1. The problem statement, all variables and given/known data a truck of mass 2000 kg traveling 20 m/s to the right carries a load of mass 500 kg positioned 3 meters behind the driver. The load is kept on the truck by friction, and the coefficient of friction between the load and the bed of the truck is 0.5. What is the shortest distance in which the truck can stop without having the load slide forward enough to hit the driver? (Note that the load can slide forward 3 meters before it hits the driver.) 2. Relevant equations 3. The attempt at a solution to solve this question, take truck moving right = +ve the change in momentum of the box = 0 - (500)(20) = -10000 kgms-1 hence, the Force (Fc) that brings about this change = -10000/t assuming the box doesn't move after the truck stops, hence Fc = Friction force of box on truck <or should it be Forward momentum force acting right = friction force acting left on box?> <because my Fc = -ve, hence it tells me its acting left. so it can't be = friction force acting left, to cancel out each other so that box doesn't move> since friction acts in the direction of the aceleration, and the box is acelerating to the left as the truck slows down. anyway, therefore Friction = (500)(9.8)(0.5) = 2450N thus t = 10000/2450 = 4.1s assume truck decelerates constantly, 0 - 20/ t = a , => a = -20/4.1 = -4.9ms-2 finally, use s=ut+1/2at^2 , => s = 20(4.1)+1/2(-4.9)(4.1^2) = 40.8m however, the question allows the box to move 3m and i realise that the final answer of 37.8m is just 40.8- 3 = 37.8m. so is this a coincidence? or is there some reason behind it? because i still don't know how to factor in the 3m allowance from my understanding, the 3m allowance would come from the net force (Fnet) between (Fc) and Friction, as (Fc) will be larger than Friction now for the box to move 3m. but i don't know how to relate Force to distance moved. i tried using 1/2mv^2 of the box = work done against friction + work done to move box 3m. but it doesn't work. thanks for the help!