Question

Va rog foarte frumos, ajutati-ma sunt corigent la fizica si am nevoie de aceasta problema maine la pregatire, multumesc anticipat

Answer 1

ΔEc_OA=L_G=>Ec_A=mg(H-h)=2*10(10-2)=20*8=160 J

E'c_A=Ec_A(1-f)

E'm_A=E'c_A+Ep_A=mg(H-h)(1-f)+mgh=20(10-8)(1-0,75)+20*2=160*0,25+40=40+40=80 J

ΔEc_AB=L_G+L_Ff+L_N

ΔEc_AB=Ec_B-Ec' ; L_N=0 ; L_G=mgh ; LFf=-Ffl

Ec_B=mg(H-h)(1-f)+mgh-Ff*l=20(10-8)(1-0,75)+20*2-8=80-8=64 J

v=√(2Ec_B/m)=√2*64/2=√64=8 m/s

Ff=μmgcosα=>μ=Ff/mgcosα

sinα=h/l=2/4=1/2=>α=30°

μ=4/(20*√3/2)=1/4,32=0,231