A box rests on top of a flat bed truck. The box has a mass of m = 23.0 kg. The c
ID: 2202343 • Letter: A
Question
A box rests on top of a flat bed truck. The box has a mass of m = 23.0 kg. The coefficient of static friction between the box and truck is ?s = 0.85 and the coefficient of kinetic friction between the box and truck is ?k = 0.67. 1)Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box? 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?Explanation / Answer
A box rests on top of a flat bed truck. The box has a mass of m = 20.0 kg. The coefficient of static friction between the box and truck is ?s = 0.86 and the coefficient of kinetic friction between the box and truck is ?k = 0.67. 1) The truck accelerates from rest to vf = 19.0 m/s in t = 12.0 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? 1) As box doesn't slide, it has the same acceleration as the truck, so Vf = Vi + a×t (19 m/s) = (0 m/s) + a × (12 s) a = (19 m/s) / (12 s) = 1.1 m/s² < - - - - - answer 1 2) The frictional force keeps the box from sliding on the truck. It must be the same as the force exerted on the mass of the box by the acceleration of the truck, so F = m × a F = (20 kg) × (1.1 m/s²) F = 24.2 N < - - - - - answer 2 3) First find the normal force of the truck on the box Fn = (20 kg) × (9.8 m/s²) = 215.6 N Then find the maximum force of friction, knowing that µs = 0.82 Ff = Fn × µs Ff = (215.6 N) × (0.86) Ff = 176.792 N Now find max acceleration Ff = m × a (176.792 N) = (20 kg) × a a = ..... m/s² 4) First you have to recalculate the max force of friction. using µk in stead of µs Ff = Fn × µk Ff = (215.6 N) × (0.67) Ff = 135.828 N Then find the force exerted on the mass of the box by the acceleration of the truck Fa = m × a Fa = (20 kg) × (8.036 m/s²) Fa = 176.792 N The net force exerted on the box will be the difference between Ff and Fa, so Fnet = (176.792 N) - (135.828 N) = (40.964 N) So the acceleration of the box will be a = F / m a = (40.964 N) / (22 kg) a = 1.862 m/s² 5) Assuming the box has re settled and is no longer sliding when braking begins and the surface remains horizontal, the maximum negative acceleration will again be a = -8 m/s²Related Questions
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