Question

work of 4 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (in newtons) that holds the spring stretched at the same distance (11 cm)

Answer 1

The work performed on the spring is

W = 1/2 k x²

so that

4 J = 1/2 k (0.11 m)² ⇒ k ≈ 660 N/m

Then by Hooke's law, the force required to hold the spring in this position is

F = k x = (660 N/m) (0.11 m) ≈ 73 N