gaborfk
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A particle is attached between two identical springs on a horizontal frictionless table. Both springs have spring constant k and are initially unstressed.
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a) If the particle is pulled a distance x along a direction perpendicular to the initial configuration of the spring, show that the force exerted by the springs on the particle F=-2kx(1-L/(sqrt(x^2+L^2)))i (where L is the vertical distance of each spring at rest)
b) Determine the amount of work done by this force in moving the particle from x=A to x=0
On part a)
I got that the 2 is there since there are two springs. The L/(sqrt(x^2+L^2) is the sin of the angle between the a spring and the vertical. Where does the (1-) comes from? Also how do I tie the equation together, after figuring out all the parts?
On part b)
Do I have to integrate f(x) as x=A goes to x=0?
Thank you
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-------A---
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a) If the particle is pulled a distance x along a direction perpendicular to the initial configuration of the spring, show that the force exerted by the springs on the particle F=-2kx(1-L/(sqrt(x^2+L^2)))i (where L is the vertical distance of each spring at rest)
b) Determine the amount of work done by this force in moving the particle from x=A to x=0
On part a)
I got that the 2 is there since there are two springs. The L/(sqrt(x^2+L^2) is the sin of the angle between the a spring and the vertical. Where does the (1-) comes from? Also how do I tie the equation together, after figuring out all the parts?
On part b)
Do I have to integrate f(x) as x=A goes to x=0?
Thank you