To solve this problem, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the extension or compression of the spring. Mathematically, it can be expressed as:
F = kx
Where: F is the force applied to the spring (in Newtons), k is the spring constant (in N/m), x is the extension or compression of the spring (in meters).
We are given the following information: Mass of the object initially suspended from the spring = 60 g = 0.06 kg Additional mass added = 15 g = 0.015 kg Increase in length of the spring = 1.2 cm = 0.012 m
Let's calculate the spring constant first:
When only the 60g mass is suspended, the force exerted by the spring (F1) is equal to the weight of the mass: F1 = m1 * g = 0.06 kg * 9.8 m/s^2 = 0.588 N
When the additional 15g mass is added, the force exerted by the spring (F2) is equal to the weight of the combined masses: F2 = (m1 + m2) * g = (0.06 kg + 0.015 kg) * 9.8 m/s^2 = 0.735 N
The change in force (ΔF) is given by: ΔF = F2 - F1 = 0.735 N - 0.588 N = 0.147 N
According to Hooke's Law, the change in force (ΔF) is equal to the spring constant (k) multiplied by the change in length (Δx): ΔF = k * Δx
Rearranging the equation, we can solve for the spring constant (k): k = ΔF / Δx = 0.147 N / 0.012 m = 12.25 N/m
Therefore, the spring constant is 12.25 N/m.
Now, let's calculate the total extension of the spring when the additional mass is added:
The extension of the spring (x) is given by: x = ΔF / k = 0.147 N / 12.25 N/m ≈ 0.012 m
Therefore, the total extension of the spring is approximately 0.012 m (or 1.2 cm).
So, the spring constant is 12.25 N/m, and the total extension of the spring is approximately 0.012 m.