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question 3 of 20
question 3
5 points
a 11.4 n force is applied to a spring which causes the spring to elongate by 0.0200 m. what is the elongation in the spring when a 31.9 n force is applied?
0.0675 m
0.0290 m
0.0560 m
0.0450 m
0.0950 m

Explanation:

Step1: Find spring constant k

Use Hooke's Law $F = kx$. Rearrange to $k = \frac{F_1}{x_1} = \frac{11.4\,\text{N}}{0.0200\,\text{m}} = 570\,\text{N/m}$.

Step2: Calculate new elongation

Rearrange Hooke's Law to $x_2 = \frac{F_2}{k} = \frac{31.9\,\text{N}}{570\,\text{N/m}} \approx 0.0560\,\text{m}$.

Answer:

0.0560 m