A box rests on top of a flat bed truck. The box has a mass of m 23 kg. The coeff
ID: 1533425 • Letter: A
Question
A box rests on top of a flat bed truck. The box has a mass of m 23 kg. The coefficient of static friction between the box and truck is mu _s = 0.78 and the coefficient of kinetic friction between the box and truck is mu_k 0.61. The truck accelerates from rest to v_f = 15 m/s in t = 14 s (which is slow enough that the box will not slide). What is the acceleration of the box? In the previous situation, what is the frictional force the truck exerts on the box? What is the maximum acceleration the truck can have before the box begins to slide? Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box? 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? Below is some space to write notes on this problemExplanation / Answer
a) Vf = vi + a t
a = vf / t
a = 15 / 14 = 1.07 m/s2
b) F = m a = 1.07 * 23 = 24.6 N
c) Normal force = m g = 9.8 * 23 = 225 N
Force = mus N
Ff = 0.78 * 225 = 176
176 = m a
max acceleratio amax =176 / 23 = 7.65 m/s2
d) Ff = muk N = 0.61 * 225 = 137.25
accelerationbox = 137.25 / 23 = 5.97 m/s2
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