Final answer:
Using Hooke's Law and the weight of the mass, the elongation of the spring is calculated to be 13.08 cm when a 2.00-kilogram mass is suspended from it.
Step-by-step explanation:
To calculate the elongation Δx of the spring produced by the suspended 2.00-kilogram mass, you can use Hooke's law, which relates the force exerted on a spring to its extension or compression, expressed as F = kΔx. In this case, the force F is due to the weight of the mass, which can be found using the gravitational force equation F = mg, where m is mass and g is the acceleration due to gravity (approximately 9.81 m/s2).
The spring constant k is given as 150 newtons per meter, and the mass m is 2.00 kilograms. Therefore, the weight of the mass is F = 2.00 kg × 9.81 m/s2 = 19.62 N.
Now, apply Hooke's law to find the elongation Δx:
19.62 N = 150 N/m × Δx
Δx = 19.62 N / 150 N/m = 0.1308 m or 13.08 cm
So, the elongation of the spring is 13.08 cm when a 2.00-kilogram mass is suspended from it.