6. Avery bad spring does not follow Hooke\'s Law at all. If one end if held stat

Question

6. Avery bad spring does not follow Hooke's Law at all. If one end if held stationary, it is found that to stretch or compress this spring by a distance x along the x-axis requires a force parallel to the x-axis to be: F, 100x 700x2 12,000x3 where the coefficients are such that the force reported will be in the proper units of Newtons. A. How much work must be done on the spring by some external force to STRETCH the spring 0.050 m from its unstretched length? B. How much work must be done on the spring by some external force to COMPRESS the spring 0.050 m from its unstretched length?

Explanation / Answer

Fx = 100x - 700x^2 + 12000x^3

when elongation in spring x then work required to stretch further dx distance.

dW = Fx.dx = (100x - 700x^2 + 12000x^3)dx

integrating,

W = 50x^2 - 233.33x^3 + 3000x^4

limit is from 0 to x.

A) work doen to stretch by 0.50 m
henc x = 0.50 m

W = 50(0.5^2) - 233.33(0.5^3) + 3000(0.5^4) = 170.83 J

b) work done to compress by 0.5 m

x = - 0.5 m

W = 50 (-0.5)^2 - 233.33 (-0.5^3) + 3000(-0.5)^4

W = 229.2 J

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