Question

Calculate the maximum acceleration (in m/s2) of a car that is heading down a 6.290 slope (one that makes an angle of 6.290 with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved-that is, the tires are not allowed to slip during the acceleration. (Ignore rolling friction. Assume the positive direction is down the slope. Indicate the direction with the signs of your answers.) (a) on dry concrete 3.802 * m/s2 (b) on wet concrete, assuming that us = 0.700 2.341 x m/s2 (c) on ice, assuming that us = 0.100, the same as for shoes on ice -0.5811 * m/s2 +

Answer 1

Given, © = 6029 ay from newton second law Fnet Emas mgsino - f. fret mg sino - Mimgcoso. :2=g (sino - Mi Coso) a = 9.81( sin (65290) - IX COS(6-290) Iol= 8.67 mise (6) on wet concretes. a: 9.81 (sin 6-29- Cool Cos (6-29° ). 191= 5.75 m/s2 c) on ice a= 9081(SinC6-29°) - (0.1) COSC6-29°)). A= 0.103 mis? (a) ada = 8.67 mise (6) AWC = 5°75 mis? co) Qi = 0.103. mis?