Question

doua discuri de mase m1=100g si m2=300g sunt prinse intre ele cu un resort.Suspendand sistemul de discul superior resortul are lungimea l1=40cm asezandu-l pe o masa cu discul inferior,resortul are lungimea l2=20cm.Care este lungimea resortului nedeformat?

Answer 1

[tex]Se~da:\\ \\ m_{1}=100g=0,1kg\\ \\ m_{2}=300g=0,3kg\\ \\ l_{1}=40cm=0,4m\\ \\ l_{2}=20cm=0,2m\\ \\ l_{0}=?m\\ \\ \\[/tex]


[tex]Formule:\\ \\ 1~caz:\\ \\ G_{2}=F_{el_{1}}\\ \\ g\times m_{2}=k\times \Delta l_{1}\\ \\ g\times m_{2}=k\times(l_{1}-l_{0})\\ \\ k=\frac{g\times m_{2}}{l_{1}-l_{0}}\\ \\ \\ 2~caz:\\ \\ G_{1}=F_{el_{2}}\\ \\ g\times m_{1}=k \times \Delta l_{2}\\ \\ g\times m_{1}=k\times(l_{0}-l_{2})\\ \\ k=\frac{g\times m_{1}}{l_{0}-l_{2}}\\ \\ \\ [/tex]


[tex]Egalam:\\ \\ \frac{g\times m_{2}}{l_{1}-l_{0}}=\frac{g\times m_{1}}{l_{0}-l_{2}}\\ \\ m_{2}\times l_{0}-m_{2}\times l_{2}=m_{1}\times l_{1}-m_{1}\times l_{0}\\ \\ m_{2}\times l_{0}+m_{1}\times l_{0}=m_{1}\times l_{1}+m_{2}\times l_{2}\\ \\ l_{0}\times(m_{2}+m_{1})=m_{1}\times l_{1}+m_{2}\times l_{2}\\ \\ l_{0}=\frac{m_{1}\times l_{1}+m_{2}\times l_{2}}{m_{2}+m_{1}}\\ \\ \\ Calcule:\\ \\ l_{0}=\frac{0,1\times 0,4+0,3\times 0,2}{0,3+0,1}=0,25m[/tex]