What is the work done in this case?

[vcsharp2003]

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 0.5kx² = 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 = 0.5kx²

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.

 

[PetSounds]

With only one significant figure in your original measurements, I don't think the difference in your answers will matter.

 

[TJGilb]

So, the work in this context is force applied dotted with the displacement. You already recognize the force of the spring is $F=kx$. We know that the displacement is 15 cm. Since the force changes as we stretch it further and further, we need to take the integral of the force with respect to x. When you do this, you'll end up with $W=\frac 1 2 kx^2$.