1. The problem statement, all variables and given/known data The coefficients of friction between the flat bed of the truck and crate are (coefficient of static friction) = 0.8 and (coefficient of kinetic friction) = 0.7. The coefficient of kinetic friction between the truck tires and the road surface is 0.9. If the truck stops from an inital speed of 15 m/s with maximum braking (wheels skidding), determine where on the bed the crate finall comes to rest or the velocity relative to the truck with which the crate strikes the wall at the forward edge of the bed. The distance from the crate to the wall at the forward edge of the bed is 3.2m Truck is going to the right. 2. Relevant equations Friction Force = Coefficient of Friction * Normal Force. X=Xinitial+v*t V=Vinitial+a*t X=Xinitial+Vinitial*t+(1/2)*a*t^2 V^2=Vinitial^2+2*a*(X-Xinitial) 3. The attempt at a solution M*Atruck=(.9)MG Atruck=(.9)(9.81)=8.829 m/s^2 going to the left V=Vinitial+A*T 15=0+(8.829)(T) T=1.69895 , time it takes for the truck to stop (coeff. of kinetic friction of the crate)*Normal Force=M*Acrate (.7)M(9.81)=M*(Acrate) Acrate=6.867 acceleration of the crate in relation to the truck = 6.827-8.829 = 1.962 m/s^s going to the right V^2 = Vinitial + 2(A)(X-Xinitial) V^2 = 0 + 2(1.962)(3.2) V = 3.54356 m/s What did I do wrong? I couldn't figure out how to use the coeff. of static friction of the crate and truck bed.