Question

ron weighs a bottle with a mass of 0.5 kilograms on a spring scale. the spring stretches by 1 centimeter. he then weighs a second bottle on the same spring scale. what is the mass of the second bottle, in kg, if the spring stretches by 2 cm? use 1 kg = 9.80 n.
1 kg
0.5 kg
9.8 kg
2 kg

Explanation:

Step1: Establish the proportional - relationship

According to Hooke's law, the force exerted on the spring is proportional to the extension of the spring, and the force \(F = mg\). For the first bottle, \(F_1=m_1g\) and the extension \(x_1\). For the second bottle, \(F_2 = m_2g\) and the extension \(x_2\). Since \(\frac{F_1}{x_1}=\frac{F_2}{x_2}\) (because the spring constant \(k\) is the same for the same spring), and \(F = mg\), we have \(\frac{m_1g}{x_1}=\frac{m_2g}{x_2}\), which simplifies to \(\frac{m_1}{x_1}=\frac{m_2}{x_2}\).

Step2: Substitute the known values

We know that \(m_1 = 0.5\space kg\), \(x_1=1\space cm\), and \(x_2 = 2\space cm\). Substituting these values into \(\frac{m_1}{x_1}=\frac{m_2}{x_2}\), we get \(m_2=\frac{m_1x_2}{x_1}\).

Step3: Calculate the mass of the second bottle

\(m_2=\frac{0.5\times2}{1}=1\space kg\).

Answer:

1 kg