Question

The stiffness of a particular spring is 730 N/m. One end of the spring is attached to a wall. When you pull on the other end of the spring and hold it stretched with a steady force of 133 N, the spring elongates to a total length of 79 cm. What was the relaxed length of the spring?

Answer 1

Answer:

The relaxed length of spring was 61 cm

Explanation:

To solve this problem, we will use Hooke's Law. Hooke's law is stated as:

F = k Δx

F = k(x - x₀)

where,

F = Force applied to the spring = 133 N

k = Spring Constant = 730 N/m

x = Total Length after elongation = 79 cm = 0.79 m

x₀ = Initial or Relaxed Length of Spring = ?

Therefore,

133 N = (730 N/m)(0.79 m - x₀)

(0.79 m - x₀) = (133 N)/(730 N/m)

(0.79 m - x₀) = 0.18 m

x₀ = 0.79 m - 0.18 m

x₀ = 0.61 m = 61 cm