A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 46 cm. What is the spring constant of the spring?

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

k = 17043.5 N/m = 17.04 KN/m