Asked by Amanda
An old car is traveling down a long, straight, dry road at 25.0 m/s when the driver slams on the brakes, locking the wheels. The car comes to a complete stop after sliding 305 m in a straight line. If the car has a mass of 755 kg, what is the coefficient of kinetic friction between the tires and the road?
vi= 25 m/s
vf=0
x=305
I used vf^2= vi^2+2ax
0=(25^2)+2a(305)
0=625+ 610 a
a=-625/610
a= -1.025 m/(s^2)
mu(k)= a/g
mu(k)= -1.025/9.8
mu(k)= -.1045
I put this in as an answer and it said I was wrong. Can anyone tell me what I'm doing wrong? Or possibly a different way to go about the problem? Help is much appreciated!
vi= 25 m/s
vf=0
x=305
I used vf^2= vi^2+2ax
0=(25^2)+2a(305)
0=625+ 610 a
a=-625/610
a= -1.025 m/(s^2)
mu(k)= a/g
mu(k)= -1.025/9.8
mu(k)= -.1045
I put this in as an answer and it said I was wrong. Can anyone tell me what I'm doing wrong? Or possibly a different way to go about the problem? Help is much appreciated!
Answers
Answered by
Iman
You did everything right. You just need to change the sign at the end. so your answer should be +0.1045
Hope this helps :)
Hope this helps :)
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