Answer to Question #272218 in Physics for Rizzykul

Question

A 1.50kg object hangs motionless from a spring with a force constant of k=250N/m.how far is the spring stretched from it equilibrium length

Answer 1

The weight of the object is:

$$W = mg$$

where $m = 1.50kg,g=9.81N/kg$. According to the Hook's law, the elongation of the spring is:

$$x = \dfrac{W}{k}$$

where $k=250N/m$. Thus, obtain:

$$x = \dfrac{mg}{k} = \dfrac{1.50kg\cdot 9.81N/kg}{250N/m}\approx 0.0589m$$

Answer. 0.0529m.