Question

determina constanta elastica a unui resort prin intermediul careia e trasa o remorca cu masa de 1 t,daca la miscarea cu viteza constanta aceasta se intinde cu 2 cm si forta de frecare reprezinta 10% din greutatea lui.

Answer 1

1t=1000kg
2cm=0.02m
G=mg=1000 kg*9.8N/kg=9800N
v=ct
Ft=Ffr
Ffr=10%G=980N
→    →
Fe=-Ffr
Fe=Ffr
K=Fe/Δl=980/0.02=49000N/m

Answer 2

m=1t=1000kg
Δl=2cm=0,02m
Ff=10/100 ·G
G=m·g
G=1000kg·10N/kg
G=10000N
Ff=10/100 ·10000N
Ff=1000N
Ff=k·Δl
1000N=k·0,02m
k=1000N/0,02m
k=50000N/m=500N/cm(constanta elastica a resortului);