How Does Spring Constant Affect Work Done on a Stretched Spring?

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SUMMARY

The discussion focuses on calculating the work done by a spring with a spring constant of 15 N/cm when stretched. The work done on the cage when the spring is stretched from its relaxed length by 7.6 mm (converted to 0.76 cm) is determined using the formula W = 0.5KX². For the first stretch of 7.6 mm, the work done is 0.5 * 15 N/cm * (0.76 cm)². Additionally, the work done for the subsequent stretch of another 7.6 mm can be calculated using the same formula, considering the total stretch of 1.52 cm.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the work-energy principle
  • Basic knowledge of unit conversions (mm to cm)
  • Proficiency in using the formula W = 0.5KX²
NEXT STEPS
  • Calculate work done for multiple stretches of a spring using W = 0.5KX²
  • Explore the implications of different spring constants on work done
  • Learn about energy conservation in elastic systems
  • Investigate real-world applications of Hooke's Law in engineering
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Students studying physics, particularly those focusing on mechanics and elasticity, as well as educators looking for practical examples of spring dynamics.

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Homework Statement



A spring with a spring constant f 15 N/cm has a cage attached to one end.
a) How much work does the spring force do on the cage when the spring is stretched from its relaxed length by 7.6 mm?
b) How much additional work is done by the spring force when the spring is stretched by additional 7.6 mm?


I converted the 7.6 mm to .76 cm but I aint sure on the question for this. Is it W - 1/s x 15 x .76 ?
 
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use W=0.5KX^2 to find the work done..
 

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