question about hooke's law and work

[hardwoodlampshade]

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we know that in general work is equal to force times distance (specifically the force vector component you use in the equation must be parallel to the displacement). but for calculating the work that is done when compressing a spring, we have to take into account the fact that the force exerted by the object compressing the spring is not constant. we know that the force it takes to compress a spring x distance is: F=kx, which implies that as x increases, F increases. so if we integrate the linear graph of force applied on spring versus x, we get work. the equation for the work done on the spring when the spring is compressed a distance x would be (excuse the bad notation):

Work = integral sign F*dx
= integral sign kx*dx
= (kx^2)/2


why can't we just substitute F=kx directly into W=Fx to get:

W=kx^2

 

[Cawolf]

Because the displacement is accounted for in the integration range, not by adding another x variable into the term.

If we say that x = b - a
W = (integral from a to b) of Fdx = F∫dx = Fx

The PE is the integral of the work which is the 0.5kx^2 term.

 

[sazerac]

W=Fx is a simplification for a constant force. You can't use that equation where it isn't meant to be used.

If you prefer, you can think of the F term as the "average F". The force in your spring is going smoothly from zero to kx. What is the average force? 0.5 kx. What is the work done by this average force? 0.5 kx^2.