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A 1.5 kg box moves back and forth on a horizontal frictionless surface between two springs.?

The box is initially compressed against the stronger spring, K=32 N/cm, 4.0 cm and released from rest. The smaller box is K=16 N/cm.
A) By how much will the box compress the weaker spring?
B) What is the maximum speed the box will reach?

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1 answer
  1. I think that you meant to write "The weaker spring

    The 4.0 cm compression of the strong spring will result in a 128 N push by the strong (compressed) spring and a 64 N pull by the other (stretched) spring. It will start compressing the weaker spring when released. The potential energy stored in springs when release occurs will be (1/2)[32 + 16]*4^2= 384 J

    (A) The weaker spring starts out stretched. When it gets maximum compression later, the kinetic energy will be zero and 384 J will again be stored in spring potential energy. The compression of the weaker spring will be 4 cm.

    (B) Solve (1/2)(M1 + M2)V^2 = 384 J to get V.

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