Question

A block of wood slides on a frictionless horizontal surface. It is attached to a spring and oscillates with a period of 0.8 s. A second block rests on top of the first. The coefficient of static friction between the two blocks is 0.25.

If the amplitude of oscillations is 1.2 cm, will the block on the top slip?

What is the greatest amplitude of oscillation for which the top block will not slip?

Answer 1

The frictional force on the top mass, Ff = s * m * g
Where s is the coefficient of static friction and m is the mass on the top.
The maximum acceleration of mass m without slipping, a_max = Ff/m
=  s * g
= 0.25 * 9.8
= 2.45 m/s²

Period of oscillation, T = 0.8 s
Angular frequency, = 2/T
= 2/0.8
= 7.85 rad/s

i)
The maximum acceleration due to oscillation, a = A * ²
Where A is the amplitude of the oscillation.
a = 1.2 * 10^-2 * 7.85²
= 0.74 m/s²
Since the maximum acceleration due to oscillation can not overcome the maximum acceleration the top mass can withstand due to friction (a_max = 2.45 m/s²), the block on top will not slip.

ii)
Equating the maximum acceleration due to friction to a_max,
A * ² = a_max
A * 7.85² = 2.45
A = 2.45 / 61.69
= 0.0397 m
= 3.97 cm