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A 0.505-kg block slides on a frictionless horizontal surface with a speed of 1.18 m>s. The...

A 0.505-kg block slides on a frictionless horizontal surface with a speed of 1.18 m>s. The block encounters an unstretched spring and compresses it 23.2 cm before coming to rest. (b) For what length of time is the block in contact with the spring before it comes to rest? (c) If the force constant of the spring is increased, does the time required to stop the block increase, decrease, or stay the same? Explain.

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Answer #1

m = 0.505 kg v= 1-18 mils = 23.21m = 0.232 m » Where the kinetic energy = spring Potenlied energs. KE = PEspring 7 m / kk myz

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