A force of 7 pounds compresses a 15 inch spring a total of 4 inches. How much work is done in compressing the spring another 3 inches?
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Originally Posted by CalcGeek31 A force of 7 pounds compresses a 15 inch spring a total of 4 inches. How much work is done in compressing the spring another 3 inches? use Hooke's law, $\displaystyle F = kx$ , ... find the spring constant $\displaystyle k$ $\displaystyle W = \int_4^7 kx \, dx$
Set the 0 to be at 15 the natural length of the spring F = kx 7 = 4k so k = 7/4 F = (7/4)x W = integral [from -4 to -7 of (7/4)xdx]
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