Question

Un resort de costanta elastica k=400N/mse suspenda un corp cu masa m=4kg.Sa se determine alungirea resolrtului

Answer 1

$k=400 \ N/m \\ m=4 \ kg \\ g=10 \ N/kg \\ \boxed{\Delta l-?} \\ \bold{Rezolvare:} \\ \boxed{k= \frac{F_e}{\Delta l} \Rightarrow \boxed{\Delta l= \frac{F_e}{k} }} \\ F_e=G=m \cdot g \\ F_e=G=4 \ kg \cdot 10 \ N/kg \\ F_e=40 \ N \\ \Delta l= \frac{40 \ N}{400 \ N/m} \\ \Rightarrow \boxed{\Delta l= 0,1 \ m}$

Answer 2

Se da:
k=400 N/m
m=4 kg
--------------
ΔL=?

     Fe
ΔL=-----
        k
Fe=G
G=m*g=4kg*10N/kg=40N
        40 N
ΔL=-----------=0,1m
        400N/m
                                       Raspuns:ΔL=0.1m