How Is Work Done by a Spring Calculated?

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Homework Help Overview

The problem involves calculating the work done by a spring when a mass is hung from it and subsequently extended further by an external force. The context is within the subject area of mechanics, specifically focusing on spring dynamics and energy considerations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the spring constant and the work done by the spring, questioning the correct limits and values to use in their equations. There is confusion regarding the potential energy changes and the correct interpretation of the displacement values.

Discussion Status

Participants are actively engaging with each other's reasoning, with some providing clarifications on the equations to use. There is an ongoing exploration of the correct approach to calculating the work done, with no explicit consensus reached but productive dialogue occurring.

Contextual Notes

There is mention of specific values and equations, including the spring constant and displacement, which are critical to the calculations. Participants are also addressing the initial conditions and how they affect the final calculations.

brunettegurl
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work done by the spring URGENT! pls. help

Homework Statement


An ideal spring is hung vertically from the ceiling. When a 2.0 kg mass hangs at rest from it, the spring us extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. while the spring is being extended by the forxe, the work done by the spring is?

Homework Equations


Ws=o.5kx^2
Fs=kx
Weight=mg

The Attempt at a Solution


so first i found k of the string by equating mg=kx with x=0.06 m where i got a value of 326.67 N/m. then i used Ws=o.5kx^2 with delta x=0.1m and tried to solve for work and my answer is coming out wrong. The answer is supposed to be -3.6 J ..i'm not getting that or close to it..pls help
 
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You need to account for the potential energy in the spring already as a result of the initial weight hanging on it don't you?
 


aare you saying that i should use this equation W=o.5kx^2 +mg where x=0.1 and k=326.67N/m
 


brunettegurl said:
aare you saying that i should use this equation W=o.5kx^2 +mg where x=0.1 and k=326.67N/m

No. I'm saying that the work that goes into increasing the potential energy in the spring is from .06 to .16, not from .00 to .10.
 


i'm sorry but I'm really confused by what you mean wouldn't delta x= 0.1[0.16-0.06] ? or do you mean that when I'm solving for k value in the equation mg=kx my x should actually be the difference between 0.06-0.16 and not 0.06?
 


Remember that

W = F*d

but in this case it's

W = ∫ Fx dx = ∫ -k*x dx

and your limits are from .06 to .16

This comes out to W = -1/2*k*x2 from .06 to .16 or

W = -1/2*k*(.162 - .12)
 
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brunettegurl said:
i'm sorry but I'm really confused by what you mean wouldn't delta x= 0.1[0.16-0.06] ? or do you mean that when I'm solving for k value in the equation mg=kx my x should actually be the difference between 0.06-0.16 and not 0.06?

You have the correct value of k by the way, so that's not the issue.
 


ok but then would the k value still be right as i used x=0.06 m?? because i used this equation W = -1/2*k*(.16^2 - .1^2) with the k value of 326.67 N and still am not getting the right answer
 


Sorry. I typed it wrong.

1/2*k*(.162 - .062)

Is what it should be.
 
  • #10


i'm getting -2.54J

W=-0.5*(326.67N)(0.16^2-0.1^2)
W=-0.5*(326.67)(0.0156)
W=-2.548026 J
 
  • #11


ok yea now I'm getting the answer thank you so much :)
 

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