Question

A coil spring stretches 4.00 cm when a mass of 500 g is suspended from it. what is the force constant of the spring?

Answer 1

Answer: 122.625 N/m

Explanation:

To find the force constant (k) of the spring, you can use Hooke's Law, which is given by the formula:

F = kx, where:

• F is the force applied to the spring (Newtons, N)
• k is the spring constant (Newtons/meter, N/m)
• x is the displacement from the equilibrium position (meters, m)

We are given the following information:

• m = 500 g = 0.5 kg
• x = 4 cm = 0.04 m

The force (F) can also be calculated by multiplying the mass (m) by the acceleration due to gravity:

F = 0.5 × 9.81

F = 4.905 N

Since F = kx, we can substitute 4.905 N for F:

4.905 = kx

Substitute our known value of x:

4.905 = k × 0.04

Solve for k:

k = 122.625

The force constant of the spring is 122.625 N/m.