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Old May 5th 2014, 09:57 AM   #1
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How to determine the energy stored in a spring

Hi there!
First of all, i would like to say special thanks to the admin to allow me being a part of this wonderful forum. It is my first post and i don't know much about the physics, just trying to learn. My question is

I want to determine the energy stored in the spring in ft/pound. The spring is fitted in the tube of an airgun. The data is given below:-
Please see this link to see how the airgun works
http://www.arld1.com/

i) The free length of spring is 10 inches
ii) The spring is compressed in the tube by a spring guide i.e 2.5 inches. so i think the initial compression is 2.5 inches
iii) When i chowk the gun the piston stock (the piston moves backward when chowk the gun) is 2.6 inches

I want to determine how many foot pound energy is stored in the spring when i chowk the gun.
In other words, i want to determine the energy stored in spring where the spring is in the tube, The tube is closed from its front. I pushed the piston in the tube and then i insert the spring i.e 10 inches long. The spring is compressed by a 2.5 inches long guide. The spring guide is attached to the open side of the tube like a bottle cap. A Carrier latch is attached to the spring, i apply some force to the carrier latch, the carrier latch move in the arch, thus the piston inside the tube moves back and compress the spring. The distance travelled by the piston inside the tube is 2.6. What is the formula to determine the energy stored in the spring. I think it relates to hook's law. But i don't know how to plot a graph for this purpose.
Any help in this regard would be highly appreciated.
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Old May 5th 2014, 10:49 AM   #2
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The potential energy stored in a spring is calculated from PE=(1/2)kx^2. Here 'k' is the spring constant, which is a measure of how stiff the spring is in pounds/inch. To determine what the value of 'k' is for your spring you need to do an experiment of measuring the force required to displace it a certain distance. In this case you want the force to compress it from rest to 2.6 inches (your description is a bit hard to follow, but it seems that the compressed spring is 2.6 inches shorter than its uncompressed length, correct?). The value of k is then k = F/x. Combining this with the equation for PE yields:

PE = (1/2)Fx

If the compression force F is given in pounds force and x is inches, this gives PE in units of inch-pounds.

Last edited by ChipB; May 5th 2014 at 12:01 PM.
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Old May 6th 2014, 09:42 AM   #3
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Thanks for the respond Chip b
There are two things

i) The spring is already compressed in the tube upto 2.5 inches from one end. I further compressed it further upto 2.6 inches.

ii) How can i practically determine the applied force. Kindly explain as it is my first practical and i m very unfamiliar with the methods.
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Old May 6th 2014, 11:09 AM   #4
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Tke the spring out of the rifle, clamp it to a table or some other structure that won't move, and use a force gauge to measure the amount of force required to compress it 2.5 inches and then another 2.6 inches. Another technique would be to place a known weight on the spring and measure how far it compresses. For example if you put a 10 pound weight on the spring and it compresses 2.5 inches under the weight then the value of 'k' is 10 pounds/2.5 inch = 4 pounds/inch.

Another idea: if you're interested in finding the energy that is imparted to the pellet when it fires you can aim the rifle straight up, pull the trigger, and see how high the pellet goes. The initial energy from the rifle is equal to mgh, where m is the mass of the pellet, g= acceleration due to gravity, and h= height the pellet reaches.

Last edited by ChipB; May 6th 2014 at 02:22 PM.
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Old May 6th 2014, 11:35 AM   #5
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Thanks a lot chipB, You gave me the strength of knowledge, for which i would be really thankful to you till my last breath.

At this weekend i am going to perform this practical. I'll inform you about the result. Hope i'll succeed what i wanna do. In case, i have to face any trouble, i'll touch u again.

The theory i read about this practical was related to hook's law and plot a graph and that was pretty tough to understand. The method u told me is pretty simple.

Bundle of thanks again
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