Answer:
Option A. 28 cm
Explanation:
Hooke's Law
If a force F is applied to a string of constant k, it stretches a distance of x. The relation between those variables is:
[tex]F=k*x[/tex]
Note that the distance is not the actual length of the spring, but the elongation, or difference between the natural length of the spring and the stretched length.
The spring described has a natural length of 10 cm and is stretched to 22 cm when a force of 4 N is applied. The elongation is x=22-10=12 cm.
Applying Hooke's Law, we can find the value of k:
[tex]4\ N=k*12\ cm[/tex]
Solving for k:
[tex]\displaystyle k=\frac{4\ N}{12\ cm}=\frac{1}{3}\ N/cm[/tex]
Thus, the equation for F is:
[tex]\displaystyle F=\frac{1}{3}*x[/tex]
Where x is in cm and F is in N.
If a force of F= 6 N is applied, we can find the new elongation:
[tex]\displaystyle 6=\frac{1}{3}*x[/tex]
Thus:
x=18 cm
The new length of the spring is 10 cm + 18 cm = 28 cm
Option A. 28 cm
