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 17 cm when the weight is 3 kg, what is the distance when the weight is 9 kg?

Assuming the distance here is the distance from the spring's equilibrium position, you have by Hooke's Law: $F=kx$where we're dealing in magnitudes only, k is the spring constant and x is the displacement from equilibrium. Now, the system is in equilibrium when the force of gravity and that of the spring are equal; that is$mg=kx$So do two different masses,$\frac{m_2g}{m_1g}=\frac{kx_2}{kx_1}\rightarrow \frac{m_2}{m_1}=\frac{x_2}{x_1}\rightarrow x_2=\frac{m_2}{m_1}x_1=\frac{9kg}{3kg}17cm$$=51cm$That is, the spring is displaced from its equilibrium position by 51 centimeters.