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Jakob Littel

Feb 20, 2025

A work W0 is required to stretch a certain spring 2 cm from its equilibrium position.?

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Hermina Ferry

Feb 20, 2025

How much work is required to stretch the spring 1 cm from equilibrium? Work = ∫F∙dx = ∫kx∙dx = ½kx² —–> Wo = ½k*0.02² = k/5000 so now we know k = 5000*Wo½kx² = 2500Wox² —–> 2500Wo*0.01² = 0.25Wo = Wo/4 <—— ½kx² evaluated from 0.02 to 0.03 = 2500*Wo(0.03² – 0.02²) = 1.25Wo = 5Wo/4... Show More
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Hermina Ferry

Feb 20, 2025

Work is equal to the change of elastic potential energy.That’s a quadratic function of stretch length (x^2): W = 0.5kx^2If you stretch it twice as much, required work quadruples (2^2 = 4)If you stretch it half as much, it’s one quarter. Etc.W = W0/4 for 1 cm.Suppose the spring is already stretched 2 cm from equilibrium. How much additional work is required to stretch it to 3 cm from equilibrium?If spring has stiffness k, then elastic potential energy at x=2 cm is W0 = 0.5*k*x^2 = 0.5 * k * 2^2 = 2k(you can turn cm to m or you don’t have to, because it wil cancel anyway)Now imagine we start here and stretch it even more to x=3 cm. We are doing work to increase the potential energy stored in it.After we’re done, new total energy = 0.5 * k * x^2 = 4.5kThe change from x=2 to x=3 was 4.5k – 2k = 2.5 kWe already said W0 was 2k.To find ratio, divide 2.5k / 2k = 1.25So from x=2 to x=3 is 1.25W0.... Show More
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