Given that :
A mass of 4 kg is hung from a spring. If the spring's elongation is .6 meters
mass = 4 kg
elongation = 0.6 m
Using hooke's law
where
F is force
K is spring constant
x is spring elongation
Also , F = mg
where g is gravitational constant
F = 4 kg × 9.8 m/s²
F = 39.2 N
Now Using hooke's law
F = -kx
39.2 = -k (0.6)
39.2N /0.6 m = -k
k = 65.33 N/m
Therefore, the spring constant of the spring is 65.33 N/m
Answer:
Approximately [tex]65\; {\rm N\cdot m^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)
Explanation:
By Hooke's Law, the force required to compress or stretch an ideal spring is proportional to the displacement of the spring from the equilibrium position. The spring constant of that spring is the ratio between the amount of force exerted on the spring and the displacement of the spring:
[tex]\displaystyle k = \frac{(\text{force on spring})}{(\text{displacement of spring})}[/tex].
The force on the spring in this question would be equal to the weight [tex]m\, g[/tex] of the object attached to the spring. Given the displacement from the equilibrium position, the spring constant of this spring would be:
[tex]\begin{aligned} k = \frac{(4\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}})}{(0.6\; {\rm m})} \approx 65\; {\rm N\cdot m^{-1}}\end{aligned}[/tex].
