calculation of force through calculus

itgirl268 · 01-03-2014 12:27PM #1

Work of 6 Joules is done in stretching a spring from its natural length to 21 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (21 cm)?
iv tried it several times but answer is incorrect..? think im calculating it incorrectly

TheBody · 01-03-2014 12:52PM

itgirl268 wrote: »

Work of 6 Joules is done in stretching a spring from its natural length to 21 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (21 cm)?
iv tried it several times but answer is incorrect..? think im calculating it incorrectly

It's been a LONG time since I've done any physics, but from my dodgy memory:

Work=Force x Distance

Rearranging this tells us that:

[latex]Force=\frac{Work}{Distance} [/latex]

From that, can you figure out the answer?

itgirl268 · 01-03-2014 05:04PM

so i done it 6J/.21 = 28.57 but this is incorrect am i leaving out something??

TheBody · 01-03-2014 07:05PM

I'm afraid, I have to bow out to superior minds on this.

I'll stick to doing maths!!

bnt · 01-03-2014 07:13PM

Another way of looking at it: the Work = Force x Distance formula assumes the force is constant but, in the case of a spring, it's not. It's proportional to the extension/compression of the spring. (F = k x, where k is the spring constant).

So, you're not just multiplying force by distance, you're integrating the force function over distance:

W = ∫ F dx
= ∫ k x dx
= ½ k x²

From that you can calculate k, then the force.

k = 272,1, F = 57.14

itgirl268 · 02-03-2014 11:36AM

would a similar idea be applied to this as above????

A chain 67 meters long whose mass is 27 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 7 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared.

dlouth15 · 06-03-2014 04:41PM

itgirl268 wrote: »

would a similar idea be applied to this as above????

A chain 67 meters long whose mass is 27 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 7 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared.

A similar approach is required here because you have a force that varies as the chain is being drawn in. You need to express that force as a function of the length of chain still hanging. You then need to derive an expression for the small bit of work required to haul the chain in by an infinitesimal length. The total work will be the sum of all these small bits which will involve an integral.

calculation of force through calculus

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