A box rests on top of a flat bed truck. The box has a mass of m = 16 kg. The coe

Question

A box rests on top of a flat bed truck. The box has a mass of m = 16 kg. The coefficient of static friction between the box and truck is ?s = 0.82 and the coefficient of kinetic friction between the box and truck is ?k = 0.66.

1)

The truck accelerates from rest to vf = 16 m/s in t = 16 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. Vf=vi + a(t)
a=16/16=1

2. F=Ma=(1*16)=16

3. Normal force= mg = (9.8)(16)=156
Force=UsN
Ff=(.82)(156)=127.72
127.72=Ma
A=127.72/16=7.96

4. Ff=UkN
Ff=(.67)(156)
Ff=102.32
102.32/16=6.436

5. Same answer as Number 3

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