A block with mass m =6.5 kg is hung from a vertical spring. When the mass...

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Physics

A block with mass m =6.5 kg is hung from a vertical spring. Whenthe mass hangs in equilibrium, the spring stretches x = 0.22 m.While at this equilibrium position, the mass is then given aninitial push downward at v = 4.1 m/s. The block oscillates on thespring without friction. 1) What is the spring constant of thespring? N/m 2) What is the oscillation frequency? Hz 3) After t =0.39 s what is the speed of the block? m/s 4) What is the magnitudeof the maximum acceleration of the block? m/s2 5) At t = 0.39 swhat is the magnitude of the net force on the block? N 6) Where isthe potential energy of the system the greatest? At the highestpoint of the oscillation. At the new equilibrium position of theoscillation. At the lowest point of the oscillation.

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4.4 Ratings (919 Votes)

1 The spring constant K is computed with the information known about the mass at rest F kx mg k022 k mg022 65981022 Spring constant k 28984 Nm 2 The frequency of oscillation is f sqrt km 2 f sqrt28984 65 2 Oscillation frequency f 1063 See Answer

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