If there is time, your instructor may have you write up a solution to this probl
ID: 1772072 • Letter: I
Question
If there is time, your instructor may have you write up a solution to this problem:A box of mass 12.0 kg rests on the flat floor of a truck. The coefficient of static friction between the box and floor is 0.19 and the coefficient of kinetic friction is 0.15. The truck stops at a stop sign and then starts to move with an acceleration of 2.20 m/s2. If the box is a distance 1.80 m from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?Explanation / Answer
The box moves only when the net force on it is greater than the static friction.
max static friction acceleration as = µs*g = 0.19 * 9.8m/s^2 = 1.862 m/s^2
The truck exceeds that, so the box moves and we have kinetic friction:
ak = µk*g = 0.15 * 9.8m/s^2 = 1.47 m/s^2
so the net acceleration a = 2.20 m/s^2 - 1.47m/s^2 = 0.73 m/s^2
s = ut + (1/2)a*t^2
s = 0 + (1/2)a*t^2
1.80 m = (1/2)*0.73*t^2
t = 2.2207 s
Truck will travel a distance :
Sd = ut + 1/2at^2
Sd = 0 + (1/2)*2.2*(2.2207)^2
Sd = 5.42 m
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