Tension of a rubber cord problem

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SUMMARY

The tension in the rubber cord when the 2kg block is at its lowest point is 39.2 N, not 19.6 N as initially calculated. This discrepancy arises because the block is released from a position where it stretches the cord beyond its new equilibrium length, necessitating the application of conservation of energy principles. The gravitational potential energy of the block and the elastic potential energy of the stretched cord must be considered to accurately determine the tension and the stretched length of the cord.

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a 2kg block hangs from a rubber cord. the block is supported so that the cord is not stretched. the unstretched length is .5m and its mass is 5g. the spring constant is 100 N/m. The block is released and stops at the lowest point. a) determine the tension in the cord when the block is at this lowest point b) hat is the length of the cord in the stretched position c) find the speed of a transverse wave in the cord if the block is held at this lowest position.

a) i said that the tension in the string is the same as mg
so T = mg
T = (2kg)(9.8m/s^2)
T = 19.6 N

the back of the book says the answer is 39.2 N, which is double mine. The teacher said something about conservation of energy but i still can't figure out where that factor of two comes in :(

i know how to do b) and c), but they depend on a) so of course they come out wrong too. :(
 
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Your mistake is that you are assuming that the lowest point is the new equilibrium length of the spring, i.e., that the mass was slowly and gently lowered and released at the point where it would just hang there without moving. That's not what is happening. This mass was being supported at the spring's old unloaded length, and then suddenly released so it falls downward and stretches the spring beyond its new loaded length to some bottom position where it momentarily stops falling and then bounces upward, etc. The question asks for the tension and position at that instantaneous lowest position.

There are several ways to solve this, but as your teacher said, the easiest way is by using potential energy -- gravitational potential energy and the potential energy of the stretched spring. See what you can do with that.
 
Doh, I didn't realize that it's FALLING even after the teacher hinted about conservation of energy! Thanks, I got the answer now. =]
 

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