Consider a large spring, hanging vertically, with spring constant k = 3220 N/m. If the spring is stretched 25.0 cm from equilibr
ium and a block is attached to the end, the block stays still, neither accelerating upward nor downward. What is the mass of the block?
1 answer:
Answer:
m = 82.1 kg
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
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