Question

ce masa are corpul suspendat de un resort cu constanta elastica k=500N/m daca acesta sa alungit cu 10 cm.....va rog ajutatima dau coroana​

Answer 1

k = Fe/Δl

Fe = G = m×g

k = m×g/Δl => m = k×Δl/g

m = 500N/m×0,1m/9,8N/kg

= 50N/9,8N/kg

≈ 5, 1 kg