a 0.1 kg mass is suspended at rest from a spring near the Earth's surface, the distance that the spring is stretched is measured to be 1.0 cm. What is the spring constant of the spring (remember the MKS units)?

A mass of 2 kg is attached to a spring with constant 18 N/m. How far is the spring stretched?

A spring with a spring constant of 28.9 N/m is attached to different masses, and is stretched 5cm. What is the force exerted on the spring?

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring. Mathematically, this can be represented as F = k * x, where F is the force exerted by the spring, k is the spring constant, and x is the displacement or stretch of the spring.

Question 1:
Given:
Mass (m) = 0.1 kg
Stretch (x) = 1.0 cm = 0.01 m

To find the spring constant (k), we can rearrange Hooke's Law formula as k = F/x.

Since the mass is suspended at rest, the force exerted by the spring is equal to the weight of the mass, which is given by F = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s² near Earth's surface).

Substituting the values into the formula, we have:
k = (m * g)/x = (0.1 kg * 9.8 m/s²)/(0.01 m) = 98 N/m

Therefore, the spring constant of the spring is 98 N/m.

Question 2:
Given:
Mass (m) = 2 kg
Spring constant (k) = 18 N/m

To find the stretch (x) of the spring, we rearrange Hooke's Law formula as x = F/k.

The force exerted by the spring is the weight of the mass, given by F = m * g.

Substituting the values into the formula, we have:
x = (m * g)/k = (2 kg * 9.8 m/s²)/(18 N/m) ≈ 1.089 m

Therefore, the spring is stretched approximately 1.089 meters.

Question 3:
Given:
Spring constant (k) = 28.9 N/m
Stretch (x) = 5 cm = 0.05 m

To find the force exerted on the spring, we can use Hooke's Law formula, F = k * x.

Substituting the given values into the formula, we have:
F = (k * x) = (28.9 N/m * 0.05 m) ≈ 1.445 N

Therefore, the force exerted on the spring is approximately 1.445 Newtons.