Question

An unstretched spring with a spring constant of 100 N/m stretches a distance of 0.4 m when a mass is attached to the spring. The size of the mass must be of:

1) 9.8 kg
2) 0.004 kg
3) 40 kg
4) 4 kg

Answer 1

Using Hooke's Law, the mass causing a spring with a constant of 100 N/m to stretch 0.4 m is calculated to be approximately 4 kg.

Step-by-step explanation:

The question deals with the relationship between a spring's extension and the mass causing the extension. According to Hooke's Law, the force exerted by a spring is equal to the spring constant (k) multiplied by the displacement from the equilibrium position (x). The force exerted by the spring is also equal to the gravitational force (mg), where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s2). Setting these equal gives the equation kx = mg, which can be rearranged to solve for the mass as m = kx/g.

In this case, with a spring constant of 100 N/m and a stretch of 0.4 m, the mass attached to the spring can be calculated as:

m = (100 N/m * 0.4 m) / 9.8 m/s2

m ≈ 4 kg.

Therefore, the mass attached to the spring is approximately 4 kg, making option 4) the correct answer.