Question

A spring with spring constant 120 N/m and unstretched length 0.4 m has one end anchored to a wall and a force F is applied to the other end.

Required:
a. If the force F does 250 J of work in stretching out the spring, what is its final length?
b. If the force F does 250 J of work in stretching out the spring, what is the magnitude of F at maximum elongation?

Answer 1

(a) The work done by F in stretching the spring a distance x is

W = 1/2 (120 N/m) x ²

so F performs 250 J of work on the spring, then

250 J = 1/2 (120 N/m) x ²   ==>   x ² ≈ 4.17 m²   ==>   x ≈ 2.04 m

(b) In order for F to perform 250 J of work, its magnitude must be

F = (120 N/m) (2.04 m) ≈ 245 N