Question

Colegul meu, Cristi , a urmarit pe Facebook o inregistrare video cu un concurs de saniute . Din varful unui deal cu inaltimea de 30 m si lungimea de 80 m este lasata libera o saniuta . Aceasta coboara la inceput pe o portiune cu zapada , de lungime 60 m , apoi patrunde pe o portiune fara zapada , care se intinde pana la baza dealului . Saniuta se opreste exact la capatul pantei , fara a intra pe suprafata orientala . Coeficientul de frecare intre saniuta si zapada este u₁=0,3 . Calculeaza viteza maxima a saniutei pe parcursul coborarii si coeficientul de frecare dintre saniuta si portiunea fara zapada .

Answer 1

[tex]\displaystyle Se~da:\\ \\ h=30m\\ \\ l=80m\\ \\ l_1=60m\\ \\ \mu_1=0,3\\ \\ v_{max}=?\frac ms\\ \\ \mu_2=?\\ \\ \\ Formule:\\ \\ Se~observa~ca~dupa~terminarea~portiunii~cu~zapada,~sania~incepe\\ sa~se~incetineze,~astfel~la~baza~are~viteza~egala~cu~0,~deci~viteza\\ maxima~se~va~inregistra~inainte~de~portiunea~fara~zapada~sau\\ din~date,~dupa~ce~corpul~va~parcurge~portiunea~l_1[/tex]

[tex]\displaystyle l_1=\frac{\Delta v^2}{2\times a}\\ \\ l_1=\frac{v^2-v_0^2}{2\times a},~unde~v_0=0,~iar~v=v_{max}\\ \\ l_1=\frac{v_{max}^2}{2\times a},~de~unde~scoatem~v_{max}:\\ \\ v_{max}=\sqrt{2\times l_1\times a}\\ \\ \\ Pe~axa~O_y:\\ \\ N-G_y=0\\ \\ N-G\times\cos\alpha=0\\ \\ N=m\times g\times\cos\alpha\\ \\ \\[/tex]


[tex]\displaystyle Presupunem~ca~dealul~este~un~triunghi~dreptunghic,~deci:\\ \\ \cos\alpha=\frac bl,~iar~din~teorema~lui~Pitagora:~b=\sqrt{l^2-h^2},~astfel:\\ \\ \cos\alpha=\frac{\sqrt{l^2-h^2}}{l},~prin~urmare:\\ \\ \\ N=m\times g\times\frac{\sqrt{l^2-h^2}}{l}\\ \\ \\ Pe~axa~O_x:\\ \\ G_x-F_{fr}=m\times a\\ \\ G\times \sin\alpha-\mu\times N=m\times a,~unde~\sin\alpha=\frac hl\\ \\ m\times g\times \frac hl-\mu\times m\times g\times\frac{\sqrt{l^2-h^2}}{l}=m\times a,~unde~m~se~taie,~iar~g\\ factor~comun\\ \\ [/tex]

[tex]\displaystyle g\times(\frac hl-\mu\times\frac{\sqrt{l^2-h^2}}{l})= a\\ \\ a=g\times\frac{h-\mu\times\sqrt{l^2-h^2}}{l}\\ \\ \\ Inlocuind~a~in~formula~vitezei~maxime,~obtinem~formula~finala:\\ \\ v_{max}=\sqrt{2\times l_1\times g\times\frac{h-\mu\times\sqrt{l^2-h^2}}{l}}\\ \\ \\ \\ Pentru~a~afla~\mu_2,~facem~practic~acelasi~lucru,~doar~ca~analizam\\ miscarea~saniei~pe~portiunea~unde~nu~este~zapada~(l_2=l-l_1)\\ \\ l-l_1=\frac{\Delta v^2}{2\times a_2}\\ \\[/tex]

[tex]\displaystyle l-l_1=\frac{v^2-v_0^2}{2\times a_2},~unde~v=0,~iar~v_0=v_{max}\\ \\ l-l_1=\frac{-v_{max}^2}{2\times a_2}\\ \\ a_2=\frac{-v^2_{max}}{2\times (l-l_1)}\\ \\ \\ Axa~O_y~contine~aceleasi~parametrii~ca~si~in~primul~punct.\\ \\ Pe~O_x:\\ \\ G_x-F_{fr_2}=m\times a_2\\ \\ G\times\sin\alpha-\mu_2\times N=m\times a_2\\ \\ m\times g\times \frac hl-\mu_2\times m\times g\times\frac{\sqrt{l^2-h^2}}{l}=m\times a_2,~se~taie~m,~se~imparte\\ la~g~si~se~inmuteste~la~l\\ \\ [/tex]

[tex]\displaystyle h-\mu_2\times\sqrt{l^2-h^2}=\frac{l\times a_2}{g}\\ \\ \mu_2\times\sqrt{l^2-h^2}=\frac{h\times g-l\times a_2}{g}\\ \\ \mu_2=\frac{h\times g-l\times a_2}{g\times \sqrt{l^2-h^2}}\\ \\ Introducind~a_2~si~punind~sub~acelasi~numitor~obtinem~formula.\\ \\ \mu_2=\frac{h\times g-l\times \frac{-v^2_{max}}{2\times (l-l_1)}}{g\times\sqrt{l^2-h^2}}\\ \\ \mu_2=\frac{2\times h\times g\times(l-l_1)+l\times v_{max}^2}{2\times g\times(l-l_1)\times\sqrt{l^2-h^2}}\\ \\ \\[/tex]


[tex]\displaystyle Calcule:\\ \\ v_{max}=\sqrt{2\times 60\times 10\times\frac{30-0,3\times\sqrt{80^2-30^2}}{80}}\approx 10,78\frac ms\\ \\ \mu_2=\frac{2\times 30\times 10\times(80-60)+80\times 10,78^2}{2\times 10\times(80-60)\times\sqrt{80^2-30^2}}\approx 0,72 [/tex]