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A massless spring has unstretched length /0 and force
Chapter 5, Problem 5.1(choose chapter or problem)
A massless spring has unstretched length /0 and force constant k. One end is now attached to the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now /1. (a) Write down the condition that determines 11. Suppose now the spring is stretched a further distance x beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is F = kx. That is, the net force obeys Hooke's law, when x is the distance from the equilibrium position a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form U (x) = const ikx2.
Questions & Answers
QUESTION:
A massless spring has unstretched length /0 and force constant k. One end is now attached to the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now /1. (a) Write down the condition that determines 11. Suppose now the spring is stretched a further distance x beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is F = kx. That is, the net force obeys Hooke's law, when x is the distance from the equilibrium position a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form U (x) = const ikx2.
ANSWER:Step 1 of 3
Part (a)
Initially at an unstretched length of spring, the spring is in equilibrium condition. The net force acting on the system is zero.
Hence, in equilibrium condition the spring force will be equal to the gravitational force but opposite in direction. i.e., .
Now, the spring is stretched further by a distance of x beyond its changed equilibrium length.
The equation for the net force acting on spring-mass system is as follows: