Question: A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22
A block with mass m =6.8 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.22 m. While at this equilibrium position, the mass is then given an initial push downward at v = 5 m/s. The block oscillates on the spring without friction.
- 1) What is the spring constant of the spring?
- 2 )What is the oscillation frequency?
- 3) After t = 0.3 s what is the speed of the block?
- 4) What is the magnitude of the maximum acceleration of the block?
- 5) At t = 0.3 s what is the magnitude of the net force on the block?
- 6) Where is the potential energy of the system the greatest?
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Lets solve these step by step 1 Spring Constant k Hookes Law states F kx where F is the force exerted by the spring k is the spring constant and x is ... View full answer
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