A 0.47-kg object connected to a light spring with a force constant of 19.2 N/m o

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Jun 26th, 2015

Thank you for the opportunity to help you with your question!

a) Ep = ½kx² = ½ * 19.2N/m * (0.04m)² = 0.01536 J 
Ek = 0.01536 J = ½mv² = ½ * 0.47kg * v² 
v = 0.256 m/s 

b) Ek = E - Ep = 0.01536J - ½ * 19.2N/m * (0.015m)² = 0.0132 J = ½ * 0.47kg * v² 
v = 0.237 m/s 

c) see b. 0.237

d) At one-half the maximum speed, the kinetic energy is 1/4 the total energy; therefore the potential energy is 3/4 of the total energy. 
0.01536J * 0.75 = ½ * 19.2N/m * x² 
x = 0.0346 m = 3.46 cm

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 26th, 2015

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