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The driver of an 1800kg car traveling at 29.0m/s slams on the brakes, locking the wheels on the dry pavement. The coefficient of kinetic friction between rubber and dry concrete is typically 0.600. Find the stopping distance

ive been trying to this problem forever and i know that Its just im missing something small but i cant figure out what.

Ive solved for the KE_initial= 756900J and i know K_final= 0 I know i have to use the constant accel. equation

vf^2=vi^2 + 2as. I just dont know how to go about starting the problem all the unknowns have got me confused.

My attempt went like this. I got the F_normal by 1800kg(9.8) to get 17640N.

17640(.600) gave me 10584N which i used as the frictional force.

a=F/m so then 17640N-10584N/1800 =3.92 i used that for acceleration but i got the wrong answer i dont know where im messing up

ive been trying to this problem forever and i know that Its just im missing something small but i cant figure out what.

Ive solved for the KE_initial= 756900J and i know K_final= 0 I know i have to use the constant accel. equation

vf^2=vi^2 + 2as. I just dont know how to go about starting the problem all the unknowns have got me confused.

My attempt went like this. I got the F_normal by 1800kg(9.8) to get 17640N.

17640(.600) gave me 10584N which i used as the frictional force.

a=F/m so then 17640N-10584N/1800 =3.92 i used that for acceleration but i got the wrong answer i dont know where im messing up

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