# Using a spring dynamometer, lift a 2kg weight vertically upward, with an acceleration of 2.5m / s2.

**Using a spring dynamometer, lift a 2kg weight vertically upward, with an acceleration of 2.5m / s2. Determine the modulus of elongation of the spring, if its stiffness is 1000N / m.**

m = 2 kg.

g = 9.8 m / s2.

a = 2.5 m / s2.

k = 1000 N / m.

x -?

When lifting it vertically upwards, 2 forces act on the load: gravity m * g,

directed vertically downwards, elastic force Fel of the spring, directed vertically upwards.

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

m * a = – m * g + Fcont – 2 Newton’s law for projections on the vertical axis.

Fcont = m * a + m * g = m * (a + g).

Let us express the elastic force according to Hooke’s law: Fel = k * x, where k is the stiffness of the spring, x is the elongation of the spring.

m * (a + g) = k * x.

x = m * (a + g) / k.

x = 2 kg * (2.5 m / s2 + 9.8 m / s2) / 1000 N / m = 0.0246 m.

Answer: the elongation of the spring is x = 0.0246 m.