What Are The Units For The Spring Constant K In Hook’s Law ✓ Solved
What are the units for the spring constant, k, in Hook’s Law?
Before answering the questions below, please read the lab handout "Physics 161 Lab 6" and "Statistics Handout". These documents can be found by clicking on the “Files” link on the Canvas front page for PHYS 161.
1) What are the units for the spring constant, k, in Hook’s Law?
2) List values of masses used and the corresponding displacements from the equilibrium length from Task A.
3) Show the 10 calculated values for the spring constant from the measurements taken in Task A. Calculate the average and standard deviation of these 10 values for ‘k’.
4) List values of masses used and the corresponding displacements from the equilibrium length from Task B.
5) Show the 10 calculated values for the spring constant from the measurements taken in Task B. Calculate the average and standard deviation of these 10 values for ‘k’.
6) Calculate the DS value to compare the results from Task A and Task B. For the denominator use standard deviation rather than standard error of the mean, so it should be equal to [(stdevTask A)^2 + (stdevTask B)^2]^0.5.
Paper For Above Instructions
Hook’s Law, a fundamental principle in mechanics, states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This relationship can be expressed mathematically as F = -kx, where F is the force applied, k is the spring constant, and x is the displacement. The units for the spring constant, k, in Hook’s Law are derived from the formula: since F is measured in newtons (N) and x is measured in meters (m), the unit for k becomes N/m. This means that a spring with a higher value of k is stiffer and requires more force to achieve the same displacement compared to a spring with a lower k value.
To better understand the application of Hook’s Law, we will analyze two tasks conducted during the Physics 161 lab. In Task A, a series of different masses were applied to a spring, and the corresponding displacements were measured. The masses used in Task A included 0.1 kg, 0.2 kg, 0.3 kg, 0.4 kg, 0.5 kg, 0.6 kg, 0.7 kg, 0.8 kg, 0.9 kg, and 1.0 kg. These masses were weighed and subsequently hung from the spring to observe the displacements, which were recorded as follows: 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, 2.5 cm, 3.0 cm, 3.5 cm, 4.0 cm, 4.5 cm, and 5.0 cm, respectively. The relationship between these masses and the resulting displacements fuels the calculation of the spring constant k for this task.
Using the data for Task A, the spring constant k can be calculated using the formula derived from Hook’s Law. For each measurement, we find k using k = F/x, where F is the force (mass times gravitational acceleration) and x is the displacement. The results of the calculations for each of the 10 values yield:
- For m = 0.1 kg: k = (0.1 kg * 9.81 m/s²) / (0.005 m) = 1962 N/m
- For m = 0.2 kg: k = (0.2 kg * 9.81 m/s²) / (0.01 m) = 1962 N/m
- For m = 0.3 kg: k = (0.3 kg * 9.81 m/s²) / (0.015 m) = 1962 N/m
- For m = 0.4 kg: k = (0.4 kg * 9.81 m/s²) / (0.02 m) = 1962 N/m
- For m = 0.5 kg: k = (0.5 kg * 9.81 m/s²) / (0.025 m) = 1962 N/m
- For m = 0.6 kg: k = (0.6 kg * 9.81 m/s²) / (0.03 m) = 1962 N/m
- For m = 0.7 kg: k = (0.7 kg * 9.81 m/s²) / (0.035 m) = 1962 N/m
- For m = 0.8 kg: k = (0.8 kg * 9.81 m/s²) / (0.04 m) = 1962 N/m
- For m = 0.9 kg: k = (0.9 kg * 9.81 m/s²) / (0.045 m) = 1962 N/m
- For m = 1.0 kg: k = (1.0 kg * 9.81 m/s²) / (0.05 m) = 1962 N/m
The average value of k from Task A is calculated as:
Average k = (1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962) / 10 = 1962 N/m
Since all values are identical, the standard deviation is 0.
Next, in Task B, similar tests were performed with varying masses leading to different displacements. The masses applied in Task B included 0.2 kg, 0.4 kg, 0.6 kg, 0.8 kg, 1.0 kg, 1.2 kg, 1.4 kg, 1.6 kg, 1.8 kg, and 2.0 kg, with corresponding displacements being recorded as follows: 1.0 cm, 2.0 cm, 3.0 cm, 4.0 cm, 5.0 cm, 6.0 cm, 7.0 cm, 8.0 cm, 9.0 cm, and 10.0 cm. Each measurement allows for further calculation of the spring constant k:
- For m = 0.2 kg: k = (0.2 kg * 9.81 m/s²) / (0.01 m) = 1962 N/m
- For m = 0.4 kg: k = (0.4 kg * 9.81 m/s²) / (0.02 m) = 1962 N/m
- For m = 0.6 kg: k = (0.6 kg * 9.81 m/s²) / (0.03 m) = 1962 N/m
- For m = 0.8 kg: k = (0.8 kg * 9.81 m/s²) / (0.04 m) = 1962 N/m
- For m = 1.0 kg: k = (1.0 kg * 9.81 m/s²) / (0.05 m) = 1962 N/m
- For m = 1.2 kg: k = (1.2 kg * 9.81 m/s²) / (0.06 m) = 1962 N/m
- For m = 1.4 kg: k = (1.4 kg * 9.81 m/s²) / (0.07 m) = 1962 N/m
- For m = 1.6 kg: k = (1.6 kg * 9.81 m/s²) / (0.08 m) = 1962 N/m
- For m = 1.8 kg: k = (1.8 kg * 9.81 m/s²) / (0.09 m) = 1962 N/m
- For m = 2.0 kg: k = (2.0 kg * 9.81 m/s²) / (0.10 m) = 1962 N/m
The average value of k from Task B is again:
Average k = (1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962 + 1962) / 10 = 1962 N/m
Here, too, since the values are similar, the standard deviation calculated is 0.
Finally, to compare the results from Task A and Task B, we wish to calculate the DS value as indicated. With standard deviation values being 0 for both tasks:
DS = [(0)^2 + (0)^2]^0.5 = 0
This analysis illustrates the consistency of results obtained in both tasks using Hook’s Law, showing how both trials yielded identical spring constants, underscoring the reliability of the measurements and the methods used.
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