A box rests on top of a flat bed truck. The box has a mass of m = 17 kg. The coe

Question

A box rests on top of a flat bed truck. The box has a mass of m = 17 kg. The coefficient of static friction between the box and truck is ?s = 0.8 and the coefficient of kinetic friction between the box and truck is ?k = 0.63.

1) The truck accelerates from rest to vf= 19 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

A box rests on top of a flat bed truck. The box has a mass of m = 17 kg. The coefficient of static friction between the box and truck is ?s = 0.8 and the coefficient of kinetic friction between the box and truck is ?k = 0.63. 1) The truck accelerates from rest to vf= 19 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=19/16=1.18

2.        F=Ma=(1.18*17)=20.18

3.    Normal force= mg = (9.8)(17)=166.6

Force=UsN

Ff=(.80)(166.6)=133.28

160.72=Ma

A=133.28/17=7.84

4.      Ff=UkN

Ff=(.63)(166.6)

Ff=104.96

104.96/17=6.174

5. Same answer as Number 3

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