Answer to Question #102746 in Classical Mechanics for Aya

Question #102746
a spring is stretched by 5 cm when a mass of 70 gm is hung on it. If total mass of 100 gm is hung on spring and the mass starts vertical oscillation. what is the spring constant k ?
1
Expert's answer
2020-02-11T10:00:47-0500

To solve the problem it is necessary to apply the Hooke Law, which is given by.


"F_{K}=K\\;X"


Where.

  • The deformation is "X=5 cm \\times \\frac{1\\;m}{100\\;cm}=0.05\\;m"
  • The constant is "K"


The restorative force is equal to the mass of the intended placement.


"F_{k}=W"


The weight placed is


"W=m\\;g"


Where.

  • The mass is "m=70 g\\times \\frac{1\\;kg}{1000\\;g}=0.07\\;Kg"
  • The acceleration of gravity is "g=9.8\\;\\frac{m}{s^{2}}"


Evaluating numerically.


"W=0.07\\;Kg\\times 9.8\\frac{m}{s^{2}}\\\\ W=0.686\\;N"


The expression for the elastic constant is


"F_{k}=K\\;X \\\\ K\\;X=F_{K} \\\\ K=\\frac{F_{k}}{X}"


Numerically evaluating to calculate the constant.


"K=\\frac{0.686\\;N}{0.05\\;m} =13.72\\;\\frac{N}{m}"


The spring constant is "\\boxed{K=13.72\\;\\frac{N}{m}}"


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