# If the length of a #20 cm# spring increases to #56 cm# when a #15 kg# weight is hanging from it, what is the spring's constant?

K= 408 N/m

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The spring constant ( k ) can be calculated using Hooke's Law: ( F = kx ), where ( F ) is the force applied, ( k ) is the spring constant, and ( x ) is the displacement from the equilibrium position. Rearranging the equation to solve for ( k ), we get: ( k = \frac{F}{x} ). Given that the weight is ( W = mg ), where ( m ) is the mass and ( g ) is the acceleration due to gravity, ( F = mg ). Substituting the values, ( F = (15 , \text{kg})(9.8 , \text{m/s}^2) = 147 , \text{N} ). The displacement ( x ) is the change in length of the spring, so ( x = 56 , \text{cm} - 20 , \text{cm} = 36 , \text{cm} = 0.36 , \text{m} ). Therefore, ( k = \frac{147 , \text{N}}{0.36 , \text{m}} = 408.33 , \text{N/m} ). So, the spring constant is approximately ( 408.33 , \text{N/m} ).

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