Question

Consider a large spring, hanging vertically, with spring constant k = 3220 N/m. If the spring is stretched 25.0 cm from equilibrium and a block is attached to the end, the block stays still, neither accelerating upward nor downward. What is the mass of the block?

Answer 1

m = 82.1 kg

Step-by-step explanation:

For this exercise we must use Hook's law that states that the force exerted by a spring is proportional to the displacement

Fe = -k x

Let's use Newton's second law to establish equilibrium, the elastic force up and the body weight down

Fe - W = 0

Fe = W = mg

k x = m g

m = k x / g

Let's reduce the distance to SI units

x = 25 cm (1 m / 100cm) = 0.250 m

m = 3220 0.250 /9.8

m = 82.1 kg