When a6 gram mass is attached to a spring hanging vertically, the spring is stretched to a total length of 24 centimeters. When a 10 gram mass is attached to the spring (and only the 10 gram mass), the spring is stretched to a total length of 36 centimeters. What is the total length of the spring when a 12 grammass is attached to it? Elimination Tool Select one answer A 18cm 15 cm B 36 cm D 30 cm E 42cm

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When a6 gram mass is attached to a spring hanging vertically, the spring is stretched to a total length of 24 centimeters. When a 10 gram mass is attached to the spring (and only the 10 gram mass), the spring is stretched to a total length of 36 centimeters. What is the total length of the spring when a 12 grammass is attached to it? Elimination Tool Select one answer A 18cm 15 cm B 36 cm D 30 cm E 42cm

Answer 1

### E. \( 42 \, \mathrm{cm} \)

Answer 2

## Step1: Identify the key relationships ### The problem involves forces and Hooke's Law, which states that the force exerted by a spring is proportional to its extension. Hooke's Law is given by \( F = k \cdot x \), where \( k \) is the spring constant and \( x \) is the extension of the spring. ## Step2: Calculate forces for given masses ### Convert masses to forces using \( F = m \cdot g \) (where \( g \approx 9.8 \, \mathrm{m/s^2} \)). For a 6 gram mass (\( 0.006 \, \mathrm{kg} \)): \[ F_1 = 0.006 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 0.0588 \, \mathrm{N} \] For a 10 gram mass (\( 0.01 \, \mathrm{kg} \)): \[ F_2 = 0.01 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 0.098 \, \mathrm{N} \] ## Step3: Determine the extension for each mass ### Subtract the original length of the spring (assumed constant) from the given total lengths. Extension for 6 gram mass: \[ x_1 = 24 \, \mathrm{cm} \] Extension for 10 gram mass: \[ x_2 = 36 \, \mathrm{cm} \] ## Step4: Calculate the spring constant (k) ### Use Hooke's Law \( F = k \cdot x \). For the 6 gram mass: \[ k = \frac{F_1}{x_1} = \frac{0.0588 \, \mathrm{N}}{24 \, \mathrm{cm}} = \frac{0.0588 \, \mathrm{N}}{0.24 \, \mathrm{m}} = 0.245 \, \mathrm{N/m} \] Ensure consistency and check with the 10-gram mass: \[ k = \frac{F_2}{x_2} = \frac{0.098 \, \mathrm{N}}{36 \, \mathrm{cm}} = \frac{0.098 \, \mathrm{N}}{0.36 \, \mathrm{m}} = 0.272 \, \mathrm{N/m} \] ## Step5: Calculate the spring constant and new extension accurately ### As the calculations vary slightly, average \( k \): \[ k = \frac{0.245 + 0.272}{2} \approx 0.259 \, \mathrm{N/m} \] ## Step6: Calculate the new extension for a 12 gram mass \[ F = 0.012 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 0.1176 \, \mathrm{N} \] \[ x = \frac{F}{k} = \frac{0.1176 \, \mathrm{N}}{0.259 \, \mathrm{N/m}} \approx 0.454 \, \mathrm{m} \] ## Step7: Calculate the total length of the spring \[ \text{Total length} = x = 0.454 \, \mathrm{m} = 45.4 \, \mathrm{cm} \]

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