How much work is required to stretch a spring by an additional 4.0 cm?

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SUMMARY

The discussion centers on calculating the work required to stretch a spring by an additional 4.0 cm after an initial stretch of 2.0 cm, which requires 5.0 J of work. The spring constant (k) was determined to be 25000 N/m. The correct approach to calculate the total work involves using the formula for work on a spring, specifically W = 1/2 k x^2, where x is the total stretch. The area under the force-displacement graph must be considered, as simply adding the initial work to the work for the additional stretch is incorrect due to the non-linear relationship between force and displacement.

PREREQUISITES
  • Understanding of Hooke's Law: F = kx
  • Familiarity with the work-energy principle in the context of springs
  • Knowledge of calculating areas under curves in physics
  • Ability to manipulate algebraic equations involving quadratic terms
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  • Study the derivation of the work formula for springs: W = 1/2 k x^2
  • Learn about the graphical representation of force vs. displacement for springs
  • Explore the implications of non-linear relationships in physics problems
  • Practice problems involving multiple stretches of springs and their corresponding work calculations
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Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for effective teaching methods for work-energy concepts.

Calpalned
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Homework Statement


If it requires 5.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additioanl 4.0 cm?

Homework Equations



Force done on spring: ##F_P = kx ##
Work: ##\frac {1}{2}kx^2 ##

The Attempt at a Solution



I solved for K and got 25000.
Plugging it into the equation for the work on a spring, I get ##\frac {1}{2}(25000)(0.06)^2 ## where 0.06 is the total length stretched. Doing it that way, I get the right answer.

However, why can't I use the original information (5 J) and simply add it to the remaining 4 cm? ##5 + \frac {1}{2}(25000)(0.04)^2 ## Work is area under the force/displacement graph. Therefore, both ways should work, but that's the case.
 
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Calpalned said:

Homework Statement


If it requires 5.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additioanl 4.0 cm?

Homework Equations



Force done on spring: ##F_P = kx ##
Work: ##\frac {1}{2}kx^2 ##

The Attempt at a Solution



I solved for K and got 25000.
Plugging it into the equation for the work on a spring, I get ##\frac {1}{2}(25000)(0.06)^2 ## where 0.06 is the total length stretched. Doing it that way, I get the right answer.

However, why can't I use the original information (5 J) and simply add it to the remaining 4 cm? ##5 + \frac {1}{2}(25000)(0.04)^2 ## Work is area under the force/displacement graph. Therefore, both ways should work, but that's the case.
That's because ##(0.06)^2\ne(0.02)^2+(0.04)^2\ ## .
 
Your force-displacement graph should be a triangle. Mark O, A, B, C at 0, 2, 4, 6 cm on the x axis, with D, E, F on the slope above A, B, C respectively. You want area ACFD. You won't get that by adding OAD to OBE.
 

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