Question

De un resort de constanta elasctica K=400 N/m se suspende un corp cu masa m=4 kg. Sa se determine alungirea resortului.

Heelp

Answer 1

$k=400 \ N/m \\ m=4 \ kg \\ g=10 \ N/kg \\ \boxed{\Delta l-?} \\ \bold{Rezolvare:} \\ \boxed{k=\frac{F_e}{\Delta l}} \ \ \ \ \ \ \ \ \boxed{F_e=G=m \cdot g} \\ F_e=G=4 \ kg \cdot 10 \ N/kg \\ F_e=40 \ N \\ \Delta l=\frac{F_e}{k} \\ \Delta l=\frac{40 \ N}{400 \ N/m} \\ \Rightarrow \boxed{\Delta l=0,1 \ m=\bold{10 \ cm}}$