Answer to Question #232780 in Mechanical Engineering for jhy

Question

Three springs with the same constant connected in series and parallel, and a 1-kg object attached at one end of a spring, as shown in figure below. Spring constant is k1 = k2 = k3 = 250 N/m. What is the change in length of the three springs? Acceleration due to gravity is g = 9.806 m.s-2

Answer 1

Solution;

Given;

$k_1=k_2=k_3=250N/m$

Object mass,$m=1kg$

Acccelaration due to gravity,$g=9.806ms^{-2}$

Hence ;

Object weight,$w=m×g=1×9.806=9.806N$

Find the equivalent constant;

$k_{12}=k_1+k_2=250+250=500N/m$

Such that $k_{12}$ and $k_3$ are in series ,and whose equivalent is ;

$\frac1k=\frac{1}{k_3}+\frac{1}{k_{12}}=\frac1{250}+\frac{1}{500}=\frac{3}{500}$

$k=\frac{500}{3}$ N/m

To find change in length in the springs,$\Delta x$ ;

$\Delta x=\frac{w}{k}=9.806×\frac3{500}$

$\Delta x=0.058836m$