Question

A box rests on top of a flat bed truck. The box has a mass of m = 20 kg. The coefficient of static friction between the box and truck is mu_s = 0.86 and the coefficient of kinetic friction between the box and truck mu_s = 0.67. The truck accelerates from rest to v_f = 19 m/s in t = 12 s (which is slow enough that the box will not slide). What is the acceleration of the box? m/s^2 In the previous situation, what is the frictional force the truck exerts on the box? N What is the maximum acceleration the truck can have before the box begins to slide? m/s^2 Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box? m/s^2 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/s^2

Answer 1

1. Vf=vi + a(t)
a=19/12=1.58

2. F=Ma=(20*1.58)=31.6

3. Normal force= mg = (9.8)(20)=196
Force=UsN
Ff=(.86)(196)=168.56
168.56=Ma
A=168.56/20=8.428

4. Ff=UkN
Ff=(.67)(196)
Ff=131.32
131.32/20=6.566

5. Same answer as Number 3