The force of 0.8m strctches an elastic spring by 2cm find the elasticconstant of the spring
We can use Hooke's Law to find the elastic constant of the spring:
F = kx
where F is the force applied to the spring, x is the displacement of the spring from its equilibrium position, and k is the elastic constant (also known as the spring constant).
In this case, we are given that the force applied is 0.8 m and the displacement is 2 cm, or 0.02 m. Plugging in these values, we get:
0.8 = k(0.02)
Solving for k, we get:
k = 0.8 / 0.02
k = 40 N/m
Therefore, the elastic constant of the spring is 40 N/m.