An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N, the spring is stretched by 15.5 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.4 cm from that position. (in J)

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is

43.8 N = k (0.155 m)   ==>   k = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

+15-35=-20

going in opposite directions