Finding Work Done by a Spring Force on a Pen

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SUMMARY

The discussion focuses on calculating the work done by a spring force in a pen with a spring constant of 221 N/m. The spring is initially compressed by 4.90 mm and requires an additional compression of 5.20 mm to extend the pen tip into writing position. The work done by the spring can be calculated using the formula W elastic = 0.5kx^2, where 'x' represents the total compression from the unstrained position. The algebraic sign of the work must be considered, indicating whether the work done is positive or negative based on the spring's potential energy change.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the formula for elastic potential energy (W elastic = 0.5kx^2)
  • Knowledge of algebraic signs in physics work calculations
  • Basic concepts of work-energy principles in mechanics
NEXT STEPS
  • Review the derivation and applications of Hooke's Law in mechanical systems
  • Study the implications of positive and negative work in energy transfer
  • Explore examples of work done by springs in various mechanical devices
  • Learn about energy conservation principles related to potential and kinetic energy
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators seeking to clarify spring force calculations and their implications.

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Homework Statement


A pen contains a spring with a spring constant of 221 N/m. Whe the tip of the pen is in its retracted position, the spring is compressed 4.90 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 5.20 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.


Homework Equations


W elastic=.5kx^2


The Attempt at a Solution


I have no idea what they mean by keep an algebraic answer and where does x final come into the equation along with the additional x value.
 
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maw07005 said:

Homework Statement


A pen contains a spring with a spring constant of 221 N/m. Whe the tip of the pen is in its retracted position, the spring is compressed 4.90 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 5.20 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.


Homework Equations

W elastic=.5kx^2
this is the Potential Energy of the spring when it is displaced a distance x from its unstretched position. Don't confuse work with Potential Energy.

The Attempt at a Solution


I have no idea what they mean by keep an algebraic answer and where does x final come into the equation along with the additional x value.
The work done by a force moving through a distance can be plus or minus (or 0). They want to know here if the work done by the spring is positive or negative (or 0), and what is its value. That's what they mean by an algebraic answer. What do you know about the relationship between the work done by a spring and its potential energy change?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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