A 5-kg block is suspended from a spring, causing the spring to stretch 10 cm from equilibrium. What is the spring constant for this spring?
Correct Answer: C
Rationale: The spring constant (k) can be calculated using Hooke's Law formula: F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from equilibrium. In this case, the force applied is equal to the weight of the block, F = mg, where m = mass of the block = 5 kg and g = acceleration due to gravity = 9.8 m/s^2. The displacement x = 10 cm = 0.1 m. Substituting the values, we have: 5 kg * 9.8 m/s^2 = k * 0.1 m. Solving for k gives k = 5 * 9.8 / 0.1 = 49 N/m. Therefore, the spring constant for this spring is 49 N/cm. Choice A (4.9 N/cm) is incorrect because it is one decimal place lower than the correct answer. Choice B (9.8 N/cm) is incorrect as it does not account for the correct calculation based on the given information. Choice D (50 N/cm) is incorrect because it is slightly higher than the accurate value obtained through the calculations.