Help with 2 Physics questions

  • Thread starter vee123
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  • #1
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My physics teacher gave me the following two problems as homework and I am completely lost on them.

1) If you cut a spring in half, the k value doubles. An object is attached to the lower end of a 100-coil spring hanging from a ceiling. The spring stretches by 0.16 m. The spring is then cut in half. By how much does each spring stretch?

I don't think he gave me enough information, so I don't know what to do. Any suggestions?

2) A small ball is attached to one end of a spring that has an unstrained length of 0.0200 m. The spring is held at one end and whirled in a horizontal circle at 3m/s. The spring stretches by 0.10m. How much would the spring stretch if the ball was attached to a vertical spring on the ceiling?

I know the spring on the ceiling will stretch less than the spring that is swung in the circle. And the two springs have the same unstrained length. Since I am given velocity, do I use this formula to find the k value?:

1/2kx^2 + mgh(initial height) + 1/2mv^2(initial veloicty) + Work increase = mgh(final height) + 1/2mv^2(final velocity) + Work decrease + 1/2kx^2

Any help will be greatly appreciated!
 

Answers and Replies

  • #2
quasar987
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vee123 said:
My physics teacher gave me the following two problems as homework and I am completely lost on them.

1) If you cut a spring in half, the k value doubles. An object is attached to the lower end of a 100-coil spring hanging from a ceiling. The spring stretches by 0.16 m. The spring is then cut in half. By how much does each spring stretch?

I don't think he gave me enough information, so I don't know what to do. Any suggestions?

There is enough info. The strategy will be to first find the k constant of the unstretched spring using all you know about spring force and forces in general (i.e. the 2nd law).

vee123 said:
2) A small ball is attached to one end of a spring that has an unstrained length of 0.0200 m. The spring is held at one end and whirled in a horizontal circle at 3m/s. The spring stretches by 0.10m. How much would the spring stretch if the ball was attached to a vertical spring on the ceiling?

I know the spring on the ceiling will stretch less than the spring that is swung in the circle. And the two springs have the same unstrained length. Since I am given velocity, do I use this formula to find the k value?:

1/2kx^2 + mgh(initial height) + 1/2mv^2(initial veloicty) + Work increase = mgh(final height) + 1/2mv^2(final velocity) + Work decrease + 1/2kx^2

Any help will be greatly appreciated!

I can't see what you did there with that big equation but it's probably wrong and if somebody here doesn't tell you why, you should ask your teacher why it's wrong.

IMO, the main startegy line here is exactly the same as in problem 1: the ball going around in circle is undergoing an acceleration, right? The centripetal acceleration. What is the cause of that force? The spring of course. Use the second law to find what that force is and find the srping constant from it.
 
  • #3
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The spring is whirled in a horizontal circle. It is in uniform circular motion. Use that to compare the forces acting on the two springs in either situation.
 
  • #4
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For the first problem do I do something like this(?):

Force normal=ma-Force weight
Force normal=ma-umg
 
  • #5
quasar987
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Draw the free body diagram. Draw all the vector forces. Since all vectors are in one dimension, we may, if we chose, drop the vector notation and work only with scalar, at the condition that we set a + direction and a - direction.

I can't really make sense of your use of the normal force, but here are some infos that you may or may not be be aware of...

1) The term "normal" means "perpendicular" and is not a synonim of the "reaction force" Newton's third law talks about.

2) If you meant "normal force" to designate the force of the spring being the reaction force to that of gravity on the block, that is incorect. Because the "reaction force" to the force of gravity from the earth to the block (mg) is a force of gravity from the block to the earth, and therefor it does not act on the block itself and therefor has nothing to do in the equation of motion of motion (F = ma) of the block.

Try again :smile:

(And what is "u" in umg ?!?)
 
  • #6
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Hm.. I'm still don't know what to do. My physics teacher told us that when an object is hanging vertically force normal and force weight are the forces we use in the free body diagram. We haven't done any problems like these in class. We've only used Hooke's law and Work=1/2kx^2.
Thanks for your help, I really do appreciate it, I think I might talk to my physics teacher tomorrow.

(The "u" in umg is the coefficent of friction. I didn't know what I was doing, so just ignore it.. Sorry about that!)
 
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