Question
Suppose a spring is constructed that DOESN'T obey Hooke's law, but instead exhibits a restoring force given by Fx =kx3 where k =8 N m3(it
Suppose a spring is constructed that DOESN'T obey Hooke's law, but instead exhibits a restoring force given by Fx =kx3 where k =8 N m3(it needs these units so that force is newtons). How much work is done to stretch this spring by The spring in the previous problem is released from that stretched position with a 0.20kg mass attached.
Assume the whole arrangement is horizontal so that gravity isn't doing work. How fast will the mass be moving when the spring once again reaches equilibrium length? 0.50m from equilibrium?
Suppose in the prior problem that the mass attached to the spring experiences a little sliding friction as it slides across a horizontal surface while being pulled by the spring. If =0.05 how fast will the mass be moving when the spring reaches equilibrium length
Step by Step Solution
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Step: 1
To find how much work is done to stretch the spring by a distance of x we need to calculate the work done by the nonHookes law spring force Fx kx3 The ...Get Instant Access to Expert-Tailored Solutions
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Step: 2
Step: 3
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