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EstoMax
11-13-2003, 10:13 PM
ok.. i have a takehome physics test and the last problem is.. Design your own problem.. so i wanted to make mine kinda interesting and hard.. something we havent done in class anyway.
so this is what i did..

A racecar on a cosmic raceway[no air resistance, we dont know how to calculate that in yet] is slowing down and has 2 seconds to go from 75 to 25 meters per second. The coefficient of friction between the car and the track is 0.980. The car weighs 500kg. The gravity on the planet is same as on earth (9.81). The slope of the straight part of the track is 8 degrees from the horizontal.

i posted the pic too :

this is what you have to do:
Find Force Applied on the X plane
Find Force Applied on the Y plane
Find Force Kinetic

ill post my calculations in next reply

edit: noticed that i had only 2 significant figures which would skew the answers too much. so i changed the coefficient of friction to 0.980 so it would be 3, it doesnt change the answers though.

EstoMax
11-13-2003, 10:22 PM
Ok,
plz tell me if i did something wrong i wanna get good grade on this test :) if u have extra time and ur a physics pro i can give u the rest of the problems on my take home test to look at as well, but im fairly confidential that i did it right.. unless i did somethign terribly wrong on this one...

Ok,
to find the acceleration, you take -50m/s and divide by 2 seconds, to find that you accelerate -25m/s
then to find the Force, use Newtons second law (F=ma). that is -25m/s x 500kg = -12500 N
Then to find force applied in the X direction, you do Force x Cos angle.
-12500 x Cos 8 = -12378 N => =>(3 Significant figures) 12400 N
Then to find force applied in the Y direction, you do Force x Sin angle.
-12500 x Sin 8 = -1740 N

Then to find Force Kinetic, you will multiply coefficient of friction x Force Normal.
Force Normal is weight x gravity (500x9.81 = 4905)
then 0.98 x 4905 = 4807 4800 N for Force Kinetic

is this correct?

tnx a bunch!

max

kaput
11-13-2003, 10:47 PM
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kaput
11-13-2003, 10:53 PM
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EstoMax
11-13-2003, 11:02 PM
well i used high coefficient of friction cause the higher the friction the better the traction right ? or they could be racers that have no wheels :) just slide on ground with rockets haha

now that im thinking i should have used a low coefficient of friction.. DOH!

max

kaput
11-13-2003, 11:10 PM
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EstoMax
11-13-2003, 11:11 PM
yea thanks man.. i think thats all good. thanks for spotting that mistake ill go fix it
so after i find the new force normal i do what i did b4 right?
since the new normal force will be 4857N instead of the 4905.
to find Force Kinetic u multiply the coefficient of friction with force normal tho.. cause in physics book it says coefficient of friction (MU-k) = Fk / Fn so if u rearrange it u get Fk = Mu-k x Fn

max

EstoMax
11-13-2003, 11:17 PM
btw.. ima chance the coefficient of friction.. to like 0.050 :) cause i understood it wrong..

max

kaput
11-13-2003, 11:22 PM
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EstoMax
11-13-2003, 11:25 PM
ok thanks.. so the right answer to Force Kinetic would be ..243 N, using the coefficient 0.050 (which should be more reasonable :))
0.050 x 4857<--( 500x9.81xCos8 )

thanks a lot man!

max