a 1600 kg car moves along a horizontal road at a speed vo = 14.6 m/s. the road is wet so the static friction between coefficient between the tires and the road is only 0.296 (static friction coefficient) & the kinetic friction coefficient is even lower (0.2072). acceleration of gravity = 9.8 m/s^2.
WHAT IS THE HIGHEST POSSIBLE DECELERATION OF THE CAR UNDER SUCH CONDITIONS. answer in m/s^2.
2006-09-30 13:57:59 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
c'mon girls, don't let one of the guys beat you to this one......
2006-09-30 14:06:03 · update #1
As long as the car is moving, the static coeff doesn't matter. Once it is stopped, there is no decel.
therefore:
decel is constant 9.8m/s^2 * .296=2.90m/s^2
2006-10-04 08:21:11 · answer #1 · answered by yupchagee 7 · 16⤊ 0⤋
That would be the deceleration due to friction. The vehicle is moving, so you have to use the coefficient of kinetic friction, 0.2072, which I'll abbreviate u. (It would normally be the Greek letter mu, which an underscore of k for kinetic.) The friction force is mgu, but F = ma and a = F/m, so the deceleration due to friction is mgu/m = gu. Just multiply the acceleration of gravity by the coefficient of kinetic friction.
2006-09-30 21:05:52 · answer #2 · answered by DavidK93 7 · 0⤊ 0⤋
maximum possible .2072g
2006-09-30 21:07:20 · answer #3 · answered by Dr M 5 · 0⤊ 0⤋