Answer:
The force applied to the spring is 35,780 N
Explanation:
Given;
extension of the spring, x = 200 mm = 0.2 m
energy stored in the spring, E = 3578 J
The force applied to the spring is calculated as;
E = ¹/₂Fx
where;
F is the applied force
F = 2E/x
F = ( 2 x 3578) / 0.2
F = 35,780 N
Therefore, the force applied to the spring is 35,780 N
