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2) calcalate The wor r heir cquilibriung posi tions, assuane Hooke's Iawis obe a74 spring that reauired 120sOf work to be Stretehed 0.4m from i o equilibriam posiition. a.l) set up a integral thaat glue) the worle 2ove | 9.2 Te omAunt of wor reauileAr

2) calcalate The wor r heir cquilibriung posi tions, assuane Hooke's Iawis obe a74 spring that reauired 120sOf work to be Stretehed 0.4m from i o equilibriam posiition. a.l) set up a integral thaat glue) the worle 2ove | 9.2 Te omAunt of wor reauileAr

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2) calcalate the worr required to Stredch 1.25 from heir cquilibriung posi tions,assuane Hooke's lawis obeyed a}4 spring that reauired 12o ĵOf work to be 6.4m from i ts equilibriam position. a.l) setup aintegral that glvej the workk doue stretehed qi2 The amucr t of wor ceauiresr DA spring Daat of 300 tobe m its equilibrinm eas iri on required a foree stretched 0;2 m from 6.1) same as al? b.2) Some awestion as a.27
Transcribed Image Text:2) calcalate the worr required to Stredch 1.25 from heir cquilibriung posi tions,assuane Hooke's lawis obeyed a}4 spring that reauired 12o ĵOf work to be 6.4m from i ts equilibriam position. a.l) setup aintegral that glvej the workk doue stretehed qi2 The amucr t of wor ceauiresr DA spring Daat of 300 tobe m its equilibrinm eas iri on required a foree stretched 0;2 m from 6.1) same as al? b.2) Some awestion as a.27
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Step 1

Since we only answer up to one question, we will answer the first question only. Please resubmit the question and specify the other question you would like to get answered. 

To determine: The work required to stretch 1.25 m from equilibrium position. 

According to Hook's law when a sample is stretched from the force becomes proportional to this extension. 

F(x)=kx

where k is constant. 

The work done to stretch a spring from equilibrium position is equal to the stored potential energy. Thus work done is

W=12kx2

The work done is given as 120 J and the spring is stretched 0.4 m from equilibrium position, so value of x is 0.4 m. Substitute these values in above equation and determine the value of spring constant (k). 

120 J=12×k×0.4 m2k=1500 J/m2

The value of k is 1500 J/m2. 

(a.1)

The equation for work done is given as force applied over distance. 

W=∫F.dx

The force is F=kx, 

W=∫kx.dx

 

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