Asked by Corey
A 2.0 kg box moves back and forth on a horizontal frictionless surface between two different springs with one 32 N/cm and the other at 16 N/cm.The box is initially pressed against the stronger spring, compressing it 3.5 cm , and then is released from rest.
By how much will the box compress the weaker spring? What is the maximum speed the box will reach?
By how much will the box compress the weaker spring? What is the maximum speed the box will reach?
Answers
Answered by
Damon
potential energy U = (1/2) k x^2
(1/2)(3200 N/m)(.035)^2
max speed = Vmax
(1/2) m Vmax^2 = U
compression of weak spring, same potential energy when stopped
U = (1/2)(1600)(x^2)
x is in meters of course
(1/2)(3200 N/m)(.035)^2
max speed = Vmax
(1/2) m Vmax^2 = U
compression of weak spring, same potential energy when stopped
U = (1/2)(1600)(x^2)
x is in meters of course
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