If a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of the spring?

0.1 N

Explanation:

Considering that Hooke's law, F=xk where k is the spring constant in N/m, x is extension in meters, F is spring force in N.

Converting the given extension from cm to m, we divide by 100 hence 5/100=0.05 m

Substituting 0.05 m for x and 2 N/m for k then the spring force will be

F=2*0.05=0.1 N