If the length of a #65 cm# spring increases to #79 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

To find the spring constant (k), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

Hooke's Law is given by:

F = k * x

Where: F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.

We know that the weight (force) applied to the spring is due to gravity, which is given by:

F = m * g

Where: m is the mass hanging from the spring (5 kg in this case), and g is the acceleration due to gravity (approximately 9.8 m/s^2).

The displacement of the spring (x) is the difference between the stretched length and the original length:

x = ΔL = L_final - L_initial

Given: L_initial = 65 cm L_final = 79 cm m = 5 kg g ≈ 9.8 m/s^2

Substituting the values into the equations:

x = 79 cm - 65 cm = 14 cm = 0.14 m

F = m * g = 5 kg * 9.8 m/s^2 = 49 N

Now, we can rearrange Hooke's Law to solve for the spring constant (k):

k = F / x

Substitute the values:

k = 49 N / 0.14 m = 350 N/m

Therefore, the spring constant is 350 N/m.