This question is concerned with calculating the work required to stretch a spring.
A certain 'engineering' spring does not obey Hooke's law. Instead, the tension T in the spring when it is extended a distance x beyond its natural length is given by ;
T=sqrt(5x+11) - sqrt(11)
Find the work done in stretching the spring from x = 14 to x = 19
2007-03-02 04:06:12 · 3 answers · asked by ozi 1 in Science & Mathematics ➔ Engineering
Substitute 14 into the equation and calculate T. Then substitute 19 into the equation and calculate T again. The difference between the two values of T represents the change in tension when moving from one extreme to the other.
Subtract 14 from 19. This is the distance actually moved.
Work is force times distance.
That should be all you need to figure it out for yourself.
2007-03-02 04:56:39 · answer #1 · answered by Diogenes 7 · 0⤊ 1⤋
Integrate the function with respect to x and solve the definite integral between the limits of 14 and 19
the integration = [(11+5x)^1.5]/7.5 - (11^.5)*x between limits 14 and 19
82.495-50.767 = 31.728
2007-03-02 15:24:44 · answer #2 · answered by minorchord2000 6 · 1⤊ 0⤋
Work is force being applied over some distance:
W = Fx
...ya he's right...
2007-03-02 12:54:53 · answer #3 · answered by vcas30 3 · 0⤊ 1⤋