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What is the force constant of a spring which is stretched 4cm by a mass
of 0.8kg? (g=10N/kg)

Sagot :

Lanuel

The force constant of this stretched spring is equal to 200 N/m.

Given the following data:

  • Extension = 4 centimeters
  • Mass = 0.8 kilograms
  • Acceleration due to gravity = 10 N/kg

Conversion:

Extension = 4 centimeters = [tex]\frac{4}{100} = 0.04\; meters[/tex]

To calculate the force constant of a stretched spring, we would apply Hooke's law:

First of all, we would determine the force acting on the spring.

[tex]Force = mg\\\\Force = 0.8 \times 10[/tex]

Force = 8 Newton

Mathematically, Hooke's law force is given by the formula:

[tex]F = ke[/tex]

Where:

  • F is Hooke's law force.
  • k is spring constant.
  • e is the extension.

Substituting the given values into the above formula, we have:

[tex]8 = k\times 0.04\\\\k = \frac{8}{0.04}[/tex]

Spring constant, k = 200 N/m

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