How do I calculate the force of an object attached to a spring?

In summary, the experiment involved attaching weights to a spring and measuring the stretched spring, as well as the highest and lowest points when the spring was released. The goal is to calculate the force in Newtons for each trial, but the equation F=ma may not apply in this case as the acceleration is not relevant. The mass of the spring may also need to be taken into account.
  • #1
seallen
7
0

Homework Statement


I gathered the following data in an experiment:
Trial Mass (kg) Stretched Spring (m) Force (N) Highest Point (m) Lowest Point (m)
1 .05 .05 ? 0 .09
2 .1 .095 ? .015 .185
3 .25 .245 ? .065 .45

In the experiment i attached different weights to a spring and measured the spring first as it was stretched and I then released the spring to see the highest and lowest points. I now am supposed to calculate the force in Newtons for each trial.

Homework Equations


F=ma I know is the equation for force but how is it applicable to this data? I understand m would be equal to the mass of the object in kg, but how would I find a for the equation?

The Attempt at a Solution


I originally thought that I would use the stretched spring as a, but now I am not so sure, would I use the change in the highest and lowest point instead? however, acceleration really doesn't apply here so I am confused.
 
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  • #2
seallen said:
In the experiment i attached different weights to a spring and measured the spring first as it was stretched and I then released the spring to see the highest and lowest points.
Please clarify that. When you released the spring, there were no weights left on it? And you measured the highest point it reached when it bounced back (and it continued to oscillate?).
If so, that's not good. The mass of the spring has become important.
What force are you trying to calculate? The force when the spring was loaded was the weight acting under gravity. Are you trying to estimate g? You'll need the spring modulus in that case.
 

1. What is Hooke's Law and how does it relate to calculating the force of an object attached to a spring?

Hooke's Law states that the force exerted by a spring is directly proportional to the distance the spring is stretched or compressed from its equilibrium position. This means that as the displacement of the spring increases, the force it exerts also increases. Therefore, to calculate the force of an object attached to a spring, we can use Hooke's Law to determine the relationship between the force and displacement.

2. How do I determine the spring constant in order to calculate the force?

The spring constant is a measure of the stiffness of a spring and is typically represented by the letter "k". It can be determined by conducting an experiment where the displacement of the spring is measured for different applied forces. The slope of the resulting graph (force vs. displacement) will equal the spring constant.

3. Can I use Hooke's Law to calculate the force of an object attached to a non-ideal spring?

Hooke's Law is only applicable to ideal springs, which means they follow the linear relationship between force and displacement. Non-ideal springs, such as ones with varying stiffness or ones that do not return to their original shape after being stretched, may not follow this relationship and therefore Hooke's Law cannot be used to calculate the force.

4. Is there a specific unit for the force calculated using Hooke's Law?

The unit for force is typically measured in Newtons (N). However, the unit for the spring constant is measured in Newtons per meter (N/m). Therefore, when using Hooke's Law to calculate the force of an object attached to a spring, the resulting unit will be N/m multiplied by the displacement of the spring.

5. Can Hooke's Law be used for both compression and tension springs?

Yes, Hooke's Law can be used for both compression and tension springs. The only difference is the direction in which the force is exerted. For compression springs, the force is exerted towards the center of the spring, while for tension springs, the force is exerted away from the center of the spring. However, the relationship between force and displacement is still the same for both types of springs.

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