SOLUTION: A force of 50 pounds stretches a spring 1.5 inches. (Hooke's Law)
(a.) Write the force F as a function of the distance x the spring is stretched.
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Question 91701This question is from textbook Elementary and Intermediate Algebra
: A force of 50 pounds stretches a spring 1.5 inches. (Hooke's Law)
(a.) Write the force F as a function of the distance x the spring is stretched.
This question is from textbook Elementary and Intermediate Algebra
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A force of 50 pounds stretches a spring 1.5 inches. (Hooke's Law)
(a.) Write the force F as a function of the distance x the spring is stretched.
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Amount of force and stretch are directly related:
F = kx
50 = k*1.5
k = 10/0.3
k = 100/3
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EQUATION:
F = (100/3)x
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Cheers,
Stan H.
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