## rnick 2 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.

1. rnick

Every 10 avenues is .1 miles

2. amistre64

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

That is exactly how I feel it only says that so I don't even know what to do

4. amistre64

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

What are our definitions of Work?

6. amistre64

|dw:1361801140841:dw|

7. amistre64

43, not 45

8. amistre64

$\frac{1}{2\Delta x}(v_f^2-v_o^2)=a$ you say from 97 to 98 is .1 miles/10; for a distance of .01 miles?

9. amistre64

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

Thank you, I atually figured it out earlier but seeing this just assured me! <3

11. rnick

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

cool :)

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