Determine the spring constant k
1. Determine the spring constant, k, of Spring 1, by using Hooke’s Law. Take three different measurements (since 3 masses) and do three calculations and average your k’s to get a more accurate answer. By applying lots of friction, you will be able to get your mass to hang still. Show a table of your data and your calculations of k.
Spring 1 with three different masses:
50 grams:
Spring stretched: 5m
F = 50gm * 9.8 m/s2 = 490.00 N
K= 490/5m
= 98
100 grams:
Spring stretched: 10m
F = 100gm * 9.8 m/s2 = 980.00 N
K = 980/10m
= 98
250 grams:
Spring stretched: 25m
F = 250gm * 9.8m/s2 = 2,450.00 N
K = 2,450/25 = 55.68
= 98
2. Determine the three unknown masses: Red, Green, and Gold. Use Hooke’s Law to determine the masses. Show all data and calculations.
The displacement seem to always be equal to a tenth of the mass. So, I was able to deduce the data by using the following formula: d = m/10, where d = displacement, m = mass in grams.
The Red block had a displacement of 30cm, which means it 30= m/10, which is equal to 300gm.
The Green block had a displacement of 7cm, which means 7=m/10, which is equal to 70gm.
The Gold block had a displacement of 15cm, which means 15=m/10, which is equal to 150gm.
3. If we look at the equations the m for mass is not a variable in Hooke’s Law. Therefore, it is safe to say that it will not affect the force that a spring exerts on a particular object. It is still a very important to know the mass as that is the way we find F = ma.
4. The displacement is much shorter when the environment is the moon and much longer when the environment is jupiter. For example with the 250gm block the displacement on Earth was 0.25m (Gravity of 9.8m/s2). It was 0.04 on the moon (Gravity of 1.62m/s2), and it was unmeasurable in Jupiter as it shot straight down (Gravity of 22.28m/s2).
5. When the spring is stiffer the numbers for k increased the force exerted as well as the numbers for displacement from equilibrium. It is the opposite
Spring 1 with three different masses:
50 grams:
Spring stretched: 5m
F = 50gm * 9.8 m/s2 = 490.00 N
K= 490/5m
= 98
100 grams:
Spring stretched: 10m
F = 100gm * 9.8 m/s2 = 980.00 N
K = 980/10m
= 98
250 grams:
Spring stretched: 25m
F = 250gm * 9.8m/s2 = 2,450.00 N
K = 2,450/25 = 55.68
= 98
2. Determine the three unknown masses: Red, Green, and Gold. Use Hooke’s Law to determine the masses. Show all data and calculations.
The displacement seem to always be equal to a tenth of the mass. So, I was able to deduce the data by using the following formula: d = m/10, where d = displacement, m = mass in grams.
The Red block had a displacement of 30cm, which means it 30= m/10, which is equal to 300gm.
The Green block had a displacement of 7cm, which means 7=m/10, which is equal to 70gm.
The Gold block had a displacement of 15cm, which means 15=m/10, which is equal to 150gm.
3. If we look at the equations the m for mass is not a variable in Hooke’s Law. Therefore, it is safe to say that it will not affect the force that a spring exerts on a particular object. It is still a very important to know the mass as that is the way we find F = ma.
4. The displacement is much shorter when the environment is the moon and much longer when the environment is jupiter. For example with the 250gm block the displacement on Earth was 0.25m (Gravity of 9.8m/s2). It was 0.04 on the moon (Gravity of 1.62m/s2), and it was unmeasurable in Jupiter as it shot straight down (Gravity of 22.28m/s2).
5. When the spring is stiffer the numbers for k increased the force exerted as well as the numbers for displacement from equilibrium. It is the opposite