Question

A box of books is initially at rest a distance D = 0.441 m from the end of a wooden board. The coefficient of static friction between the box and the board is μ_s = 0.331, and the coefficient of kinetic friction is μ_k = 0.296. The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board.

Answer 1

Let θ = angle of elevation of the board.

 The frictional forces are:
Static Friction -
 ƒs = (Us) • (Fɴ)      
 ƒs = (0.331) • ( m • g • cos[θ] ) 

Kinetic Friction -
ƒк = (Uк) • (Fɴ)
ƒк = (0.296) • ( m • g • cos[θ] )

Fi = initial force in the direction of motion > ƒs

For Motion to begin-
 Fi =  m • g • sin[θ] >= ƒs
 m • g • sin[θ] >= (0.331) • ( m • g • cos[θ] )
 tan[θ] >= 0.331
θ = tan^-1(0.331)
 θ = 18.31º 

 Once motion begins the net force on the box (F) in the direction of motion is:

  F = Fi − ƒк
  F = m • g • sin[θ] − (0.296) • ( m • g • cos[θ] )
F = m • g • sin[18.31] − (0.296) • ( m • g • cos[18.31] )

 We Know F = m • a 
 a = F ⁄ m 
a =  g • sin[18.31] − (0.296) • (  g • cos[18.31] )
a = 9.8 * sin[18.31] - (0.296) • (  9.8* cos[18.31] )
a = 0.323 m/s^2

 V² − (Vi)² = 2 • a • d  ...  initial velocity = Vi = 0
V² = 2  *  0.323 * 0.441
V = sqrt( 2  *  0.323 * 0.441)
V = 0.534 m/s
Speed of the box as it reaches the end of the board V = 0.534 m/s