Question

if a force of 256 n is applied to a spring, the spring is stretched a distance of 9.50 cm from its equilibrium position. what is the elastic constant of this spring?
2.69 × 10³ n/m
24.3 n/m
2.43 × 10³ n/m
26.9 n/m

Explanation:

Step1: Recall Hooke's Law

$F = kx$

Step2: Rearrange for spring constant $k$

$k = \frac{F}{x}$

Step3: Convert distance to meters

$x = 9.50\ \text{cm} = 0.0950\ \text{m}$

Step4: Substitute values and calculate

$k = \frac{256\ \text{N}}{0.0950\ \text{m}} \approx 2694.7\ \text{N/m} = 2.69 \times 10^3\ \text{N/m}$

Answer:

2.69 X 10³ N/m