Question 2.) A box rests on top of a flat bed truck. The box has a mass of m = 1
ID: 2131015 • Letter: Q
Question
Question 2.)
A box rests on top of a flat bed truck. The box has a mass of m = 18 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 = 0.61.
1)
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m/s2
A box rests on top of a flat bed truck. The box has a mass of m = 18 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 = 0.61. The truck accelerates from rest to vf = 17 m/s in t = 13 s (which is slow enough that the box will not slide). What is the acceleration of the box? m/s2 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/s2 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 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)
a = v/t = 17/13 = 1.3077 = 1.31 m/s2
2)
F = m a = 18 * 1.3077 = 23.5 N
3)
a_max = us g = 0.8*9.8 = 7.84 m/s2
4)
ma = us m g
==> a = uk g = 0.61 *9.8 = 5.978 m/s2
5)
7.84 m/s2
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