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# 6 - homework 05 JACOBS AMBER Due 1:00 am Therefore Question...

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homework 05 – JACOBS, AMBER – Due: Feb 26 2008, 1:00 am 2 x st x eq correct 5. x st x eq 6. x st x eq Explanation: The car starts at x st with no kinetic energy, and so the spring’s potential energy is a max- imum. Once released, the cart accelerates to the right and its kinetic energy increases as the potential energy of the spring is converted into kinetic energy of the cart. As the cart passes the equilibrium position, its kinetic en- ergy is a maximum and so the spring’s poten- tial energy is a minimum. Once to the right of x equ , the car starts to compress the spring and it slows down as its kinetic energy is converted back to potential energy of the recompressed spring. At the rightmost point it reaches, the cart reverses its direction of travel. At that in- stant, it has no kinetic energy and the spring again has maximum potential energy. Question 3, chap 8, sect 6. part 2 of 2 10 points The forces vector F st and vector F eq acting at x st and x eq are related as follows 1. vector F st > 0 ˆ ı , vector F eq < 0 ˆ ı . 2. vector F st > 0 ˆ ı , vector F eq > 0 ˆ ı . 3. vector F st < 0 ˆ ı , vector F eq > 0 ˆ ı . 4. vector F st > 0 ˆ ı , vector F eq = 0 ˆ ı . correct 5. vector F st < 0 ˆ ı , vector F eq < 0 ˆ ı . 6. vector F st = 0 ˆ ı , vector F eq > 0 ˆ ı . 7. vector F st = 0 ˆ ı , vector F eq < 0 ˆ ı . 8. vector F st < 0 ˆ ı , vector F eq = 0 ˆ ı .
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6 - homework 05 JACOBS AMBER Due 1:00 am Therefore Question...

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