A block with mass m =7.5 kg is hung from a vertical spring. When the mass...

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Physics

A block with mass m =7.5 kg is hung from a vertical spring. Whenthe mass hangs in equilibrium, the spring stretches x = 0.25 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.
After t = 0.3 s what is the speed of the block?
What is the magnitude of the maximum acceleration of theblock?
At t = 0.3 s what is the magnitude of the net force on theblock?
Where is the potential energy of the system the greatest?
At the highest point of the oscillation.
At the new equilibrium position of the oscillation.
At the lowest point of the oscillation.

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4.0 Ratings (703 Votes)

First of all determine the spring constant k of the spring Use the expression F kx mg k025 k mg025 75981025 2943 Nm So the frequency of oscillation is f sqrt km 2 sqrt 2943 75 2 10 Hz Now the mass is given an See Answer

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A block with mass m =7.5 kg is hung from a vertical spring. When the mass...

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Physics

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