A box rests on top of a flat bed truck. The box has a mass of m = 16 kg. The coe
ID: 1871138 • Letter: A
Question
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 s = 0.82 and the coefficient of kinetic friction between the box and truck is k = 0.64. (a) The truck accelerates from rest to vf = 19 m/s in t = 13 s (which is slow enough that the box will not slide). What is the acceleration of the box? (b) In the previous situation, what is the frictional force the truck exerts on the box? (c) What is the maximum acceleration the truck can have before the box begins to slide? (d) Now the acceleration of the truck remains at that value, and the box begins to slide. What is the acceleration of the box?
Explanation / Answer
Apply kinematic equation
vf = vi + at
a = vf- vi/t
= 19-0/13
=1.46 m/s^2
(2)
the frictional force the truck exerts on the box is
f= ma = 16 ( 1.46) = 23.36 N
(3)
the maximum acceleration the truck can have before the box begins to slide
Fnet = f max
= usN
= usm g
ma= us mg
a = us g
=0.82 (9.8)
=8.036 m/s^2
(4)
F net = F- fk
ma' = ma- uk mg
a' =a- uk g
= 8.036 - (0.64) (9.8)
=1.764 m/s^2
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