Question

m A box rests on top of a flat bed truck. The box has a mass of m = 16 kg. The coefficient of static friction between the box and truck is is = 0.82 and the coefficient of kinetic friction between the box and truck is Hk = 0.64. 1) The truck accelerates from rest to Vp = 19 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 Submit + 2) In the previous situation, what is the frictional force the truck exerts on the box? N Submit 3) What is the maximum acceleration the truck can have before the box begins to slide? m/s2 Submit 4) 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 Submit 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? Im/s2 Submit +

Answer 1

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от: 16 -V Т) ц,= 0.82- = 0-69 y = 19 m/s 1. 13s using v=utat - 1$ = +134 | 9 1.Ч. 52- * a. 13 2. fama - Ј6 х ,че 25:38 м Psudo force а 3 . for max force or accelarken JE N mas "] 2 مي "1 "] = 0-02 х 4. в ас а) = 8.03 Фls И. Now So е fa la N we "] = 049x 9.8 This is same as part 3 a = -0. 036 mls ) 6-2-12 ) о S. So Х