Question

A vertically hung spring has a spring constant of 150. newtons per meter. A

2.00-kilogram mass is suspended from the spring and allowed to come to rest.

Calculate the elongation of the spring produced by the suspended 2.00-kilogram mass. [Show all
work, including the equation and substitution with units.]

Answer 1

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.

Answer 2

Based on Hooke's law, the force a spring exerts is negative based on the elongation length. Therefore, since a 2.00 kilogram weight exerts 20 Newtons (gravitation acceleration constant = 10), than the spring must pull back with -20 Newtons. This means that -20 = -kx, where k = spring constant and x = elongation length. k = 150, so that means that x = 20/150, or 2/15 meters.