# If the length of a #2 cm# spring increases to #6 cm# when a #3 kg# weight is hanging from it, what is the spring's constant?

So easy. The answers are above.

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The spring constant is 60 N/m.

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To find the spring constant ( k ), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, it can be expressed as:

[ F = kx ]

Where:

- ( F ) is the force exerted by the spring (in Newtons),
- ( k ) is the spring constant (in Newtons per meter), and
- ( x ) is the displacement from the equilibrium position (in meters).

Given that the length of the spring increases from 2 cm to 6 cm when a 3 kg weight is hanging from it, the displacement ( x ) is ( 6 , \text{cm} - 2 , \text{cm} = 4 , \text{cm} = 0.04 , \text{m} ) (since 1 cm = 0.01 m).

We also know that the force exerted by the weight (gravity) is given by ( F = mg ), where ( m = 3 , \text{kg} ) is the mass and ( g = 9.8 , \text{m/s}^2 ) is the acceleration due to gravity.

So, ( F = (3 , \text{kg})(9.8 , \text{m/s}^2) = 29.4 , \text{N} ).

Now, we can use Hooke's Law to find the spring constant:

[ k = \frac{F}{x} = \frac{29.4 , \text{N}}{0.04 , \text{m}} = 735 , \text{N/m} ]

Therefore, the spring constant is ( 735 , \text{N/m} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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