A) 392 N/m
The spring constant can be found by applying Hooke's Law:
F = kx
where
F is the force applied to the spring
k is the spring constant
x is the stretching of the spring
Here we have:
F is the weight of the block hanging from the spring, which is
[tex]F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N[/tex]
The stretching of the spring is
[tex]x=15 cm - 10 cm = 5 cm = 0.05 m[/tex]
Therefore its spring constant is
[tex]k=\frac{F}{x}=\frac{19.6 N}{0.05 m}=392 N/m[/tex]
B) 17.5 cm
Now that we know the value of the spring constant, we can calculate the new stretching of the spring when a mass of m=3.0 kg is applied to it. In this case, the force applied on the spring is
[tex]F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N[/tex]
Therefore the stretching of the spring is
[tex]x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m = 7.5 cm[/tex]
And since the natural length of the spring is 10 cm, the new length will be
L = 10 cm + 7.5 cm = 17.5 cm
