Question

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De un resort elastic de constanta k=100N/m este suspendat un corp cu masa de 2kg .Se considera g=10N/kg .Sa se afle forta deformatoare si alungire a resortului'

Answer 1

Se da:
m=2 kg
g=10 N/kg
k=100N/m
Rezolvare:
Fe=G=m x g=2 kg x 10 N/kg=20N
alungirea=Fe:k=20N:100N/m=0,2m
Raspuns:Fe=20N;alungirea=0,2m

Answer 2

$k=100 \ N/m \\ m=2\ kg \\ g=10 \ N/kg \\ \boxed{F_e-? } \ \ \ \ \ \boxed{\Delta l-?} \\ \bold{Rezolvare:} \\ \boxed{k=\frac{F_e}{\Delta l}} \ \ \ \ \boxed{F_e=G=m \cdot } \\ F_e=G=2 \ kg \cdot 10 \ N/kg \\ \Rightarrow \boxed{F_e=20 \ N} \\ \Delta l=\frac{F_e}{k} \\ \Delta l=\frac{20 \ N}{100 \ N/m} \\ \Rightarrow \boxed{\Delta l=0,2 \ m}$