I have to use the graph of force against extension for an elastic string obeying Hooke's Law. A force F produces an extension x, then the work done on the string is (1/2)kx^2.
2007-01-05 23:41:46 · 6 answers · asked by jeanpace89 1 in Education & Reference ➔ Homework Help
If the string obeys Hooke's Law then F = kx
Work done = force times distance
In this case the distance is x, and the average force will be (1/2)kx, because when the extension is 0 it is 0 and when it's x it's kx.
So work done = (1/2)kx times x = (1/2)kx^2 QED
2007-01-06 00:16:14 · answer #1 · answered by Mike 2 · 0⤊ 0⤋
If the string obeys Hooke's Law then F = kx
Work done = force times distance
In this case the distance is x, and the average force will be (1/2)kx, because when the extension is 0 it is 0 and when it's x it's kx.
So work done = (1/2)kx times x = (1/2)kx^2 QED
2007-01-06 07:49:38 · answer #2 · answered by Richard 3 · 0⤊ 0⤋
That's the hooke's law part that talks about Potential Energy.
Regular Hooke's law is
Force on Spring = (Spring Constant ) * (Elongation of the spring)
This basically says that the elongation of the spring is directly proportional to the Force acted on the spring. This is because the other fact is a constant, so it might as well be writeen as y=2x or something. It's directly proportional
So if you have a force v. extension graph- it should be a line.
2007-01-06 07:47:13 · answer #3 · answered by Anonymous · 1⤊ 1⤋
Work done = Force x distance moved ,av force =1/2kx
WD =0 .5kx*x= 0.5kx^2 Then (as you mentioned your graph)
Use your graph to find values for extension and force,then use them in the equation.I can not show you without seeing the graph.
2007-01-06 12:23:14 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Keep Trying
2007-01-06 07:50:34 · answer #5 · answered by Jewel 6 · 0⤊ 1⤋
i knew the answer 2 years ago.....
2007-01-06 08:26:50 · answer #6 · answered by Immortal 4 · 0⤊ 1⤋