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rnick

  • 3 years ago

You are standing at a red light on SW 8th Street at 97th Avenue. The light turns green, you start speeding up, and the moment you reach 98th Avenue you reach a speed of 43 mph. What is the work done on your 1,277-kg car by the frictional force excerted from the road on the tires.

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  1. rnick
    • 3 years ago
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    Every 10 avenues is .1 miles

  2. amistre64
    • 3 years ago
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    im not sure that i understand the question as stated. With the information given, we could possibly determine the acceleration of the car and Force = mass x accel. Im thinking work is equal to Force x distance. But there is not coeffs of friction given or total energy of the system/ loss with which to compare with. So im pretty much stymied ....

  3. rnick
    • 3 years ago
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    That is exactly how I feel it only says that so I don't even know what to do

  4. amistre64
    • 3 years ago
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    since a tire is round, would we have to play with the circle versions? i dont see that being a viable option since we dint know the radius .... I think its just asking us the work done to move the car

  5. amistre64
    • 3 years ago
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    What are our definitions of Work?

  6. amistre64
    • 3 years ago
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    |dw:1361801140841:dw|

  7. amistre64
    • 3 years ago
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    43, not 45

  8. amistre64
    • 3 years ago
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    \[\frac{1}{2\Delta x}(v_f^2-v_o^2)=a\] you say from 97 to 98 is .1 miles/10; for a distance of .01 miles?

  9. amistre64
    • 3 years ago
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    the track at the high school is .25 miles oval so the straight run is about .1 miles. and they have 10 intersection in that length?

  10. rnick
    • 3 years ago
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    Thank you, I atually figured it out earlier but seeing this just assured me! <3

  11. rnick
    • 3 years ago
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    it's actually just the change in kinetic energy, we use the conservation of energy and assume that,\[\frac{ 1 }{ 2 }m(v_f-v_i)^2=Friction\]

  12. amistre64
    • 3 years ago
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    cool :)

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