Question

Work of 3 Joules done in stretching a spring from its natural length to 18 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (18 cm)?

Answer 1

\[W_{spring} = \frac{ 1 }{ 2 }k x^2\]\[F_{spring} = -kx\]

Answer 2

\[3 Joules = \frac{ 1 }{ 2 }k (0.18)^2\] from this you get k\[F = -k(0.18)\]