A weight of 2 kg is supplied on a spring with a stiffness of 200N / m. Determine the length of the stretched spring

A weight of 2 kg is supplied on a spring with a stiffness of 200N / m. Determine the length of the stretched spring if its unstrained length is 0.3 m, and the spring with the load rises with an acceleration of 5 m / s directed upwards.

m = 2 kg.
k = 200 N / m.
g = 9.8 m / s2.
l0 = 0.3 m.
a = 5 m / s2.
l -?
Let’s write Newton’s 2 law for a body when moving with acceleration up: m * a = F – m * g, where F is the elastic force of the spring, m * g is the force of gravity.
We express the elastic force according to Hooke’s law: F = k * (l – l0) = k * l – k * l0.
m * a = k * l – k * l0 – m * g.
m * a + k * l0 + m * g = k * l.
The formula for determining the length of the spring will take the form: l = (m * a + k * l0 + m * g) / k.
l = (2 kg * 5 m / s2 + 200 N / m * 0.3 m + 2 kg * 9.8 m / s2) / 200 N / m = 0.448 m.
Answer: when moving, the length of the spring will become l = 0.448 m.