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In the figure, a slab of mass m1 = 40 kg rests on a frictionless floor,...

In the figure, a slab of mass m1 = 40 kg rests on a frictionless floor, and a block of mass m2 = 11 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. A horizontal force ModifyingAbove Upper F With right-arrow of magnitude 107 N begins to pull directly on the block, as shown. In unit-vector notation, what are the resulting accelerations of (a) the block and (b) the slab?

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Answer #1

slab mass m1 = 40 kg

block of mass m2 = 11 kg

coefficient of static friction = 0.60

coefficient of kinetic friction = 0.40

F on block = 107 N

Normal = 11*9.8

N = 107.8 N

Static Ff = 107.8*0.6 = 64.68 N which is less than 107 N

11 kg block will slide on 40 kg slab with net Force FN

FN = 107 N - kinetic Ff
FN = 107 – (107.8*0.4)

FN = 63.88 N

11 kg block will slide

a = 63.88/11

a = 5.8 m/s/s

acceleration of 40 kg slab

a = Kinetic_Ff/mb

a = 43.12/40

a = 1.078 m/s/s

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