a)F+N+Ff+G(vectori pe toate)=m*a(vector pe a)
(Ox):F*cosα-Ff=m*a
(Oy):F*sinα+N-m*g=0=>Ff=μ*N=μ*(m*g-F*sinα)=>F*cosα+-μ*m*g+μ*F*sinα=m*a=>F=[m(a+μ*g)]/(cosα+μ*sinα)
LF=F*d1*cosα=[m(a+μ*g)d1]*cosα/(cosα+μ*sinα)=[m(a+μ*g)*d1]/(1+μ*tgα)=[1(1+0,1*10)*5]/(1+0,1*tg30)=10/(1+0,1*√3/3)=10/(1+0,1*1,73/3)=10/(1+0,1*0,57)=10/(1+0,057)=10/1,057=9,46 J
b)LN=0 , deoarece forta normala este in permanenta perpendiculara pe directia miscarii
c)LFf=Ff*d3*cos180=-μ*N*d3
N=m*g-F*sinα=m*g-[m(a+μ*g)sinα]/(cosα+μ*sinα)=[m(g*cosα-a*sinα)]/(cosα+μ*sinα)
LFf=-[μ*m(g*cosα-a*sinα)d3]/(cosα+μ*sinα)=-[0,1*1(10*√3/2-1*1/2)3]/(√3/2+0,1*1/2)=-0,1(10*0,865-1*0,5)3/(0,865+0,1*0,5)=-0,3(8,65-0,5)/(0,865+0,05)=-(0,3*8,15)/0,915=-2,445/0,915=-2,67 J
