A box rests on top of a flat bed truck. The box has a mass of m = 15 kg. The coe

Question

A box rests on top of a flat bed truck. The box has a mass of m = 15 kg. The coefficient of static friction between the box and truck is ?s = 0.81 and the coefficient of kinetic friction between the box and truck is ?k = 0.63.

1)The truck accelerates from rest to vf = 15 m/s in t = 14 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?

Explanation / Answer

1. Vf=vi + a(t)
a=15/14=1.071 m/s^2

2. F=Ma=(1.071*15)=16.07 N

3. Normal force= mg = (9.8)(15)=147
Force=UsN
Ff=(.81)(147)=119.07
119.07=Ma
A=119.07/15=7.938 m/s^2

4. Ff=UkN
Ff=(.63)(147)
Ff=92.61
==>92.61/15=6.174 m/s^2

5. Same answer as Number 3

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