A physics book is placed on th

A physics book is placed on the middle of a horizontal tabletopand gets pushed with an initial speed of 3m/s. It then slides tothe edge of the table and falls off with a speed of 1.10m/s,striking the floor 0.350s later. The static friction coefficientbetween the book and table is 0.320, and the kinectic frictioncoefficient is 0.250. Ignore air resistance. Find a) the distancethe book was sliding on the table b)the height of the tabletopabove the floor c)the horizontal distance from the edge of thetable to the point where the book hits the floor d)the magnitudeand the direction of the books velocity, just before it reaches thefloor.



Vi = initial speed of the book = 3 m/s

Vf = final speed of the book = 1.10 m/s

acceleration due to frictional force is given as

a = – g = -(0.250) (9.8) = – 2.45 m/s2

d = distance travelled on the table

Using the equation

Vf2 = Vi2 + 2 ad

1.102 = 32 + 2 (- 2.45) d

d = 1.6 m


consider the motion along the vertical direction after fallingoff te table

Voy = initial velocity = 0 m/s

Y = vertical displacement = h = height of the table = ?

a = acceleration = 9.8 m/s2

t = time of travel = 0.35 sec

using the equation

Y = Voy t + (0.5) a t2

Y = 0 (0.35) + (0.5) (9.8) (0.35)2

Y = 0.6 m


consider the motion along the horizontal direction

X = horizontal distance travelled

Vox = horizontal velocity = 1.10 m/s

t = time of travel = 0.35 sec

horizontal distance is given as

X = Vox t since there is no acceleration alongX-direction

X = (1.10) (0.35) = 0.385 m


Along the horizontal direction

Vfx = Vox = 1.10 m/s

Along the vertical direction

Vfy = Voy + at

Vfy = 0 + (9.8) (0.35)

Vfy = 3.43 m/s

net velocity is given as

V = sqrt(Vfx2 +Vfy2)

V = sqrt((1.10)2 + (3.43)2) = 3.6 m/s

direction is given as

=tan-1(3.43/1.10) = 72.2 degree below horizontal

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