SOLUTION: The distance D that a spring is stretched by hanging object varies directly as the weight W of the object. If a 6-kg object streches a spring 19cm, find the distance the spring is

Question 587885: The distance D that a spring is stretched by hanging object varies directly as the weight W of the object. If a 6-kg object streches a spring 19cm, find the distance the spring is streched when the weight is 19kg.
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
): The distance D that a spring is stretched by hanging object varies directly as the weight W of the object.
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D = kW
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Solve for "k" using "a 6-kg object streches a spring 19cm"
19 = k*6
k = 19/6
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Equation:
D = (19/6)W
Find the distance the spring is stretched when the weight is 19kg
D = (19/6)*19
D = 60.2 cm
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Cheers,
Stan H.
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Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
The distance D that a spring is stretched by hanging object varies directly as the weight W of the object. If a 6-kg object streches a spring 19cm, find the distance the spring is streched when the weight is 19kg.
**
D=(19/6)*19==≈60.17 cm
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