A box rests on top of a flat bed truck. The box has a mass of m = 20 kg. The coe

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 ?s = 0.77 and the coefficient of kinetic friction between the box and truck is ?k = 0.6.

1.The truck accelerates from rest to vf = 16 m/s in t = 15 s (which is slow enough that the box will not slide). What is the acceleration of the box?

2. In the previous situation, what is the frictional force the truck exerts on the box?

3. What is the maximum acceleration the truck can have before the box begins to slide?

4. Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box?

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

given,

mass of the box = 20 kg

coefficient of static friction = 0.77

coefficient of kinetic friction = 0.6

acceleration = velocity / time

acceleration = 16 / 15

acceleration of truck = 1.0666 m / s^2

a) the acceleration of the box = 1.0666

force = mass * acceleration

force = 20 * 1.0666

b) frictional force the truck exerts on the box = 21.332 N

box will slide when force due to acceleration overcomes force due to static friction

so k * mg = ma

k * g = a

0.77 * 9.8 = a

a = 7.546 m/s^2

c) maximum acceleration the truck can have before the box begins to slide = 7.546 m/s^2

now box begins to slide kinetic friction will act on the box

m * acceletation of truck - kinetic friction = ma

m * acceleration of truck - k *mg = ma

acceletation of truck - k * g = a

7.546 - 9.8 * 0.6 = a

a = 1.666 m / s^2

acceleration = 7.546 - 1.666

d) acceleration of the box = 5.88 m / s^2

e) maximum deceleration the truck can have without the box sliding = 7.546 m/s^2

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