Question

A cubical block of mass M vibrates horizontally
with amplitude of 4.0 cm and a frequency of
2.0 Hz. A small block of mass m is placed on the bigger
block. In order that the smaller block does not slide on
the bigger block, the minimum value of the coefficient of
static friction between the two blocks is
(1) 0.36
(2) 0.40
(3) 0.64
(4) 0.72​

Answer 1

0.40 0.40

That block does not slide on the bigger block the maximum value of the confidence of static frequency between the two block is 0.40. If is wrong that your answer 0.64.

Answer 2

Answer:

The minimum value of the coefficient of static friction between the two blocks is 0.64.

Explanation:

Given a block of mass M vibrating with amplitude,  $x= 4.0 cm=0.04m$

             and frequency, $f=2.0 Hz$

A small block of mass m is placed on the block of mass M.

Acceleration of the system in simple harmonic motion is

                            $a=\omega^{2}x =(2\pi f)^{2}x$

                                          $=(2\pi \times 2)^{2} \times 0.04$

                                          $=6.31rad/s^{2}$  

In order to keep the smaller block in contact with the bigger block, the friction force between the two blocks should be more than the acceleration.

                             $\mu g \ge 6.31$

                               $\mu > \frac{6.31}{10} =0.631\approx 0.64$

Therefore, the minimum value of the coefficient of static friction is 0.64.