The figure below shows an initially stationary block of mass m on a floor. A for
ID: 1282343 • Letter: T
Question
The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F = 0.540mg is then applied at upward angle theta = 18 degree. (a) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are mu s = 0.595 and mu k = 0.505? 0 m/s^2 (b) What is the magnitude of the acceleration of the block across the floor if the friction coefficients are mu s = 0.410 and mu k = 0.305? m/s^2 What is the vertical component and horizontal component of the applied force? What is the gravitational force? What is the vertical acceleration? From Newton's second law for vertical motion, what is the normal force? What is f s,max? How does that value compare with the horizontal component of the applied force?Explanation / Answer
a)Fx==0.54mgcos18=0.5135mg
Fy==0.54mgsin18=Fx==0.54mgcos18=0.1668mg
normal force=mg-0.1668mg=0.8331mg
static friction maximum=us N=0.595*0.8331mg=0.4957mg
0.5135mg is greater then 0.4957mg so,it will move.
kinetic friction maximum=uk N=0.505*0.8331mg=0.4207mg
a=(0.5135mg-0.4207mg)/m=0.9103m/s2
b)Fx==0.54mgcos18=0.5135mg
Fy==0.54mgsin18=Fx==0.54mgcos18=0.1668mg
normal force=mg-0.1668mg=0.8331mg
static friction maximum=us N=0.410*0.8331mg=0.3415mg
0.5135mg is greater then 0.3415mg so,it will not move.
kinetic friction maximum=uk N=0.305*0.8331mg=0.2540mg
a=(0.5135mg-0.2540mg)/m=2.5456m/s2
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