A spring that does not conform to Hooke\'s law has a force given by F=50x^3 N, w

Question

A spring that does not conform to Hooke's law has a force given by F=50x^3 N, where x is the distance stretched from its normal length in m. The direction of the force is opposite the displacement. A) One end of the spring is connected to an overhead support and the other end is connected to a 1.5 kg mass. The mass is slowly lowered to its resting position. How far has the spring stretched to reach equilibrium? B) The mass is now raised so that the spring is un stretched and the released from rest. What is the maximum distance the mass falls below the equilibrium position? C) What is the maximum speed of the mass during its fall?

Explanation / Answer

A)here we have only potential energy.

force=k*x^2/2

50*x^3=k*x^2/2

x=k/100 where x is the distance and k is force constant

B) mass will fall half the distance of equilibrium distance below equilibrium.

C)here we have only kinetic energy.

F=m*v^2/2

50*x^3=m*v^2/2

V=(k^3/m)*10^-4

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