Work and changing forces - Integration

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urbanyoung
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Hi,

I'm doing some mechanics study and one thing keeps confusing me. In the textbook (Serway/Jewett Physics for scientists and engineers, 7th ed) they introduce i into integral equations. I've put a picture of it below to save me from trying to type the symbols. The equation is for the work done by a spring.

Handwritten-style equation: W_s = integral of F_s dot dr from x_i to x_f of (-kx i-hat) dot (dx i-hat), equals integral from -x_max to 0 of (-kx) dx, equals one half k x_max squared.


I'm hoping someone can explain to me what is being done. I thought it would be dot product, but doesn't that introduce a cosine of the angle? I'm also not sure what the dr is, or well, what the r is.

Thanks.
 
In this case, i is a unit vector (vector with magnitude 1) that points in the +x direction. i dot i = (1)(1)cos(0) = 1. In this case, the book apparently assumes that both the force and the displacement are along the x-axis.
 
The equation for work indeed involves the dot product. The integral used to calculate work does so by analyzing the vector force F and the path of its application. The path is split into tiny elements dr. For the particular application, namely compressing a spring, the force is acting along the x-axis, which is why dr changes to dxi. Similarly, the force compressing the spring is a function of the spring constant k and the amount the spring is compressed, xi. Remember, when a vector is dotted with itself, like i*i is in the integral, the resulting scalar quantity is the square of the length of the original vector, which is 1 for i. Since the angle between a vector and itself is zero, the cosine is equal to 1.
 
The equation is for the work done by a spring.

Since this is a spring its a pretty fair assumption that only one axis need be considered.
You would have to go into 3D for the general case.