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anonymous

  • 5 years ago

Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If the distance is 23 cm when the weight is 3 kg, what is the distance when the weight is 5 kg?

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  1. anonymous
    • 5 years ago
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    F = -k*x where k is the spring constant, F is the force applied, and x is the distance. In this case, it can be equated to m*g (mass*gravitational acceleration). You know the first F and m, as well as g, so you can find the spring constant. Then you solve for what x would be when m=5.

  2. anonymous
    • 5 years ago
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    so the answer is...

  3. anonymous
    • 5 years ago
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    ...work it out?

  4. anonymous
    • 5 years ago
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    i dont understand it

  5. anonymous
    • 5 years ago
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    Okay, so in this case, the force on the spring is the same as the gravitational force. For the 3 kg weight, you have k * 23 cm = 3 kg * 9.8 m/s^2. Solve for k, the spring's constant. Now, do that same equation again for the 5 kg mass, but instead solving for the length: x.

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