Answer to Question #232780 in Mechanical Engineering for jhy

Question #232780

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


1
Expert's answer
2021-09-04T01:00:54-0400

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\u00d7g=1\u00d79.806=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\u00d7\\frac3{500}"

"\\Delta x=0.058836m"








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