Question

The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold the spring elongated a total length 17 cm. What was the relaxed length of the spring? Answer in units of m.

Answer 1

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

Answer 2

The relaxed length of the spring is 12.24 cm.

How do you calculate the relaxed length of the spring?

Given that the force F on the spring is 2 N is required to hold the spring elongated a total length x of 17 cm. The spring constant k is 42 N/m.

The relaxed length of the spring can be calculated by Hooke's law which is given below.

$F = k(x-x_0)$

Where x is the stretched length and x0 is the relaxed length of the spring.

Substituting the values in the above equation, we get the relaxed length of the spring.

$2 = 42 ( 0.17 -x_0)$

$0.0476 = 0.17 - x_0$

$x_0 = 0.1224 \;\rm m$

$x_0 = 12.24 \;\rm cm$

Hence we can conclude that the relaxed length of the spring is 12.24 cm.

To know more about Hooke's law, follow the link given below.

https://brainly.com/question/3355345.