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To determine the force constant of a spiral spring using the given data, you must first find the weights corresponding to each mass and then calculate the force constant using Hooke's law. Here is a step-by-step solution to your problem:
1. List the given data:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Mass } m \, (g) & \text{Weight } W \, (N) & \text{Extension } e \, (cm) \\ \hline 50 & & 6.5 \\ \hline 100 & & 11.0 \\ \hline 150 & & 15.0 \\ \hline 200 & & 20.0 \\ \hline 250 & & 25.0 \\ \hline \end{array} \][/tex]
2. Convert masses to kilograms (since the weight is measured in Newtons and we need to maintain consistency in units):
[tex]\[ m_{\text{kg}} = \left[ \frac{50}{1000}, \frac{100}{1000}, \frac{150}{1000}, \frac{200}{1000}, \frac{250}{1000} \right] = [0.05, 0.1, 0.15, 0.2, 0.25] \, \text{kg} \][/tex]
3. Calculate weights using the formula [tex]\( W = m \times g \)[/tex], where [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]:
[tex]\[ W = \left[ 0.05 \times 9.81, 0.1 \times 9.81, 0.15 \times 9.81, 0.2 \times 9.81, 0.25 \times 9.81 \right] \][/tex]
The calculated weights are:
[tex]\[ W = [0.4905, 0.9810, 1.4715, 1.9620, 2.4525] \, \text{N} \][/tex]
4. Convert extensions from centimeters to meters for consistency:
[tex]\[ e_{\text{m}} = \left[ \frac{6.5}{100}, \frac{11.0}{100}, \frac{15.0}{100}, \frac{20.0}{100}, \frac{25.0}{100} \right] = [0.065, 0.11, 0.15, 0.2, 0.25] \, \text{m} \][/tex]
5. Use Hooke's law to calculate the force constant [tex]\( k \)[/tex], which states:
[tex]\[ F = k \cdot e \Rightarrow k = \frac{F}{e} \][/tex]
Calculate the force constant for each pair of weight and extension:
[tex]\[ k = \left[ \frac{0.4905}{0.065}, \frac{0.9810}{0.11}, \frac{1.4715}{0.15}, \frac{1.9620}{0.2}, \frac{2.4525}{0.25} \right] \][/tex]
Which simplifies to:
[tex]\[ k = [7.546, 8.918, 9.81, 9.81, 9.81] \, \text{N/m} \][/tex]
6. Calculate the average force constant to get an overall value for [tex]\( k \)[/tex]:
[tex]\[ k_{\text{average}} = \frac{7.546 + 8.918 + 9.81 + 9.81 + 9.81}{5} = \frac{45.894}{5} = 9.179 \, \text{N/m} \][/tex]
So based on the experimental data and calculations, the table with the weights filled in would look like this:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Mass } m \, (g) & \text{Weight } W \, (N) & \text{Extension } e \, (cm) \\ \hline 50 & 0.4905 & 6.5 \\ \hline 100 & 0.9810 & 11.0 \\ \hline 150 & 1.4715 & 15.0 \\ \hline 200 & 1.9620 & 20.0 \\ \hline 250 & 2.4525 & 25.0 \\ \hline \end{array} \][/tex]
Finally, the average force constant, [tex]\( k \)[/tex], determined from the experiment is:
[tex]\[ k_{\text{average}} = 9.179 \, \text{N/m} \][/tex]
1. List the given data:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Mass } m \, (g) & \text{Weight } W \, (N) & \text{Extension } e \, (cm) \\ \hline 50 & & 6.5 \\ \hline 100 & & 11.0 \\ \hline 150 & & 15.0 \\ \hline 200 & & 20.0 \\ \hline 250 & & 25.0 \\ \hline \end{array} \][/tex]
2. Convert masses to kilograms (since the weight is measured in Newtons and we need to maintain consistency in units):
[tex]\[ m_{\text{kg}} = \left[ \frac{50}{1000}, \frac{100}{1000}, \frac{150}{1000}, \frac{200}{1000}, \frac{250}{1000} \right] = [0.05, 0.1, 0.15, 0.2, 0.25] \, \text{kg} \][/tex]
3. Calculate weights using the formula [tex]\( W = m \times g \)[/tex], where [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]:
[tex]\[ W = \left[ 0.05 \times 9.81, 0.1 \times 9.81, 0.15 \times 9.81, 0.2 \times 9.81, 0.25 \times 9.81 \right] \][/tex]
The calculated weights are:
[tex]\[ W = [0.4905, 0.9810, 1.4715, 1.9620, 2.4525] \, \text{N} \][/tex]
4. Convert extensions from centimeters to meters for consistency:
[tex]\[ e_{\text{m}} = \left[ \frac{6.5}{100}, \frac{11.0}{100}, \frac{15.0}{100}, \frac{20.0}{100}, \frac{25.0}{100} \right] = [0.065, 0.11, 0.15, 0.2, 0.25] \, \text{m} \][/tex]
5. Use Hooke's law to calculate the force constant [tex]\( k \)[/tex], which states:
[tex]\[ F = k \cdot e \Rightarrow k = \frac{F}{e} \][/tex]
Calculate the force constant for each pair of weight and extension:
[tex]\[ k = \left[ \frac{0.4905}{0.065}, \frac{0.9810}{0.11}, \frac{1.4715}{0.15}, \frac{1.9620}{0.2}, \frac{2.4525}{0.25} \right] \][/tex]
Which simplifies to:
[tex]\[ k = [7.546, 8.918, 9.81, 9.81, 9.81] \, \text{N/m} \][/tex]
6. Calculate the average force constant to get an overall value for [tex]\( k \)[/tex]:
[tex]\[ k_{\text{average}} = \frac{7.546 + 8.918 + 9.81 + 9.81 + 9.81}{5} = \frac{45.894}{5} = 9.179 \, \text{N/m} \][/tex]
So based on the experimental data and calculations, the table with the weights filled in would look like this:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Mass } m \, (g) & \text{Weight } W \, (N) & \text{Extension } e \, (cm) \\ \hline 50 & 0.4905 & 6.5 \\ \hline 100 & 0.9810 & 11.0 \\ \hline 150 & 1.4715 & 15.0 \\ \hline 200 & 1.9620 & 20.0 \\ \hline 250 & 2.4525 & 25.0 \\ \hline \end{array} \][/tex]
Finally, the average force constant, [tex]\( k \)[/tex], determined from the experiment is:
[tex]\[ k_{\text{average}} = 9.179 \, \text{N/m} \][/tex]
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