Question

A spring has a spring constant of 25 newtons per meter. Calculate the magnitude of the minimum force required to stretch the spring 0.25 meter from its equilibrium position

Answer 1

The minimum force required to stretch a spring with a spring constant of 25 N/m by 0.25 meter from its equilibrium position is calculated using Hooke's Law, resulting in a force of 6.25 newtons.

Step-by-step explanation:

The question is asking to calculate the minimum force required to stretch a spring 0.25 meter from its equilibrium position given that the spring has a spring constant of 25 newtons per meter. According to Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance, the formula can be expressed as:

F = k × x

Where F is the force in newtons, k is the spring constant in newtons per meter, and x is the displacement from the equilibrium position in meters.

Therefore, we can calculate the force as follows:

F = 25 N/m × 0.25 m = 6.25 N

Thus, the minimum force required to stretch the spring 0.25 meter from its equilibrium position is 6.25 newtons.

Answer 2

Given data:

* The spring constant of the given spring is k = 25 N/m.

* The stretched length of the spring is x = 0.25 m.

Solution:

The magnitude of minimum force required to stretch the wire is,

`\begin{gathered} F=kx \\ F=25*0.25 \\ F=6.25\text{ N} \end{gathered}`

Thus, the magnitude of minimum force required in the given case is 6.25 N.