Question

A box of books is initially at rest a distance D = 0.493 m from the end of a wooden board. The coefficient of static friction between the box and the board is μ_s = 0.357, and the coefficient of kinetic friction is μ_k = 0.280. 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 angle of elevation of the board is θ

Static frictional force:

ƒ_s = (Us) • (Fɴ)

= (0.357) • ( m • g • cos[θ] )

Kinetic frictional force:

ƒк = (Uк) • (Fɴ)

= (0.280) • ( m • g • cos[θ] )

Let F_i is initial force in the direction of motion ( > ƒs )

 F_i =  m • g • sin[θ] > ƒs
     m • g • sin[θ] > (0.357) • ( m • g • cos[θ] )
      tan[θ] > 0.357
        θ > 19.64º

This is the condition to start motion.

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As motion begins the net force on the box (F) in the direction of motion

F = Fi − ƒк

   = m • g • sin[θ] − (0.280) • ( m • g • cos[θ] )

   = m • g • sin[19.64º] − (0.280) • ( m • g • cos[19.64º] )

Now find acceleration by using Newton's second law

a = F / m

= ( m • g • sin[19.64º] − (0.280) • ( m • g • cos[19.64º] ) /m

=  g • sin[19.64º] − (0.280)  g • cos[19.64º]

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From kinematic equation of motion

V² − (Vi)² = 2 • a • s

In the given case, the initital speed is zero

V² − (0 )² = 2 • a • D

Solve for V

   V = ( 2 • a • D ) ^1/2

Substitute acceleration, a

   V = { 2 ( g • sin[19.64º] − (0.280)  g • cos[19.64º] •D } ^1/2

= { 2 ( (9.8) • sin[19.64º] − (0.280)(9.8) cos[19.64º] • 0.493 m } ^1/2

   = 2.53 m/s