SOLUTION: Hooke's Law says that the force exerted by the spring in a spring scale varies directly with the distance that the spring is stretched. If a 39 pound mass suspended on a spring sc

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Question 1111502: Hooke's Law says that the force exerted by the spring in a spring scale varies directly with the distance that the spring is stretched. If a 39 pound mass suspended on a spring scale stretches the spring 8 inches, how far will a 50 pound mass stretch the spring?
Found 2 solutions by ikleyn, amalm06:
Answer by ikleyn(52794) About Me  (Show Source):
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Direct proportion

39%2F8 = 50%2Fx.


Find x = %288%2A50%29%2F39.


Use your calculator.

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Answer by amalm06(224) About Me  (Show Source):
You can put this solution on YOUR website!
Hooke's law can be stated as follows: F=kx, where F is the force exerted on the spring, k is the spring constant, and x is the distance the spring is stretched or compressed.

In this problem, the weight of the object exerts a force F that causes the spring to stretch.

The spring constant, k, can be found as follows: mg=kx

39=(k)(8), k=39/8

For the 50 pound mass,

50=(39/8)x

x=10.25 in (Answer)