How Is Work Done by a Spring Calculated?

AI Thread Summary
The discussion focuses on calculating the work done by a spring when a mass is hung from it and further extended by an external force. The user initially struggles with the correct application of the work formula, Ws = 0.5kx^2, and the spring constant, k, which was correctly calculated as 326.67 N/m. Clarification is provided that the limits for calculating work should reflect the total extension from the initial position to the final position, specifically from 0.06 m to 0.16 m. After correcting the calculations and using the appropriate limits, the user successfully arrives at the expected answer of approximately -2.54 J. The conversation emphasizes understanding the relationship between potential energy and the work done on the spring.
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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