A box rests on top of a flat bed truck. The box has a mass of m = 22 kg. The coe

Question

A box rests on top of a flat bed truck. The box has a mass of m = 22 kg. The coefficient of static friction between the box and truck is s = 0.82 and the coefficient of kinetic friction between the box and truck is k = 0.63.

1) What is the maximum acceleration the truck can have before the box begins to slide? My answer is 8.036.

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

3) 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

part 1.   
the maximum force of friction, knowing that s = 0.82 , Ff = Fn × s ,

where the normal force of the truck on the box, Fn = (22 kg) × (9.8 m/s²) = 215.6 N

Ff = (215.6 N) × (0.82)
Ff = 176.792 N
Now find max acceleration
Ff = m × a
(176.792 N) = (22 kg) × a
a = 8.036 m/s²

part 2. F = m*a and Ff = UkN
m*a = ukN
m*a = uk*m*g
a = uk*g
a = .63 * 9.81
a = 6.18 m/s^2 for the acceleration of the box on top of the vehicle.

part 3. Assuming the box is no longer sliding , the maximum negative acceleration will again be
a = -8.036 m/s².

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