What Spring Constant Fires a 10g Mass to 100m When Compressed to 10cm?

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Homework Help Overview

The discussion revolves around determining the spring constant required to launch a 10 g mass to a height of 100 m when the spring is compressed to 10 cm. The problem involves concepts from energy conservation and potential energy in the context of mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the potential energy stored in the spring and the potential energy of the mass at its maximum height. There are discussions on using conservation of energy and the formula for potential energy of a spring.

Discussion Status

Some participants have provided guidance on relating the potential energies, while others have raised questions about the necessary parameters, such as the spring constant and the minimum velocity required for the mass to reach the desired height. There is an ongoing exploration of the correct application of energy concepts.

Contextual Notes

Participants note the importance of correctly interpreting the compression distance of the spring, emphasizing that it is 10 cm, not 1 m. There are also mentions of assumptions regarding the absence of friction and the nature of the gravitational field.

NoHeart
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with no friction- what spring constant would fire a 10 g mass to a height of 100 m. the spring is compressed to 10 cm.

i know k=F/x but i don't know how to find the force for this problem, because i need to find the acceleration and there is no time unit given. any help?
 
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You don't really need to consider the force. What is the potential energy stored in the spring before it is released? What is the potential energy of the mass when it reaches its maximum height? What is the relationship between those two quantities? I hope you see what I'm getting at.
 
Use conservation of energy. The potential energy of a spring that has been compressed over a distance x is kx²/2

You also know that at the 'end' of the motion, the object must be 100m above the horizontal. This corresponds to a potential energy of mg100 (due to gravity)...

regards
marlon
 
Read the question in terms of energy. What must be the force F=kx such that the work

[tex]W = \int_{0}^{0.10} kx dx[/tex]

done by it over 10cm is equal to the potential energy associated with an height of 100m above the surface of the earth?
 
Last edited:
how do i find the potential energy of the spring if i don't know the spring constant?
 
NoHeart said:
how do i find the potential energy of the spring if i don't know the spring constant?
like i said, the potential energy is kx²/2. You can prove this formula by integrating the force wtr displacement x, but that's not necessary.

You will need this formula to find k...

marlon
 
U need to know the minimum velocity needed for the body to reach 100 m in a constant gravity field.

Daniel.
 
dextercioby said:
U need to know the minimum velocity needed for the body to reach 100 m in a constant gravity field.

Daniel.

Nope, that is not necessary...

marlon
 
  • #10
dextercioby said:
What?

Daniel.

You don't need to know the minimal velocity.

marlon
 
  • #11
thanks so much for the help everyone, i hope I'm understanding now...
so the potential energy of the object when it reaches a height of 100m is mgh, .01kg x 10m/s^2 x 100m=10Nm=10J
is the potential energy of the object equal to the potential energy of the spring? if so, i have 10Nm=(k(1m)^2)/2 and k=20N/m
please tell me I'm at least on the right track...
 
  • #12
NoHeart said:
so the potential energy of the object when it reaches a height of 100m is mgh, .01kg x 10m/s^2 x 100m=10Nm=10J
That is correct

is the potential energy of the object equal to the potential energy of the spring?

yes indeed, that's the clue of the story

if so, i have 10Nm=(k(1m)^2)/2 and k=20N/m
please tell me I'm at least on the right track...

Beware, the compression is 10 cm NOT 1m !

marlon
 
  • #13
Okay,Marlon,gravity field is conservative,so you could use the PE,too.Okay.

Daniel.
 
  • #14
thank you so much, marlon. 10cm is .1 m, jeez! at least i know how to do this problem now...
 

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