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

To find the relaxed length of a spring, subtract the extension caused by the force from the total stretched length. Applying Hooke's Law results in a relaxed length of 60.8 cm for the given spring.

Step-by-step explanation:

The question is concerned with finding the relaxed length of a spring given its stiffness (spring constant) and the force applied to elongate it. The spring's stiffness is 730 N/m, and it's elongated to a total length of 79 cm when a force of 133 N is applied. Hooke's Law, which states that force (F) = spring constant (k) × extension (x), can be used to solve for the extension and consequently the relaxed length.

Using Hooke's Law, we get:

F = k × x

133 N = 730 N/m × x

x = 133 N / 730 N/m

x = 0.182 m

Given the total elongated length is 79 cm (0.79 m), we subtract the extension (0.182 m) to find the relaxed length:

Relaxed length = Total length - Extension

Relaxed length = 0.79 m - 0.182 m

Relaxed length = 0.608 m or 60.8 cm

Answer 2

The relaxed length of spring was 61 cm

Step-by-step 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