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question 18
problem reference 10.5
when a force of 19 n is applied to a spring, it elongates 0.085 m.
what is the spring constant?
1.62 n/m
4.47 x 10⁻³ n/m
224 n/m
448 n/m

Explanation:

Step1: Recall Hooke's Law

Hooke's Law: $F = kx$ (solve for $k$: $k = \frac{F}{x}$)

Step2: Substitute values

$F=19\,\text{N}$, $x=0.085\,\text{m}$ → $k = \frac{19}{0.085}$

Step3: Calculate result

$\frac{19}{0.085} \approx 224\,\text{N/m}$

Answer:

C. 224 N/m