Question

De un resort cu constanta elastica k=100N/m si lungimea initiala lo= 8 cm se suspenda un corp cu masa m=0,0002 t. Cunoscand g=9,8N/kg aflati lungimea finala a resortului R:/=9,96

Answer 1

$k=100 \ N/m \\ l_0=8 \ cm=0,08 \ m \\ m=0,000 \ 2 \ t=0,2 \ kg \\ g=9,8 \ N/kg \\ \boxed{l_f-?} \\ \bold{Rezolvare:} \\ \boxed{\Delta l=l_f-l_0} \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{k=\frac{F_e}{\Delta l}} \\ F_e=G=m \cdot g \\ F_e=G=0,2 \ kg \cdot 9,8 \ N/kg \\ F_e=1,96 \ N \\ \Delta l= \frac{F_e}{k}= \frac{1,96 \ N}{100 \ N/m} \\ \Delta l=0,019 \ 6 \ m \\ l_f=l_0+\Delta l=0,08 \ m+0,019 \ 6 \ m \\ \Rightarrow \boxed{l_f=0,099 \ 6 \ m=\boxed{9,96 \ cm}}$