Work done by a force on a spring

[kolleamm]

Homework Statement

The unstretched length of a spring with 'k' = 250 N/m is 20 cm. A force 'F' is applied to stretch it to a length of 24 cm. How much work was done by 'F'?

Answer : 0.2 Nm

Homework Equations

F = k * delta x
Work = F * delta x

The Attempt at a Solution

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change in spring length is 4cm = 0.04m
F = 250 N/m * (0.04m)

F = 10 N/m

Work = 10 N/m * 0.04m = 0.4

I'm not sure what I'm doing wrong

Thanks in advance

 

[gneill]

Hint: Consider whether or not the force remained constant throughout the stretching of the spring.

 

[kolleamm]

Hint: Consider whether or not the force remained constant throughout the stretching of the spring.

ah your referring to gravity?

 

[gneill]

ah your referring to gravity?

No, say the spring was horizontal on a frictionless surface and anchored at one end. Now you start pulling on the free end to stretch it. Initally the force required is zero since the spring starts out relaxed. By the time you reach 4 cm of stretch, how much force is required? How about at somewhere between those two points?

 

[kolleamm]

No, say the spring was horizontal on a frictionless surface and anchored at one end. Now you start pulling on the free end to stretch it. Initally the force required is zero since the spring starts out relaxed. By the time you reach 4 cm of stretch, how much force is required? How about at somewhere between those two points?

Ah I see, I did the integral of the force from 0 to 0.04 and I got the answer.

int 0 : 0.04 , 250x = 250/2(x^2)
125 (0.04)^2 = 0.2

Although I'm still unsure why the integral of force equals the work done

 

[gneill]

Although I'm still unsure why the integral of force equals the work done

work = force x distance

You integrated the force over the distance. $Work = \int \, F\, dx$

 

[haruspex]

Although I'm still unsure why the integral of force equals the work done

"Work=force x distance" is only valid when the force is constant over the distance. (If it were not, how would you know which value of the force to use?)
If the force is varying we can break the distance into small pieces, ds, and take the force to be approximately constant over each. For each piece, the work done is approximately the product F(s).ds. Adding these up and taking the limit as we make the pieces smaller and smaller yields the integral ∫F.ds.

 

[gneill]

Thanks for the expansion of the explanation @haruspex , I really shouldn't be so lazy with my replies