To find the work done in extending a spring

AI Thread Summary
The discussion revolves around calculating the work done in extending a spring using the correct formulas. The original poster, Mitch, initially calculates the work done as 1/5 but later realizes the correct approach involves using the formula W = 1/2 k x^2. Another participant suggests using the relationship F = kx and integrating to derive the work done. Mitch acknowledges this simpler method and thanks the contributors for their help. The conversation highlights the importance of using the appropriate formulas for accurate calculations in physics.
gnits
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Homework Statement
To find the work done in extending a spring
Relevant Equations
W.D. = F * d
Hi,

Could I please ask where I am going wrong with this very simple question:

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Here's my answer (units implied):

A force of 20 extends the spring by 1/100 and so the Work Done in performing this extension is 20 * 1/100 = 1/ 5

Now, the work done in extending a spring is given by the formula W.D. = (Y * x^2) / (2a) where 'Y' is the Modulus Of Elasticity of the spring, 'x' the Extension and 'a' the Natural Length of the spirng.

So we have 1/5 = (Y * 1/10000) / (2a) and so from this we have:

Y/(2a) = 2000

Now, using the same formula, the work done in extending the spring by b is:

(Y * b^2) / (2a) and substituting for Y/(2a) from the previous formula gives:

W.D. = 2000 * b^2 = 2x10^3 * b^2 which is twice the book answer of 1x10^3 * b^2

Thanks for any help,
Mitch.
 
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Let me see
F/A =Y x/l ==>x is elongation
So F= kx
Now work
dW = F .dx
dW = kx .dx
Integrating both we get,
W=1/2(kx^2)Hope that helps.
 
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gnits said:
Homework Statement:: To find the work done in extending a spring
Homework Equations:: W.D. = F * d

Hi,

Could I please ask where I am going wrong with this very simple question:

View attachment 253923

Here's my answer (units implied):

A force of 20 extends the spring by 1/100 and so the Work Done in performing this extension is 20 * 1/100 = 1/ 5

Now, the work done in extending a spring is given by the formula W.D. = (Y * x^2) / (2a) where 'Y' is the Modulus Of Elasticity of the spring, 'x' the Extension and 'a' the Natural Length of the spirng.

So we have 1/5 = (Y * 1/10000) / (2a) and so from this we have:

Y/(2a) = 2000

Now, using the same formula, the work done in extending the spring by b is:

(Y * b^2) / (2a) and substituting for Y/(2a) from the previous formula gives:

W.D. = 2000 * b^2 = 2x10^3 * b^2 which is twice the book answer of 1x10^3 * b^2

Thanks for any help,
Mitch.

Why not simply use ##F = kx## and ##W = \frac 1 2 k x^2##?
 
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Thanks all for your help, I see it now. Thanks, Mitch.
 
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