Finding the force on a spring (Fs)?

In summary, the conversation was about finding the force of a spring by measuring the mass of weights and the extension of the spring. The lab results showed that with increasing weight, the spring stretched further. However, there were some measurement errors leading to discrepancies in the data. The equation Fs=-kx was not helpful, but the lab partner suggested using Fs=mass X gravity to find the force of the spring. Further clarification and help was requested to understand the discrepancies in the data.
  • #1
MuffinMerc
2
0

Homework Statement


I'm trying to find the force of a spring (Fs) from the mass of the weights, and the extension of the spring. The springs length without a weight attached was 5 cm. The lab we did had us put 5 different weights on the same spring, and graph the results.

My results from the lab:
0.05 kg, spring stretched 0.021 meters.
0.1 kg, spring stretched 0.055 meters.
0.2 kg, spring stretched 0.115 meters.
0.3 kg, spring stretched 0.18 meters.
0.4 kg, spring stretched 0.25 meters.

Now, I don't know how to figure out the Fs (force on the spring) on each of these 5 results.

Homework Equations



I'm asking for an equation that can help me. So far, Fs=-kx has proven useless at the moment.

The Attempt at a Solution


My lab partner used Fs=mass X gravity to find force spring. I'm not sure if this is correct, because on the graph (Fs vs extension (x) graph) the line is supposed to be linear and straight, but my graph with the plotted points looks nothing like that when the dots are connected.

Any help would be appreciated.
 
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  • #2
Looks like you have some measurement errors. If it stretched 0.115 m with .2 kg, then with .1 kg, it ought to stretch half as much, or about 0.055 m or so. Did you miss a decimal point?
 
  • #3
PhanthomJay said:
Looks like you have some measurement errors. If it stretched 0.115 m with .2 kg, then with .1 kg, it ought to stretch half as much, or about 0.055 m or so. Did you miss a decimal point?
Oh yes, you're right! Typing error, I fixed that now.

Should make a little bit more sense now.
 

Related to Finding the force on a spring (Fs)?

What is the equation for finding the force on a spring?

The equation for finding the force on a spring is Fs = -kx, where Fs is the force on the spring in Newtons, k is the spring constant in Newtons per meter, and x is the displacement from equilibrium in meters.

How do you calculate the spring constant (k)?

The spring constant can be calculated by dividing the force applied to the spring by the change in length of the spring. This can be expressed as k = F/x, where F is the force applied to the spring and x is the displacement from equilibrium.

What is the unit of measurement for force on a spring?

The unit of measurement for force on a spring is Newtons (N). This is the standard unit for measuring force in the International System of Units (SI).

Can the force on a spring ever be negative?

Yes, the force on a spring can be negative. This occurs when the displacement from equilibrium is in the opposite direction of the force applied to the spring. It is important to note that the negative sign in the equation represents the direction of the force, not the magnitude.

How does the force on a spring change as the displacement increases?

The force on a spring is directly proportional to the displacement from equilibrium according to Hooke's Law. This means that as the displacement increases, the force on the spring also increases. However, once the spring reaches its elastic limit, the relationship between force and displacement becomes nonlinear and the force may no longer increase at the same rate.

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