Question

De un resort de R=400N/m se suspenda un corp cu m=4kg .Determinati alungirea resortului.(∆l)
dau coroana

Answer 1

k= 400N/m
m= 4kg

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Fe= G ⇒Fe=mg⇒Fe=4•10⇒Fe=40N
Fe=k∆l⇒40N=400N/m∆l⇒∆l=40N:400N/m⇒∆l=40/400m⇒∆l=0,1m

Answer 2

corpul suspendat are o greutate G = m x g 

forta de greutate determina aparitia unei forte elastice in resort, care echilibreaza  greutatea. La echilibru,

 Fe = G 

G = forta elastica = k Δl 
G = m x g = 4 kg x 10 N/kg = 40 Newton 

Δl = G / k = 40 N / 400 N/m  = 0,1 m = 10 cm