frac{cos10° (\frac{20+50}{20-50} +1)}{\frac{20+50}{20-50}}\)
\(v_3=\frac{cos10° (-1,33)}{-2,33}\)
\(v_3=°\frac{1,33cos10}{2,33}=0,56 \frac{m}{s}\)
kaj je tuki narobe? moralo bi priti \(28,14 \frac{m}{s}\)
hvala
tanges v števcu je malo zašel aneRokerda napisal/-a:neki mi ne pride prav....noro
\(P_2 = P_3\)
\(F_2 v_2 = F_3 v_3\)
\(mg(k_t cos \alpha - sin \alpha) v_2 = mgk_t v_3\)
\(( \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)} cos \alpha - sin \alpha) v_2 = \frac{-sin\alpha\left (v_2 + v_1\right )}{cos\alpha\left (v_1 - v_2\right)} v_3\)
\(( \frac{-sin \alpha (v_1 + v_2)}{v_1 - v_2} - sin\alpha) v_2 = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(-sin\alpha \frac{v_1+v_2}{v_1-v_2}v_2 - sin\alpha v_2 = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(-sin \alpha \left v_2 \left (\frac{v_1+v_2}{v_1-v_2} + 1) = -tg \frac{v_1 + v_2}{v_1-v_2} v_3\)
\(\frac{-sin \alpha \left v_2 \left (\frac{v_1+v_2}{v_1-v_2} + 1)}{-tg \frac{v_1 + v_2}{v_1-v_2}}=v_3\)
\(v_3=\frac{cos\alpha \left (\frac{v_1+v_2}{v_1-v_2} + 1)}{\frac{v_1 + v_2}{v_1-v_2}}\)
\(v_3=\frac{cos10° (\frac{20+50}{20-50} +1)}{\frac{20+50}{20-50}}\)
\(v_3=\frac{cos10° (-1,33)}{-2,33}\)
\(v_3=°\frac{1,33cos10}{2,33}=0,56 \frac{m}{s}\)
kaj je tuki narobe? moralo bi priti \(28,14 \frac{m}{s}\)
hvala
lp