What is the spring's length and constant in Hooke's law with buoyancy force?

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SUMMARY

The discussion focuses on calculating the spring constant (k) and the original length (Li) of a spring when a 250 g ball is submerged in water. The spring's length changes from 0.690 m to 0.620 m due to buoyancy. The correct values for the spring constant are K=20.6 N/m and the original length is Li=0.561 m, as opposed to the incorrect values of K=17.65 N/m and Li=0.551 m derived by the user. The discrepancy arises from an error in the second equation used to calculate the buoyancy force.

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  • Understanding of Hooke's Law and spring constants
  • Knowledge of buoyancy and Archimedes' principle
  • Familiarity with basic physics equations involving force and mass
  • Ability to manipulate algebraic equations for problem-solving
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  • Review the derivation of Hooke's Law and its applications in physics
  • Study the principles of buoyancy and how they affect submerged objects
  • Practice solving systems of equations in physics problems
  • Explore the implications of spring constants in mechanical systems
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of applying Hooke's Law and buoyancy in real-world scenarios.

vinamas
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Homework Statement


A ball with a mass of 250 g and a volume of 126 cm^3 is hung vertically to a spring the springs length becomes 0.690 m.Then the ball is slowly submerged in water with a density of 1000 kg /m^3 then the spring's length becomes 0.620 m.

Find Buoyancy force of the water on the ball

What was the spring's original length

What is the constant (k) of the spring

Homework Equations



F=K(Lf-Li)
Fb=Fg-Wapp
Fb=density x volume x gravity

The Attempt at a Solution


I attempted to solve it by creating a the following system of equations
0.250 x 9.81 = K(0.690-Li) and 0.250 x 9.81 - 1000 x 126^-6 x 9.81 = K(0.620-Li)
Every time I solve it I get that K=17.65 and Li= 0.551 m
The answer sheet says that K=20.6 and Li = 0.561
Please help I suspect that my mistake is in the second equation but am not too sure
 
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vinamas said:

Homework Statement


A ball with a mass of 250 g and a volume of 126 cm^3 is hung vertically to a spring the springs length becomes 0.690 m.Then the ball is slowly submerged in water with a density of 1000 kg /m^3 then the spring's length becomes 0.620 m.

Find Buoyancy force of the water on the ball

What was the spring's original length

What is the constant (k) of the spring

Homework Equations



F=K(Lf-Li)
Fb=Fg-Wapp
Fb=density x volume x gravity

The Attempt at a Solution


I attempted to solve it by creating a the following system of equations
0.250 x 9.81 = K(0.690-Li) and 0.250 x 9.81 - 1000 x 126^-6 x 9.81 = K(0.620-Li)
Every time I solve it I get that K=17.65 and Li= 0.551 m
The answer sheet says that K=20.6 and Li = 0.561
Please help I suspect that my mistake is in the second equation but am not too sure
I obtain your equations and your answers.
 
Thanks!
 

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