The load is suspended by a spring with a stiffness of 100 N / m from the elevator ceiling.

The load is suspended by a spring with a stiffness of 100 N / m from the elevator ceiling. The elevator descends uniformly downward to a distance of 5 m for 2 s. What is the mass of the load if the elongation of the spring with a steady motion of the load is 1.5 cm?

k = 100 N / m.
g = 9.8 m / s ^ 2.
S = 5 m.
t = 2 s.
x = 1.5 cm = 0.015 m.
m -?
Two forces act on the load in the elevator: gravity sieves Ft directed vertically downward, and the tension force of the spring N directed vertically upward.
Let’s write 2 Newton’s law: m * a = N – Fт.
The force of gravity Ft is determined by the formula: Ft = m * g.
m * a = N – m * g.
The spring tension N will be determined as follows: N = m * a + m * g = m * (a + g).
According to Hooke’s law, the force of a stretched spring N is determined by the formula: N = k * x.
m * (a + g) = k * x.
m = k * x / (a ​​+ g).
S = a * t ^ 2/2.
a = 2 * S / t ^ 2.
m = k * x / (2 * S / t ^ 2 + g).
m = 100 N / m * 0.015 m / (2 * 5 m / (2 s) ^ 2 + 9.8 m / s ^ 2) = 0.122 kg.
Answer: the mass of the cargo is m = 0.122 kg.