Question

18.0cm 20.5cm 25.0cm No Load A spring is stretched by a force f from its natural length of 18 cm. When it is stretched to 20.5 cm, a force of 50.17 N is applied. What is the spring constant of this spring? Answer in 2 decimal places and in N/m.

Answer 1

For a spring force $F= k(\Delta x)$

where F is the applied force and $(\Delta x)$ is the displacement due to that force, and k is called spring constant.

So,

$k= \frac{F}{(\Delta x)}= \frac{50.17}{(0.205-0.180)}= 2006.80$

Thus spring constant = 2006.80 N/m