6 - homework 05 JACOBS, AMBER Due: Feb 26 2008, 1:00 am 1...

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Unformatted text preview: homework 05 JACOBS, AMBER Due: Feb 26 2008, 1:00 am 1 Question 1, chap 8, sect 6. part 1 of 1 10 points The following graph represents a hypothet- ical potential energy curve for a particle of mass m . r U ( r ) O 3 U 2 U U r 2 r If the particle is released from rest at posi- tion r , its speed bardbl vectorv bardbl at position 2 r is most nearly 1. bardbl vectorv bardbl = radicalbigg U m . 2. bardbl vectorv bardbl = radicalbigg U 6 m . 3. bardbl vectorv bardbl = radicalbigg 8 U m . 4. bardbl vectorv bardbl = radicalbigg 2 U m . 5. bardbl vectorv bardbl = radicalbigg U 2 m . 6. bardbl vectorv bardbl = radicalbigg U 8 m . 7. bardbl vectorv bardbl = radicalbigg U 4 m . 8. bardbl vectorv bardbl = radicalbigg 4 U m . correct 9. bardbl vectorv bardbl = radicalbigg 6 U m . Explanation: The total energy of the particle is con- served. So the change of the potential en- ergy is converted into the kinetic energy of the particle, which gives 1 2 mv 2 = 3 U U Therefore v = radicalbigg 4 U m . Question 2, chap 8, sect 6. part 1 of 2 10 points Given: The positive x direction is to the right. In part (a) of the figure, an air track cart attached to a spring rests on the track at the position x eq and the spring is relaxed. (a) x st x eq (b) In (b), the cart is pulled to the position x st and released. It then oscillates about x eq . Which graph correctly represents the po- tential energy of the spring as a function of the position of the cart? 1. x st x eq 2. x st x eq 3. x st x eq 4. 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 springs 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 springs 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 , vector F eq < 0 . 2. vector F st > 0 , vector F eq > 0 . 3. vector F st < 0 , vector F eq > 0 . 4. vector F st > 0 , vector F eq = 0 . correct 5....
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This note was uploaded on 04/28/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas at Austin.

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6 - homework 05 JACOBS, AMBER Due: Feb 26 2008, 1:00 am 1...

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