Question

A 1200 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 54 cm . What is the spring constant.

Answer 1

the spring constant is 200kN/m.

The work done on the spring by the the free falling safe is equal to the kinetic energy of the spring: 

W1 = $\frac{1}{2}$k$x^{2}$

The work done on the safe by gravity is the potential energy of the spring: 

W2 = (mass of safe) x g x (height)
     = 1000kg x 9.8 m/s^2 x (2 + 0.54)m
     =1000kg x 9.8 m/s^2 x (2.54)m
     = 24,892 N-m


W2 = W1 =24892 N-m
 ⇒ $\frac{1}{2}$k$(0.5)^{2}$ = 24892 N-m 

Therefore, k = $\frac{(24892)*2}{0.5*0.5}$= 199136 kN/m. 

The spring constant is 199,136 N/m ≈ 200 kN/m