Force extension graph displacememt to Kenetic energy

[Roberto2391]

I am trying to work out the efficiency for a self built acrylic bow, but I am having some problems working out a predicted height my arrow will reach. For the experiment i fired the arrow into the air and recorded the height it travelled, I have also recorded the weight of the arrow and the force applied to the sting when firing the arrow.

I then proceeded to do this for ten set forces 1N- 10N recording the results, after recording the results i worked out a way in which I could convert the area under a force extension graph into kinetic energy and then into GPE. Once I had the GPE I would divide the total by gravity and then by the mass of the arrow, giving me a predicted height.


To work out the height the arrow would travel, I measured the distance the arrow traveled for each of my forces; I then drew a force extension graph, using my method I worked out that 2N my bow was 262% efficient and for 9N my bow was 0.8% efficient.

A table of my results has been uploaded

I worked out the predicted height by using this chain of thought

Displacement under the graph = Kinetic Energy
Kinetic Energy = Gravitational potential energy
Gravitational potential energy / Mass and gravity

I don’t think I am working out the predicted height correctly but I can't think of another way to do it, all help welcome.

 

[Defennder]

Please, please upload a PDF version of your doc. Many people here would not open a DOC file for fear of virus.

 

[Topher925]

Please, please upload a PDF version of your doc. Many people here would not open a DOC file for fear of virus.

I'm on my work comp so I'm not afraid.

Theres obviously something seriously wrong with your calculations. Since you haven't shown any of your calculations I don't know what that is. What function are you integrating to determine your kinetic energy? And BTW, you don't have to calculate the kinetic energy, if you know the work done by pulling the bow, then you can easily determine its ideal potential energy at its max height.

Work = Force of pull x distance of pull = Potential energy

In other words, your ideal (no losses) height will be:

height = \frac{Fd}{mg}

 

[Roberto2391]

Please, please upload a PDF version of your doc. Many people here would not open a DOC file for fear of virus.

I will remember that for the future.

Work = Force of pull x distance of pull = Potential energy

In other words, your ideal (no losses) height will be:

height = \frac{Fd}{mg}

And i had not though of working it out like that thank you very much