An object is attached to the lower end of a 32-coil spring that is hanging from the ceiling. fie spring stretches by 0.160 m. The spring is then cut into two identical springs of 16 coils each. As the drawing shows, each spring is attached between the ceiling and the object. By how much does each spring stretch

0.080 m

Explanation:

According to Hooke's law, a spring with stiffness k will stretch a distance of Δx when a force F is applied:

F = k Δx

If we say the weight of the object is W, then the stiffness of the original spring is:

W = k (0.160 m)

k = W / 0.160

When the spring is cut in half, the stiffness of each new spring is the same as the original.  This time, the weight of the object is evenly distributed between each spring, so the force on each is W/2.

F = k Δx

W/2 = (W/0.160) Δx

1/2 = Δx / 0.160

Δx = 0.080

Each spring stretches 0.080 meters.