A mass 1 kg is attaced to the hook of a spring and the spring is suspended vertically from a ceiling (see Fig. 5.12a). The spring is displaced from its equilibrium position by a distance x. The spring constant of the spring is `2.0xx10^(2)" N"//"m"`. Calculate the displacement x.

Figure 5-12 (a) Mass of 1 kg attached to a spring is suspended vertically. (b) Spring displace from equilibrium position by a distance x.

Explanation

(1) In Fig. 5-12, mass attached to the spring is shown. Figure 5-12a shows the relaxed position of spring. Figure 5-12b shows the distance x by which the spring is stretched due to displacement from equilibrium position. (2) The force on the mass is balanced by its weight acting downwards and spring force acting upwards.
Calculation: At equilibrium, spring force = weight, that is,
`vec(F)_(s)=mg`
That is, `" "|vec(F)_(s)|=kx=mg`
Substituting m = 1 kg, `k=2.0xx10^(2)" N"//"m"` and `g=10" m"//"s"^(2)`, we obtain the displacement x as
`x=(mg)/(x)=((1" kg")(10" m"//"s"^(2)))/((2.0xx10^(2)" N"//"m"))`
`=5xx10^(-2)m=5" cm"`