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The force exerted by a spring when it's displaced by x from its natural length is given by the equation F(x)=-kx, where k is a positive constant. What is the work done by a spring as it pushes out from x=-x_(2) to x=-x_(1) (where x_(2) gt x_(1))? |
Answer» Solution :We sketch the GRAPH of F(X)=-kx and calculate the area under the graph from `x=-x_(2)` to `x=-x_(1)`.  Here, the REGION is a TRAPEZOID with area `A=(1)/(2)("base"_(1)+"base"_(2))xx"HEIGHT"`, so `W=A=(1)/(2)(kx_(2)+kx_(1))(x_(2)-x_(1))` `=(1)/(2)k(x_(2)+x_(1))(x_(2)-x_(1))` `=(1)/(2)k(x_(2)^(2)-x_(1)^(2))`. |
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