Question

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

Answer 1

a)

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

V_f = 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

V_f² = V_i² + 2 a d

1.10² = 3² + 2 (- 2.45) d

d = 1.6 m

b)

consider the motion along the vertical direction after falling off te table

V_oy = 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 = V_oy t + (0.5) a t²

Y = 0 (0.35) + (0.5) (9.8) (0.35)²

Y = 0.6 m

c)

consider the motion along the horizontal direction

X = horizontal distance travelled

V_ox = horizontal velocity = 1.10 m/s

t = time of travel = 0.35 sec

horizontal distance is given as

X = V_ox t since there is no acceleration along X-direction

X = (1.10) (0.35) = 0.385 m

d)

Along the horizontal direction

V_fx = V_ox = 1.10 m/s

Along the vertical direction

V_fy = V_oy + at

V_fy = 0 + (9.8) (0.35)

V_fy = 3.43 m/s

net velocity is given as

V = sqrt(V_fx² + V_fy²)

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