Se da:
[tex]m=100dag=1kg\\
h=5m\\
l=5m\\
F_{fr}= \frac{1}{10}G=\ \textgreater \ u=0,1\\
g=10N/kg\\
a)Ec_{B}=?J\\
b)Ec_{C}=?J\\
c)h'=?m\\
[/tex]
Rezolvare:
Planurile sint lipsite de frecare, astfel putem folosi legea conservarii energiei mecanice.
[tex]a)Ep_{max_{A}}=Ec_{max_{B}}\\
Ec_{B}=mgh\\[/tex]
b)Pe portiunea BC, ne va fi comod sa folosim transformarea energiei mecanice in lucru.
[tex]deltaEc=L_{N}+L_{G}+L_{F_{fr}}\\
Ec_{C}-Ec_{B}=L_{F_{fr}}\\
Ec_{C}=Ec_{B}+L_{F_{fr}}\\
Ec_{C}=Ec_{B}-F_{fr}*l\\
Ec_{C}=Ec_{B}-u*N*l\\
Ec_{C}=Ec_{B}-u*G*l\\
Ec_{C}=Ec_{B}-u*m*g*l\\[/tex]
[tex]c)Ec_{max_{C}=Ep_{max_{D}}\\
Ec_{C}=m*g*h'\\
h'= \frac{Ec_{C}}{m*g} \\[/tex]
Calcule:
[tex]a)Ec_{B}=1*10*5=50J\\
b)Ec_{C}=50-0,1*1*10*5=45J\\
c)h'= \frac{45}{1*10}=4,5m \\[/tex]
