Final answer:
The spring constant of a spring stretched by 3.0 cm due to a 0.50 kg mass is 163.33 N/m. This is found using Hooke's Law, with the weight of the mass providing the force on the spring.
Step-by-step explanation:
The student asked: What is the spring constant of a spring where a 0.50kg mass causes it to stretch by 3.0cm? To calculate the spring constant (k), we can use Hooke's Law, which states that the force (F) exerted by a spring is proportional to the displacement (x) of the spring from its equilibrium position: F = kx. Here, the force exerted by the mass is due to gravity (weight of the mass), which can be calculated using F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).
The displacement should be converted from centimeters to meters to maintain consistency in units. Thus, x = 3.0 cm = 0.03 m. The force exerted by a 0.50 kg mass due to gravity is F = 0.50 kg × 9.8 m/s² = 4.9 N.
Now, applying Hooke's Law, we solve for k:
k = F / x = 4.9 N / 0.03 m = 163.33 N/m.
The spring constant is 163.33 N/m.