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How much (in m) will a spring that has a force constant of 44.5 N/m be stretched by an object with mass 0.490 kg when hung motionless from t
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Answers ( )
Answer:
0.108 m
Explanation:
From Hooke’s law, we have that Force applied to a spring and the corresponding extension is given as:
F = ke
where k = spring constant
We also know that weight (force) is given as:
F = mg
where m = mass and g = acceleration due to gravity
=> ke = mg
=> e = mg / k
Therefore, the extension on the spring is:
e = (0.49 * 9.80) / 44.5 = 0.108 m
It will be stretched by 0.108 m