Question

A spring has a spring constant of 300 N/m. How much force is in thespring if it is stretched 0.25 m?

Answer 1

75 Newtons

Step-by-step explanation

The extension of an elastic object when a force is applied, such as a spring, is described by Hooke's law:

force = spring constant × extension

`\begin{gathered} f=kx \\ \text{where } \end{gathered}`

force (F) is measured in newtons (N)

spring constant (k) is measured in newtons per metre (N/m)

extension, or increase in length (x), is measured in metres (m)

then

Step 1

let

`\begin{gathered} k=300(N)/(m) \\ x=0.25\text{ m} \end{gathered}`

now, replace in the formula

`\begin{gathered} f=kx \\ f=300(N)/(m)\cdot0.25\text{ m} \\ f=75\text{ N} \end{gathered}`

therefore, the answer is 75 Newtons

I hope this helps you