In the spring, one end to which the load is fixed and suspended, the spring lengthened by 4 cm

In the spring, one end to which the load is fixed and suspended, the spring lengthened by 4 cm, find the mass of the load if k 200 Newtons per meter.

x = 4 cm = 0.04 m.

k = 200 N / m.

g = 10 m / s2.

m -?

Consider the condition of the balance of the load on a stretched spring.

The load is acted upon by the force of gravity Ft directed vertically downward and the tension force of the spring F directed vertically upward: Ft = F.

Let us express the force of gravity Ft by the formula: Ft = m * g, where m is the mass of the load, g is the acceleration of gravity.

We express the tension force of the spring F by Hooke’s law: F = k * x, where k is the stiffness of the spring, x is the change in the length of the spring.

m * g = k * x.

m = k * x / g.

m = 200 N / m * 0.04 m / 10 m / s2 = 0.8 kg.

Answer: the weight on the spring has a mass of m = 0.8 kg.