URGENT!!O caramida cu masa m=2 kg aluneca accelerat pe o scandura inclinata fata de planul orizontal cu unghiul alfa 45 de grade.Daca caramida este lasata de jos in sus de-a lungul scandurii cu viteza vo=4m/s, aceasta urca cu o acceleratia dubla in modul. Sa se afle.
a)coeficientul de frecare la alunecare
b)spatiul parcurs de caramida pana la oprirea pe scandura
c)randamentul scandurii inclinate

Răspuns :

a) -Gt-Ff=ma=>-mgsinα-Ff=ma

N-Gn=0=>N=mgcosα

Ff=μN=μmgcosα

-mgsinα-μmgcosα=ma|:m=>a=-gsinα-gμcosα=-g(sinα+μcosα)

Gt-Ff=ma'=>mgsinα-Ff=ma'

N-Gn=0=>N=mgcosα

Ff=μN=μmgcosα

mgsinα-μmgcosα=ma'|:m=>a'=gsinα-gμcosα=g(sinα-μcosα)

|a|=2a'=>|-g(sinα+μcosα)|=2g(sinα-μcosα) =>g(sinα+μcosα)=2g(sinα-μcosα)|:g=>-sinα=-3μcosα=>μ=tgα/3=1/3=0,33

b) ΔEc=L_G+L_Ff+L_N

ΔEc=mv²/2-mv₀²/2=-mv₀²/2

L_N=0;L_G=-mgh=-mgxsinα

L_Ff=-Ff*x=-μNx=-μmgxcosα

-mv₀²/2=-mgxsinα-μmgxcosα|:m=>v₀²=2gxsinα+2μgxcosα=2gx(gsinα+μcosα)

v₀²=2gx(sinα+μcosα)<=>4²=2*10x(0,7+0,33*0,7)<=>16=20x*0,931<=>16=18,62x=>x=16/18,62=0,85 m

c) η=1/(1+μctgα)=1/(1+0,33*1)=1/1,33=0,75=>η=75%