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vcsharp2003

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## Homework Statement

A spring (k = 500 N/m) supports a 400 g mass in a vertical position. The spring is now stretched 15 cm and maintained at the stretched position. How much work is done in stretching the spring by 15 cm? Is this work dependent on the magnitude of force applied while stretching the spring?

I am not sure if the work done would be just 0.5kx

^{2}= 0.5 x 500 x 0.15[SUP}2[/SUP] = 5.625 J or one should use a more rigorous approach outlined under

## The Attempt at a Solution

.[/B]## Homework Equations

For a spring,

F= kx

PE

_{spring}= 0.5kx

^{2}

Work energy Theorem

ΔKE + Σ ΔPE = Work done

## The Attempt at a Solution

Initial stretch of spring by 400 g mass = (0.400 x 9.8) / 500 = 0.00784 m

The initial stretch is used to determine initial spring PE.

Since the initial and final positions are at rest, so ΔKE = 0.

Σ ΔPE = ΔPE

_{spring}+ ΔPE

_{gravity}= (0.5 x 500 x (0.15 + 0.00784)

^{2}- 0.5 x 500 x (0.00784)

^{2}) + (0 - 0.400 x 9.8 x 0.15 )

= (6.23 - 0.015) + (- 0.588)

= 5.627 J

∴ Work done = ΔKE + Σ ΔPE = 0 + 5.627 = 5.627 J

This work would be independent of the magnitude of force applied since work done depends on only ΔKE and Σ ΔPE, both of which are independent of force applied.