Answer:
The spring stretches or compresses 0.8 m (80 cm)
Explanation:
Hooke's Law
The force (F) needed to stretch or compress a spring by some distance Δx is given by
[tex]F=-k\Delta x[/tex]
Where k is a constant factor characteristic of the spring.
We are given the constant of a spring k=50 N/m and it's required to calculate the displacement Δx produced by a force of F=40 N. Solving for Δx:
[tex]\displaystyle \Delta x=\frac{F}{k}[/tex]
[tex]\displaystyle \Delta x=\frac{40}{50}[/tex]
Calculating:
[tex]\displaystyle \Delta x=0.8\ m[/tex]
The spring stretches or compresses 0.8 m (80 cm)
