Question

Rubber tires on dry concrete pavement has a  coefficient of kinetic friction of 0.62, what is the shortest distance (in meters) in which you can stop an automobile with an initial velocity of  46.9 m/s?  

Answer 1

Solution Coefficient of kinetic friction (MR) = 0.62 - Automobile intial speed (u) = 46.9 misec Final speed ) - omisec Kinetic friction force (fe) = MKEN Where FN is normal force is equal to - mg , mis mass of wo automobile and g- 9.8mis? let d is the distance travelled by automobile before coming into rest due to kinetic frictional force. . We = - FID = -(My mald L-ve sign indicates. opposite direction Pob FK and d - This work done should be equal to change. in kinetic Energy i.e. We a I my I mo? TASV = omis ² ZHRd -)-(My mg/d = mcv ² u² 3 2 Migd = uz = d = a) d = 146.995 2x0.62x9.8 - d = 181.01m a 2199.61 12.152 d = 181.01m the automobile before distance coming covered by to rest -