Here's the question you clicked on:
rnick
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.
Every 10 avenues is .1 miles
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 ....
That is exactly how I feel it only says that so I don't even know what to do
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
What are our definitions of Work?
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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?
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?
Thank you, I atually figured it out earlier but seeing this just assured me! <3
it's actually just the change in kinetic energy, we use the conservation of energy and assume that,\[\frac{ 1 }{ 2 }m(v_f-v_i)^2=Friction\]