2 A spring is stretched 0.9 m longer using a force of 294.3 newtons. What would the springs \( \mathrm{N} / \mathrm{cm} \) elastic constant be?

To find the elastic constant of the spring, we can use Hooke's Law, which states that the force required to stretch a spring by a distance \( x \) is given by: \[ F = kx \] where: - \( F \) is the force applied to the spring, - \( k \) is the elastic constant of the spring, and - \( x \) is the distance the spring is stretched. Given: - Force applied, \( F = 294.3 \) newtons, - Distance stretched, \( x = 0.9 \) meters. We need to find the elastic constant \( k \) in units of newtons per centimeter (N/cm). First, let's convert the distance from meters to centimeters: \[ x = 0.9 \, \text{m} = 0.9 \times 100 \, \text{cm} = 90 \, \text{cm} \] Now, we can use Hooke's Law to find the elastic constant \( k \): \[ k = \frac{F}{x} \] Substitute the given values: \[ k = \frac{294.3 \, \text{N}}{90 \, \text{cm}} \] Calculate the elastic constant \( k \) in N/cm. Calculate the value by following steps: - step0: Calculate: \(\frac{294.3}{90}\) - step1: Convert the expressions: \(\frac{\frac{2943}{10}}{90}\) - step2: Multiply by the reciprocal: \(\frac{2943}{10}\times \frac{1}{90}\) - step3: Reduce the numbers: \(\frac{327}{10}\times \frac{1}{10}\) - step4: Multiply the fractions: \(\frac{327}{10\times 10}\) - step5: Multiply: \(\frac{327}{100}\) The elastic constant of the spring is \( k = 3.27 \) N/cm.