Question

Resortul unui diametru cu constanta elastica 40N/m s-a alungit cu 5cm.Determinati forta elastica

Answer 1

F=k*Δl
k=40N/m
Δl=5cm=0,05m
F=40*0,05=2N
Spor!

Answer 2

$\Delta l=5 \ cm=0,05 \ m\\ k=40 \ N/m\\ \boxed{F_e-?} \\ \bold{Rezolvare:} \\ \boxed{k=\frac{F_e}{\Delta l}} \\ \Rightarrow F_e=k \cdot \Delta l \\ F_e=40 \ N/m \cdot 0,05 \ m \\ \Rightarrow \boxed{F_e=2 \ N}$