Answer:
350 N/m
Explanation:
If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m
k = 70 / 0.2 = 350 N/m
The spring constant is 350 N/m
Answer:
350 N/M
Explanation:
Here for a force of 70 N a elongation of 20 cm is observed
⇒
force = k × extension
k =[tex]\frac{force}{extension}[/tex]
extension = 20 cm = 0.2 m.
k = [tex]\frac{70}{0.2}[/tex] = 350 N/M