Share

## 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 a

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?

in progress
0

Physics
3 years
2021-09-05T11:52:20+00:00
2021-09-05T11:52:20+00:00 1 Answers
23 views
0
## Answers ( )

Answer:The relaxed length of spring was

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

F = k ΔxF = 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 mx₀ = 0.79 m – 0.18 mx₀ = 0.61 m = 61 cm