A box rests on top of a flat bed truck. The box has a mass of m = 24.0 kg. The c

Question

A box rests on top of a flat bed truck. The box has a mass of m = 24.0 kg. The coefficient of static friction between the box and truck is ?s = 0.85 and the coefficient of kinetic friction between the box and truck is ?k = 0.67.
1) The truck accelerates from rest to vf = 17.0 m/s in t = 14.0 s (which is slow enough that the box will not slide). What is the acceleration of the box?
2) In the previous situation, what is the frictional force the truck exerts on the box?
3) What is the maximum acceleration the truck can have before the box begins to slide?
4) Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box?
5) With the box still on the truck, the truck attains its maximum velocity. As the truck comes to a stop at the next stop light, what is the magnitude of the maximum deceleration the truck can have without the box sliding?

Explanation / Answer

1) acceleration of block is same as that of truck = 19/12 = 1.58
2) frictional force truck exert on block = mass of block * acceleration of block = 20*1.58 = 31.67 N
3) Force required to start sliding = maximum frictional force = s * m*g = 0.86* 20*9.81 = 168.732

acceleration = 168.732/20 = 8.4366 m/sec2

4) when box begin to slide,

force acting on block = force due to friction = k* m* g

acceleration of block =  k* g = 0.67*9.81 = 6.5727 m/sec2

5) magnitude of maximum deceleration truck can have is same as the magnitude of maximum acceleration truck can have to prevent sliding = 8.4366

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