jaden places potatoes with a mass of 0.55 kilograms on a grocers scale. the spring is displaced by 1.2…

jaden places potatoes with a mass of 0.55 kilograms on a grocers scale. the spring is displaced by 1.2 centimeters. he then places some onions on the scale that displace the spring by 0.67 cm. what is the mass of the onions (1 kg = 9.8 n)? 5.39 kg 0.31 kg 3.01 kg 0.67 kg

jaden places potatoes with a mass of 0.55 kilograms on a grocers scale. the spring is displaced by 1.2 centimeters. he then places some onions on the scale that displace the spring by 0.67 cm. what is the mass of the onions (1 kg = 9.8 n)? 5.39 kg 0.31 kg 3.01 kg 0.67 kg

Answer

Explanation:

Step1: Calculate the force of the potatoes

The force of the potatoes $F_1$ is given by $F_1 = m_1g$, where $m_1 = 0.55$ kg and $g=9.8$ N/kg. So $F_1=0.55\times9.8 = 5.39$ N.

Step2: Use Hooke's law to find the spring - constant

According to Hooke's law $F = kx$. For the potatoes, $F_1 = kx_1$, where $x_1 = 1.2$ cm $=0.012$ m. We can find the spring - constant $k=\frac{F_1}{x_1}=\frac{5.39}{0.012}\text{ N/m}$.

Step3: Calculate the force of the onions

For the onions, $x_2 = 0.67$ cm $= 0.0067$ m. Using Hooke's law again $F_2=kx_2$. Substituting $k = \frac{5.39}{0.012}$ into $F_2=kx_2$, we get $F_2=\frac{5.39}{0.012}\times0.0067$ N.

Step4: Calculate the mass of the onions

Since $F_2 = m_2g$, then $m_2=\frac{F_2}{g}$. Substituting $F_2=\frac{5.39}{0.012}\times0.0067$ and $g = 9.8$ N/kg, we have $m_2=\frac{\frac{5.39}{0.012}\times0.0067}{9.8}$. $m_2=\frac{5.39\times0.0067}{0.012\times9.8}=\frac{0.036113}{0.1176}\approx0.31$ kg.

Answer:

0.31 kg