One end of a light spring (force constant k) is attached tothe ceiling, the othe

Question

One end of a light spring (force constant k) is attached tothe ceiling, the other end is attached to an object of mass m. Thespring is initially vertical and unstressed. You then ease theobject down to an equilibrium position a distance h below itsinitial position. Next, you repeat the experiment , but instead ofeasing the object down, you release it, with the result that itfalls a distance H below the initial position before momentariallystopping. a) Show that h = mg/k. b) Use the work-kinetic energytheorm to show that H = 2h. Okaayyy...for (a) it would make sense that h = mass timesgravity, but why must we divide that by the kinetic energy? Atheight h the spring is basically resting right? For (b), the work-kinetic energy theorm is Wtotal =K = 1/2mv2 - 1/2mv2. What do I use forthe velocity? One end of a light spring (force constant k) is attached tothe ceiling, the other end is attached to an object of mass m. Thespring is initially vertical and unstressed. You then ease theobject down to an equilibrium position a distance h below itsinitial position. Next, you repeat the experiment , but instead ofeasing the object down, you release it, with the result that itfalls a distance H below the initial position before momentariallystopping. a) Show that h = mg/k. b) Use the work-kinetic energytheorm to show that H = 2h. Okaayyy...for (a) it would make sense that h = mass timesgravity, but why must we divide that by the kinetic energy? Atheight h the spring is basically resting right? For (b), the work-kinetic energy theorm is Wtotal =K = 1/2mv2 - 1/2mv2. What do I use forthe velocity?

Explanation / Answer

a) Let k be the force constant of the light spring. then wehave              F = kh The weight of the mass acting downward and we have             Fg = mg Therefore along the vertical direction we have             F = Fg             kh = mg        ==> h = mg /k       .........1          b) We know that           Kineticenergy = 0.5 kH2         Workdone = change inkinetic energy           Fg.H = 0.5 kH2            mg H =0.5 kH2            mg H =0.5 (mg / h) * H2              2* h =   H

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