A box rests on top of a flat bed truck. The box has a mass of m = 20.0 kg. The c

Question

A box rests on top of a flat bed truck. The box has a mass of m = 20.0 kg. The coefficient of static friction between the box and truck is ?s = 0.78 and the coefficient of kinetic friction between the box and truck is ?k = 0.63.
1) The truck accelerates from rest to vf = 17.0 m/s in t = 13.0 s (which is slow enough that the box will not slide). What is the acceleration of the box?
m/s2
2) In the previous situation, what is the frictional force the truck exerts on the box?
N
3) What is the maximum acceleration the truck can have before the box begins to slide?
m/s2
4) Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box?
m/s2
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?
m/s2

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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