1) 12 cm
2) 3 N
Explanation:
1)
The relationship between force and elongation in a spring is given by Hooke's law:
[tex]F=kx[/tex]
where
F is the force applied
k is the spring constant
x is the elongation
For the spring in this problem, at the beginning we have:
[tex]F=2 N[/tex]
[tex]x=4 cm[/tex]
So the spring constant is
[tex]k=\frac{F}{x}=\frac{2N}{4 cm}=0.5 N/cm[/tex]
Later, the force is tripled, so the new force is
[tex]F'=3F=3(2)=6 N[/tex]
Therefore, the new elongation is
[tex]x'=\frac{F'}{k}=\frac{6}{0.5}=12 cm[/tex]
2)
In this second problem, we know that the elongation of the spring now is
[tex]x=6 cm[/tex]
From part a), we know that the spring constant is
[tex]k=0.5 N/cm[/tex]
Therefore, we can use the following equation to find the force:
[tex]F=kx[/tex]
And substituting k and x, we find:
[tex]F=(0.5)(6)=3 N[/tex]
So, the force to produce an elongation of 6 cm must be 3 N.
