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

A weight of 2 kg is suspended on a spring with a stiffness of 200 N / m. Determine the length of the stretched spring if its unstrained length is 0.3 m, and the spring with a load rises with an acceleration of 5 m / s2 directed upwards.

m = 2 kg.

g = 9.8 m / s ^ 2.

k = 200 N / m.

l0 = 0.3 m.

a = 5 m / s ^ 2.

l -?

Let’s write 2 Newton’s law in vector form: m * a = Fcont + m * g.

Let’s write 2 Newton’s zon for projections on the vertical axis directed upwards: m * a = Fcont – m * g.

We find the elastic force of the spring according to Hooke’s law: F = k * (l – l0).

m * a = k * (l – l0) – m * g.

m * a = k * l – k * l0 – m * g.

k * l = m * a + k * l0 + m * g.

l = (m * a + k * l0 + m * g) / k.

l = (2 kg * 5 m / s ^ 2 + 200 N / m * 0.3 m + 2 kg * 9.8 m / s ^ 2) / 200 N / m = 0.45 m.

Answer: l = 0.45 m.